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Most votes on xml questions 5

Most votes on xml questions 5. #41 Android Replace "…" with ellipsis character #42 frequent issues arising in android view, Error parsing XML: unbound prefix #43 Manifest Merger failed with multiple errors in Android Studio #44 How to convert array to SimpleXML #45 How to convert JSON to XML or XML to JSON? #46 Android: remove left margin from actionbar's custom layout #47 Extract value of attribute node via XPath #48 What does the 'standalone' directive mean in XML? #49 How to read XML using XPath in Java #50 JSON and XML comparison

Read all the top votes questions and answers in a single page.

#41: Android Replace "…" with ellipsis character (Score: 313)

Created: 2011-12-16 Last updated: 2019-06-12

Tags: android, xml, encoding, adt

Since AVD tools 16 I’m getting this warning:

Replace "..." with ellipsis character (..., …) ?

in my strings.xml

at this line

 <string name="searching">Searching...</string>

How do I replace ...? Is it just literally &#8230;?

Could someone explain this encoding?

#41 Best answer 1 of Android Replace "…" with ellipsis character (Score: 546)

Created: 2011-12-16 Last updated: 2015-08-12

&#8230; is the unicode for “” so just replace it. It’s better to have it as one char/symbol than three dots.

#41 Best answer 2 of Android Replace "…" with ellipsis character(Score: 15)

Created: 2012-06-29 Last updated: 2012-06-29

To make thing short just put &#x2026; in place …

Link to XML character Entities List

  • Look at Unicode column of HTML for row named hellip

See also original question in stackoverflow

#42: frequent issues arising in android view, Error parsing XML: unbound prefix (Score: 305)

Created: 2010-02-08 Last updated: 2015-07-22

Tags: xml, android, eclipse, android-linearlayout

I have frequent problem in android view, Error parsing XML: unbound prefix on Line 2.

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout android:orientation="vertical" android:id="@+id/myScrollLayout" 
android:layout_width="fill_parent"  android:layout_height="wrap_content">
    <TextView android:layout_height="wrap_content" android:layout_width="fill_parent" 
    android:text="Family" android:id="@+id/Family" 
    android:textSize="16px" android:padding="5px" 
    android:textStyle="bold" android:gravity="center_horizontal">
    </TextView>
	
    <ScrollView xmlns:android="http://schemas.android.com/apk/res/android"
        android:layout_width="fill_parent" android:layout_height="wrap_content"
        android:orientation="vertical" android:scrollbars="vertical">
        <LinearLayout android:orientation="vertical" android:id="@+id/myMainLayout" 
        android:layout_width="fill_parent"  android:layout_height="wrap_content">
        </LinearLayout>
    </ScrollView>

</LinearLayout>

#42 Best answer 1 of frequent issues arising in android view, Error parsing XML: unbound prefix (Score: 564)

Created: 2010-02-08 Last updated: 2018-11-10

A couple of reasons that this can happen:

  1. You see this error with an incorrect namespace, or a typo in the attribute. Like ‘xmlns’ is wrong, it should be xmlns:android

  2. First node needs to contain: xmlns:android="http://schemas.android.com/apk/res/android"

  3. If you are integrating AdMob, check custom parameters like ads:adSize, you need

xmlns:ads="http://schemas.android.com/apk/lib/com.google.ads"

  1. If you are using LinearLayout you might have to define tools:

xmlns:tools="http://schemas.android.com/tools"

#42 Best answer 2 of frequent issues arising in android view, Error parsing XML: unbound prefix(Score: 101)

Created: 2012-12-16

I’m going to add a separate answer just because I don’t see it here. It’s not 100% what Pentium10 asked for, but I ended up here by searching for Error parsing XML: unbound prefix

Turns out I was using custom parameters for AdMob ads like ads:adSize, but I hadn’t added

    xmlns:ads="http://schemas.android.com/apk/lib/com.google.ads"

to the layout. Once I added it it worked great.

