Most votes on xml questions 2

Most votes on xml questions 2. #11 How can I save application settings in a Windows Forms application? #12 Can the Android layout folder contain subfolders? #13 How to pretty print XML from the command line? #14 How to make layout with rounded corners..? #15 How do I escape ampersands in XML so they are rendered as entities in HTML? #16 XDocument or XmlDocument #17 What's the difference between text/xml vs application/xml for webservice response #18 How to Deserialize XML document #19 How does one parse XML files? #20 Declaring a custom android UI element using XML

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

#11: How can I save application settings in a Windows Forms application? (Score: 607)

Created: 2009-01-17 Last updated: 2020-01-03

Tags: c#, xml, winforms, configuration-files, application-settings

What I want to achieve is very simple: I have a Windows Forms (.NET 3.5) application that uses a path for reading information. This path can be modified by the user, by using the options form I provide.

Now, I want to save the path value to a file for later use. This would be one of the many settings saved to this file. This file would sit directly in the application folder.

I understand three options are available:

  • ConfigurationSettings file (appname.exe.config)
  • Registry
  • Custom XML file

I read that the .NET configuration file is not foreseen for saving values back to it. As for the registry, I would like to get as far away from it as possible.

Does this mean that I should use a custom XML file to save configuration settings?

If so, I would like to see code example of that (C#).

I have seen other discussions on this subject, but it is still not clear to me.

#11 Best answer 1 of How can I save application settings in a Windows Forms application? (Score: 618)

Created: 2009-01-17 Last updated: 2020-02-21

If you work with Visual Studio then it is pretty easy to get persistable settings. Right click on the project in Solution Explorer and choose Properties. Select the Settings tab and click on the hyperlink if settings doesn’t exist.

Use the Settings tab to create application settings. Visual Studio creates the files Settings.settings and Settings.Designer.settings that contain the singleton class Settings inherited from ApplicationSettingsBase. You can access this class from your code to read/write application settings:

Properties.Settings.Default["SomeProperty"] = "Some Value";
Properties.Settings.Default.Save(); // Saves settings in application configuration file

This technique is applicable both for console, Windows Forms, and other project types.

Note that you need to set the scope property of your settings. If you select Application scope then Settings.Default.<your property> will be read-only.

Reference: How To: Write User Settings at Run Time with C# - Microsoft Docs

#11 Best answer 2 of How can I save application settings in a Windows Forms application?(Score: 98)

Created: 2011-06-30 Last updated: 2016-08-02

If you are planning on saving to a file within the same directory as your executable, here’s a nice solution that uses the JSON format:

using System;
using System.IO;
using System.Web.Script.Serialization;

namespace MiscConsole
    class Program
        static void Main(string[] args)
            MySettings settings = MySettings.Load();
            Console.WriteLine("Current value of 'myInteger': " + settings.myInteger);
            Console.WriteLine("Incrementing 'myInteger'...");
            Console.WriteLine("Saving settings...");

        class MySettings : AppSettings<MySettings>
            public string myString = "Hello World";
            public int myInteger = 1;

    public class AppSettings<T> where T : new()
        private const string DEFAULT_FILENAME = "settings.json";

        public void Save(string fileName = DEFAULT_FILENAME)
            File.WriteAllText(fileName, (new JavaScriptSerializer()).Serialize(this));

        public static void Save(T pSettings, string fileName = DEFAULT_FILENAME)
            File.WriteAllText(fileName, (new JavaScriptSerializer()).Serialize(pSettings));

        public static T Load(string fileName = DEFAULT_FILENAME)
            T t = new T();
                t = (new JavaScriptSerializer()).Deserialize<T>(File.ReadAllText(fileName));
            return t;

See also original question in stackoverflow

#12: Can the Android layout folder contain subfolders? (Score: 592)

Created: 2011-02-08 Last updated: 2020-08-20

Tags: android, xml, android-layout, gradle, build.gradle

Right now, I’m storing every XML layout file inside the ‘res/layout