See also original question in stackoverflow

#43: Manifest Merger failed with multiple errors in Android Studio (Score: 299)

Created: 2016-03-07 Last updated: 2019-08-20

Tags: android, xml

So, I am a beginner into Android and Java. I just began learning. While I was experimenting with Intent today, I incurred an error.

Error:Execution failed for task ':app:processDebugManifest'.
> Manifest merger failed with multiple errors, see logs

I found some solutions here and tried to implement them, but it did not work.

This is my build.gradle :

apply plugin: 'com.android.application'

android {
compileSdkVersion 23
buildToolsVersion "23.0.0"

defaultConfig {
    applicationId "com.example.rohan.petadoptionthing"
    minSdkVersion 10
    targetSdkVersion 23
    versionCode 1
    versionName "1.0"
}
buildTypes {
    release {
        minifyEnabled false
        proguardFiles getDefaultProguardFile('proguard-android.txt'), 'proguard-rules.pro'
    }
}
}

dependencies {
compile fileTree(dir: 'libs', include: ['*.jar'])
compile 'com.android.support:appcompat-v7:23.0.0'
}

This is my AndroidManifest :

<?xml version="1.0" encoding="utf-8"?>

<manifest xmlns:android=“http://schemas.android.com/apk/res/android"

package="com.example.rohan.petadoptionthing" >

<application

    android:allowBackup="true"
    android:icon="@mipmap/ic_launcher"
    android:label="@string/app_name"
    android:theme="@style/AppTheme" >
    <activity
        android:name=".MainActivity"
        android:label="@string/app_name" >
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />

            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>

    <activity android:name=".Second"
        />

    <activity android:name=".third"/>
    <activity android:name=".MainActivity"/>


</application>

This is my first week with coding, I am sorry if this is a really silly thing. I am really new to this and did not find any other place to ask. Sorry if I broke any rules

#43 Best answer 1 of Manifest Merger failed with multiple errors in Android Studio (Score: 1111)

Created: 2017-02-03 Last updated: 2020-09-21

Open application manifest (AndroidManifest.xml) and click on Merged Manifest tab on bottom of your edit pane. Check the image below:

enter image description here

From image you can see Error in the right column, try to solve the error. It may help some one with the same problem. Read more here.

Also, once you found the error and if you get that error from external library that you are using, You have to let compiler to ignore the attribute from the external library. //add this attribute in application tag in the manifest

   tools:replace="android:allowBackup" 
                                                                                                                                          
   //Add this in the manifest tag at the top

   xmlns:tools="http://schemas.android.com/tools"

#43 Best answer 2 of Manifest Merger failed with multiple errors in Android Studio(Score: 123)

Created: 2016-03-07

Remove <activity android:name=".MainActivity"/> from your mainfest file. As you have already defined it as:

 <activity
        android:name=".MainActivity"
        android:label="@string/app_name" >
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />

            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>

So, Manifest file showing ambiguity.

See also original question in stackoverflow

#44: How to convert array to SimpleXML (Score: 297)

Created: 2009-09-09 Last updated: 2011-08-13

Tags: php, xml, arrays, simplexml

How can I convert an array to a SimpleXML object in PHP?

#44 Best answer 1 of How to convert array to SimpleXML (Score: 405)

Created: 2011-05-11 Last updated: 2020-01-04

Here is php 5.2 code which will convert array of any depth to xml document:

Array
(
    ['total_stud']=> 500
    [0] => Array
        (
            [student] => Array
                (
                    [id] => 1
                    [name] => abc
                    [address] => Array
                        (
                            [city]=>Pune
                            [zip]=>411006
                        )						
                )
        )
    [1] => Array
        (
            [student] => Array
                (
                    [id] => 2
                    [name] => xyz
                    [address] => Array
                        (
                            [city]=>Mumbai
                            [zip]=>400906
                        )	
                )
				
        )
)

generated XML would be as:

<?xml version="1.0"?>
<student_info>
    <total_stud>500</total_stud>
    <student>
        <id>1</id>
        <name>abc</name>
        <address>
            <city>Pune</city>
            <zip>411006</zip>
        </address>
    </student>
    <student>
        <id>1</id>
        <name>abc</name>
        <address>
            <city>Mumbai</city>
            <zip>400906</zip>
        </address>
    </student>
</student_info>

PHP snippet

<?php
// function defination to convert array to xml
function array_to_xml( $data, &$xml_data ) {
    foreach( $data as $key => $value ) {
        if( is_array($value) ) {
            if( is_numeric($key) ){
                $key = 'item'.$key; //dealing with <0/>..<n/> issues
            }
            $subnode = $xml_data->addChild($key);
            array_to_xml($value, $subnode);
        } else {
            $xml_data->addChild("$key",htmlspecialchars("$value"));
        }
     }
}
      
// initializing or creating array
$data = array('total_stud' => 500);

// creating object of SimpleXMLElement
$xml_data = new SimpleXMLElement('<?xml version="1.0"?><data></data>');

// function call to convert array to xml
array_to_xml($data,$xml_data);

//saving generated xml file; 
$result = $xml_data->asXML('/file/path/name.xml');

?>

Documentation on SimpleXMLElement::asXML used in this snippet

#44 Best answer 2 of How to convert array to SimpleXML(Score: 221)

Created: 2009-09-09 Last updated: 2010-02-12

a short one:

<?php

$test_array = array (
  'bla' => 'blub',
  'foo' => 'bar',
  'another_array' => array (
    'stack' => 'overflow',
  ),
);
$xml = new SimpleXMLElement('<root/>');
array_walk_recursive($test_array, array ($xml, 'addChild'));
print $xml->asXML();

results in

<?xml version="1.0"?>
<root>
  <blub>bla</blub>
  <bar>foo</bar>
  <overflow>stack</overflow>
</root>

keys and values are swapped - you could fix that with array_flip() before the array_walk. array_walk_recursive requires PHP 5. you could use array_walk instead, but you won’t get 'stack' => 'overflow' in the xml then.

See also original question in stackoverflow

#45: How to convert JSON to XML or XML to JSON? (Score: 296)

Created: 2009-05-02 Last updated: 2016-11-03

Tags: c#, json, xml, json.net

I started to use Json.NET to convert a string in JSON format to object or viceversa. I am not sure in the Json.NET framework, is it possible to convert a string in JSON to XML format and viceversa?

#45 Best answer 1 of How to convert JSON to XML or XML to JSON? (Score: 439)

Created: 2009-05-02 Last updated: 2014-02-05

Yes. Using the JsonConvert class which contains helper methods for this precise purpose:

// To convert an XML node contained in string xml into a JSON string   
XmlDocument doc = new XmlDocument();
doc.LoadXml(xml);
string jsonText = JsonConvert.SerializeXmlNode(doc);

// To convert JSON text contained in string json into an XML node
XmlDocument doc = JsonConvert.DeserializeXmlNode(json);

Documentation here: Converting between JSON and XML with Json.NET

#45 Best answer 2 of How to convert JSON to XML or XML to JSON?(Score: 48)

Created: 2012-08-13 Last updated: 2020-11-22

Yes, you can do it (I do) but Be aware of some paradoxes when converting, and handle appropriately. You cannot automatically conform to all interface possibilities, and there is limited built-in support in controlling the conversion- many JSON structures and values cannot automatically be converted both ways. Keep in mind I am using the default settings with Newtonsoft JSON library and MS XML library, so your mileage may vary:

XML -> JSON

  1. All data becomes string data (for example you will always get “false” not false or “0” not 0) Obviously JavaScript treats these differently in certain cases.
  2. Children elements can become nested-object {} OR nested-array [ {} {} ...] depending if there is only one or more than one XML child-element. You would consume these two differently in JavaScript, etc. Different examples of XML conforming to the same schema can produce actually different JSON structures this way. You can add the attribute json:Array=‘true’ to your element to workaround this in some (but not necessarily all) cases.
  3. Your XML must be fairly well-formed, I have noticed it doesn’t need to perfectly conform to W3C standard, but 1. you must have a root element and 2. you cannot start element names with numbers are two of the enforced XML standards I have found when using Newtonsoft and MS libraries.
  4. In older versions, Blank elements do not convert to JSON. They are ignored. A blank element does not become “element”:null

A new update changes how null can be handled (Thanks to Jon Story for pointing it out): https://www.newtonsoft.com/json/help/html/T_Newtonsoft_Json_NullValueHandling.htm

JSON -> XML

  1. You need a top level object that will convert to a root XML element or the parser will fail.
  2. Your object names cannot start with a number, as they cannot be converted to elements (XML is technically even more strict than this) but I can ‘get away’ with breaking some of the other element naming rules.

Please feel free to mention any other issues you have noticed, I have developed my own custom routines for preparing and cleaning the strings as I convert back and forth. Your situation may or may not call for prep/cleanup. As StaxMan mentions, your situation may actually require that you convert between objects…this could entail appropriate interfaces and a bunch of case statements/etc to handle the caveats I mention above.

See also original question in stackoverflow

#46: Android: remove left margin from actionbar's custom layout (Score: 294)

Created: 2014-12-08 Last updated: 2017-05-23

Tags: android, xml, android-actionbar-compat

I am using a custom actionbar view, and as you can see in the screenshot below, there is a blank gray space in the actionbar. I want to remove it.

enter image description here

What have I done:

res/values-v11/styles.xml

<style name="AppBaseTheme" parent="@style/Theme.AppCompat.Light">
        <item name="android:actionBarStyle">@style/ActionBarStyle</item>
        <item name="actionBarStyle">@style/ActionBarStyle</item>
</style>

res/values/my_custom_actionbar.xml

<resources xmlns:android="http://schemas.android.com/apk/res/android">
    <style name="ActionBarStyle" parent="@style/Widget.AppCompat.Light.ActionBar.Solid">
        <item name="android:height">60dp</item>
    </style>
</resources>

Manifest

<uses-sdk
        android:minSdkVersion="10"
        android:targetSdkVersion="19" />

<application
            android:icon="@drawable/ic_launcher"
            android:label="@string/AppName"
            android:theme="@style/AppBaseTheme" >
    <!-- activities... etc -->
</application>

MainActivity

public void onCreate(Bundle bundle) {
    super.onCreate(bundle);

    ActionBar actionbar = getSupportActionBar();
		
    actionbar.setDefaultDisplayHomeAsUpEnabled(false);
    actionbar.setDisplayHomeAsUpEnabled(false);
    actionbar.setDisplayShowCustomEnabled(true);
    actionbar.setDisplayShowHomeEnabled(false);
    actionbar.setDisplayShowTitleEnabled(false);
    actionbar.setDisplayUseLogoEnabled(false);
    actionbar.setHomeButtonEnabled(false);

    // Add the custom layout
    View view = LayoutInflater.from(this).inflate(R.layout.actionbar, null, false);
    actionbar.setCustomView(view);
}

I have found a recent post, that is pointing out that there is an issue with the latest release. I have also updated ADT and SDK to Android 5.

https://stackoverflow.com/questions/27298282/android-actionbars-custom-view-not-filling-parent

I don’t know what should I do.

Edit (partial solution):

Not working on Android <= API 10.

https://stackoverflow.com/questions/26433409/android-lollipop-appcompat-actionbar-custom-view-doesnt-take-up-whole-screen-w

What have I changed:

Use the latest sdk version:

<uses-sdk
        android:minSdkVersion="10"
        android:targetSdkVersion="21" />

Add a toolbarStyle:

<style name="AppBaseTheme" parent="@style/Theme.AppCompat.Light">
        <item name="android:actionBarStyle">@style/ActionBarStyle</item>
        <item name="actionBarStyle">@style/ActionBarStyle</item>
        
        <item name="android:toolbarStyle">@style/ToolbarStyle</item>
        <item name="toolbarStyle">@style/ToolbarStyle</item>
</style>

<style name="ToolbarStyle" parent="@style/Widget.AppCompat.Toolbar">
    <item name="contentInsetStart">0dp</item>
    <item name="android:contentInsetStart">0dp</item>
</style>

#46 Best answer 1 of Android: remove left margin from actionbar's custom layout (Score: 660)

Created: 2015-08-31 Last updated: 2016-02-15

If you are adding the Toolbar via XML, you can simply add XML attributes to remove content insets.

<android.support.v7.widget.Toolbar
    xmlns:app="schemas.android.com/apk/res-auto"
    android:layout_width="match_parent"
    android:layout_height="wrap_content"
    android:background="@color/primaryColor"
    android:contentInsetLeft="0dp"
    android:contentInsetStart="0dp"
    app:contentInsetLeft="0dp"
    app:contentInsetStart="0dp"
    android:contentInsetRight="0dp"
    android:contentInsetEnd="0dp"
    app:contentInsetRight="0dp"
    app:contentInsetEnd="0dp" />

#46 Best answer 2 of Android: remove left margin from actionbar's custom layout(Score: 203)

Created: 2015-01-16 Last updated: 2016-09-20

try this:

    ActionBar actionBar = getSupportActionBar();
    actionBar.setDisplayShowHomeEnabled(false);
    actionBar.setDisplayShowCustomEnabled(true);
    actionBar.setDisplayShowTitleEnabled(false);
    View customView = getLayoutInflater().inflate(R.layout.main_action_bar, null);
    actionBar.setCustomView(customView);
    Toolbar parent =(Toolbar) customView.getParent();
    parent.setPadding(0,0,0,0);//for tab otherwise give space in tab
    parent.setContentInsetsAbsolute(0,0);

I used this code in my project,good luck;

See also original question in stackoverflow

#47: Extract value of attribute node via XPath (Score: 293)

Created: 2011-01-29 Last updated: 2015-08-04

Tags: xml, xpath

How can I extract the value of an attribute node via XPath?

A sample XML file is:

<parents name='Parents'>
  <Parent id='1' name='Parent_1'>
    <Children name='Children'>
      <child name='Child_2' id='2'>child2_Parent_1</child>
      <child name='Child_4' id='4'>child4_Parent_1</child>
      <child name='Child_1' id='3'>child1_Parent_1</child>
      <child name='Child_3' id='1'>child3_Parent_1</child>
    </Children>
  </Parent>
  <Parent id='2' name='Parent_2'>
    <Children name='Children'>
      <child name='Child_1' id='8'>child1_parent2</child>
      <child name='Child_2' id='7'>child2_parent2</child>
      <child name='Child_4' id='6'>child4_parent2</child>
      <child name='Child_3' id='5'>child3_parent2</child>
    </Children>
  </Parent>
</parents>

So far I have this XPath string:

//Parent[@id='1']/Children/child[@name]  

It returns only child elements, but I would like to have the value of the name attribute.

For my sample XML file, here’s what I’d like the output to be:

Child_2
Child_4
Child_1
Child_3

#47 Best answer 1 of Extract value of attribute node via XPath (Score: 376)

Created: 2011-01-29 Last updated: 2013-01-02

//Parent[@id='1']/Children/child/@name 

Your original child[@name] means an element child which has an attribute name. You want child/@name.

#47 Best answer 2 of Extract value of attribute node via XPath(Score: 157)

Created: 2013-06-05 Last updated: 2017-12-11

To get just the value (without attribute names), use string():

string(//Parent[@id=‘1’]/Children/child/@name)

The fn:string() fucntion will return the value of its argument as xs:string. In case its argument is an attribute, it will therefore return the attribute’s value as xs:string.

See also original question in stackoverflow

#48: What does the 'standalone' directive mean in XML? (Score: 291)

Created: 2011-04-07

Tags: xml

What does the ‘standalone’ directive mean in an XML document?

#48 Best answer 1 of What does the 'standalone' directive mean in XML? (Score: 210)

Created: 2013-02-04 Last updated: 2015-05-07

The standalone declaration is a way of telling the parser to ignore any markup declarations in the DTD. The DTD is thereafter used for validation only.

As an example, consider the humble <img> tag. If you look at the XHTML 1.0 DTD, you see a markup declaration telling the parser that <img> tags must be EMPTY and possess src and alt attributes. When a browser is going through an XHTML 1.0 document and finds an <img> tag, it should notice that the DTD requires src and alt attributes and add them if they are not present. It will also self-close the <img> tag since it is supposed to be EMPTY. This is what the XML specification means by “markup declarations can affect the content of the document.” You can then use the standalone declaration to tell the parser to ignore these rules.

Whether or not your parser actually does this is another question, but a standards-compliant validating parser (like a browser) should.

Note that if you do not specify a DTD, then the standalone declaration “has no meaning,” so there’s no reason to use it unless you also specify a DTD.

#48 Best answer 2 of What does the 'standalone' directive mean in XML?(Score: 118)

Created: 2013-06-26 Last updated: 2018-02-14

  • The standalone directive is an optional attribute on the XML declaration.
  • Valid values are yes and no, where no is the default value.
  • The attribute is only relevant when a DTD is used. (The attribute is irrelevant when using a schema instead of a DTD.)
  • standalone="yes" means that the XML processor must use the DTD for validation only. In that case it will not be used for:
    • default values for attributes
    • entity declarations
    • normalization
  • Note that standalone="yes" may add validity constraints if the document uses an external DTD. When the document contains things that would require modification of the XML, such as default values for attributes, and standalone="yes" is used then the document is invalid.
  • standalone="yes" may help to optimize performance of document processing.

Source: The standalone pseudo-attribute is only relevant if a DTD is used

See also original question in stackoverflow

#49: How to read XML using XPath in Java (Score: 283)

Created: 2010-05-11 Last updated: 2018-10-31

Tags: java, xml, parsing, xpath

I want to read XML data using XPath in Java, so for the information I have gathered I am not able to parse XML according to my requirement.

here is what I want to do:

Get XML file from online via its URL, then use XPath to parse it, I want to create two methods in it. One is in which I enter a specific node attribute id, and I get all the child nodes as result, and second is suppose I just want to get a specific child node value only

<?xml version="1.0"?>
<howto>
  <topic name="Java">
      <url>http://www.rgagnonjavahowto.htm</url>
  <car>taxi</car>
  </topic>
  <topic name="PowerBuilder">
       <url>http://www.rgagnon/pbhowto.htm</url>
       <url>http://www.rgagnon/pbhowtonew.htm</url>
  </topic>
  <topic name="Javascript">
        <url>http://www.rgagnon/jshowto.htm</url>
  </topic>
 <topic name="VBScript">
       <url>http://www.rgagnon/vbshowto.htm</url>
 </topic>
 </howto>

In above example I want to read all the elements if I search via @name and also one function in which I just want the url from @name ‘Javascript’ only return one node element.

#49 Best answer 1 of How to read XML using XPath in Java (Score: 405)

Created: 2010-05-11 Last updated: 2011-11-17

You need something along the lines of this:

DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = factory.newDocumentBuilder();
Document doc = builder.parse(<uri_as_string>);
XPathFactory xPathfactory = XPathFactory.newInstance();
XPath xpath = xPathfactory.newXPath();
XPathExpression expr = xpath.compile(<xpath_expression>);

Then you call expr.evaluate() passing in the document defined in that code and the return type you are expecting, and cast the result to the object type of the result.

If you need help with a specific XPath expressions, you should probably ask it as separate questions (unless that was your question in the first place here - I understood your question to be how to use the API in Java).

Edit: (Response to comment): This XPath expression will get you the text of the first URL element under PowerBuilder:

/howto/topic[@name='PowerBuilder']/url/text()

This will get you the second:

/howto/topic[@name='PowerBuilder']/url[2]/text()

You get that with this code:

expr.evaluate(doc, XPathConstants.STRING);

If you don’t know how many URLs are in a given node, then you should rather do something like this:

XPathExpression expr = xpath.compile("/howto/topic[@name='PowerBuilder']/url");
NodeList nl = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);

And then loop over the NodeList.

#49 Best answer 2 of How to read XML using XPath in Java(Score: 31)

Created: 2014-07-23 Last updated: 2014-08-28

You can try this.

XML Document

Save as employees.xml.

<?xml version="1.0" encoding="UTF-8"?>
<Employees>
    <Employee id="1">
        <age>29</age>
        <name>Pankaj</name>
        <gender>Male</gender>
        <role>Java Developer</role>
    </Employee>
    <Employee id="2">
        <age>35</age>
        <name>Lisa</name>
        <gender>Female</gender>
        <role>CEO</role>
    </Employee>
    <Employee id="3">
        <age>40</age>
        <name>Tom</name>
        <gender>Male</gender>
        <role>Manager</role>
    </Employee>
    <Employee id="4">
        <age>25</age>
        <name>Meghan</name>
        <gender>Female</gender>
        <role>Manager</role>
    </Employee>
</Employees>

Parser class

The class have following methods

  • List item
  • A Method that will return the Employee Name for input ID.
  • A Method that will return list of Employees Name with age greater than the input age.
  • A Method that will return list of Female Employees Name.

Source Code

import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
 
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathExpression;
import javax.xml.xpath.XPathExpressionException;
import javax.xml.xpath.XPathFactory;
 
import org.w3c.dom.Document;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;
 
 
public class Parser {
 
    public static void main(String[] args) {
        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        factory.setNamespaceAware(true);
        DocumentBuilder builder;
        Document doc = null;
        try {
            builder = factory.newDocumentBuilder();
            doc = builder.parse("employees.xml");
 
            // Create XPathFactory object
            XPathFactory xpathFactory = XPathFactory.newInstance();
 
            // Create XPath object
            XPath xpath = xpathFactory.newXPath();
 
            String name = getEmployeeNameById(doc, xpath, 4);
            System.out.println("Employee Name with ID 4: " + name);
 
            List<String> names = getEmployeeNameWithAge(doc, xpath, 30);
            System.out.println("Employees with 'age>30' are:" + Arrays.toString(names.toArray()));
 
            List<String> femaleEmps = getFemaleEmployeesName(doc, xpath);
            System.out.println("Female Employees names are:" +
                    Arrays.toString(femaleEmps.toArray()));
 
        } catch (ParserConfigurationException | SAXException | IOException e) {
            e.printStackTrace();
        }
 
    }
 
 
    private static List<String> getFemaleEmployeesName(Document doc, XPath xpath) {
        List<String> list = new ArrayList<>();
        try {
            //create XPathExpression object
            XPathExpression expr =
                xpath.compile("/Employees/Employee[gender='Female']/name/text()");
            //evaluate expression result on XML document
            NodeList nodes = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);
            for (int i = 0; i < nodes.getLength(); i++)
                list.add(nodes.item(i).getNodeValue());
        } catch (XPathExpressionException e) {
            e.printStackTrace();
        }
        return list;
    }
 
 
    private static List<String> getEmployeeNameWithAge(Document doc, XPath xpath, int age) {
        List<String> list = new ArrayList<>();
        try {
            XPathExpression expr =
                xpath.compile("/Employees/Employee[age>" + age + "]/name/text()");
            NodeList nodes = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);
            for (int i = 0; i < nodes.getLength(); i++)
                list.add(nodes.item(i).getNodeValue());
        } catch (XPathExpressionException e) {
            e.printStackTrace();
        }
        return list;
    }
 
 
    private static String getEmployeeNameById(Document doc, XPath xpath, int id) {
        String name = null;
        try {
            XPathExpression expr =
                xpath.compile("/Employees/Employee[@id='" + id + "']/name/text()");
            name = (String) expr.evaluate(doc, XPathConstants.STRING);
        } catch (XPathExpressionException e) {
            e.printStackTrace();
        }
 
        return name;
    }
 
}

See also original question in stackoverflow

#50: JSON and XML comparison (Score: 278)

Created: 2011-02-01 Last updated: 2012-09-09

Tags: xml, json

I want to know which is faster: XML and JSON? When to use which one ?

#50 Best answer 1 of JSON and XML comparison (Score: 247)

Created: 2011-02-01 Last updated: 2018-10-22

Faster is not an attribute of JSON or XML or a result that a comparison between those would yield. If any, then it is an attribute of the parsers or the bandwidth with which you transmit the data.

Here is (the beginning of) a list of advantages and disadvantages of JSON and XML:


JSON

Pro:

  • Simple syntax, which results in less “markup” overhead compared to XML.
  • Easy to use with JavaScript as the markup is a subset of JS object literal notation and has the same basic data types as JavaScript.
  • JSON Schema for description and datatype and structure validation
  • JsonPath for extracting information in deeply nested structures

Con:

  • Simple syntax, only a handful of different data types are supported.

  • No support for comments.


XML

Pro:

  • Generalized markup; it is possible to create “dialects” for any kind of purpose
  • XML Schema for datatype, structure validation. Makes it also possible to create new datatypes
  • XSLT for transformation into different output formats
  • XPath/XQuery for extracting information in deeply nested structures
  • built in support for namespaces

Con:

  • Relatively wordy compared to JSON (results in more data for the same amount of information).

So in the end you have to decide what you need. Obviously both formats have their legitimate use cases. If you are mostly going to use JavaScript then you should go with JSON.

Please feel free to add pros and cons. I’m not an XML expert ;)

#50 Best answer 2 of JSON and XML comparison(Score: 223)

Created: 2011-02-01 Last updated: 2017-01-25

Before answering when to use which one, a little background:

edit: I should mention that this comparison is really from the perspective of using them in a browser with JavaScript. It’s not the way either data format has to be used, and there are plenty of good parsers which will change the details to make what I’m saying not quite valid.

JSON is both more compact and (in my view) more readable - in transmission it can be “faster” simply because less data is transferred.

In parsing, it depends on your parser. A parser turning the code (be it JSON or XML) into a data structure (like a map) may benefit from the strict nature of XML (XML Schemas disambiguate the data structure nicely) - however in JSON the type of an item (String/Number/Nested JSON Object) can be inferred syntactically, e.g:

myJSON = {"age" : 12,
          "name" : "Danielle"}

The parser doesn’t need to be intelligent enough to realise that 12 represents a number, (and Danielle is a string like any other). So in javascript we can do:

anObject = JSON.parse(myJSON);
anObject.age === 12 // True
anObject.name == "Danielle" // True
anObject.age === "12" // False

In XML we’d have to do something like the following:

<person>
    <age>12</age>
    <name>Danielle</name>
</person>

(as an aside, this illustrates the point that XML is rather more verbose; a concern for data transmission). To use this data, we’d run it through a parser, then we’d have to call something like:

myObject = parseThatXMLPlease();
thePeople = myObject.getChildren("person");
thePerson = thePeople[0];
thePerson.getChildren("name")[0].value() == "Danielle" // True
thePerson.getChildren("age")[0].value() == "12" // True

Actually, a good parser might well type the age for you (on the other hand, you might well not want it to). What’s going on when we access this data is - instead of doing an attribute lookup like in the JSON example above - we’re doing a map lookup on the key name. It might be more intuitive to form the XML like this:

<person name="Danielle" age="12" />

But we’d still have to do map lookups to access our data:

myObject = parseThatXMLPlease();
age = myObject.getChildren("person")[0].getAttr("age");

EDIT: Original:

In most programming languages (not all, by any stretch) a map lookup such as this will be more costly than an attribute lookup (like we got above when we parsed the JSON).

This is misleading: remember that in JavaScript (and other dynamic languages) there’s no difference between a map lookup and a field lookup. In fact, a field lookup is just a map lookup.

If you want a really worthwhile comparison, the best is to benchmark it - do the benchmarks in the context where you plan to use the data.

As I have been typing, Felix Kling has already put up a fairly succinct answer comparing them in terms of when to use each one, so I won’t go on any further.

See also original question in stackoverflow


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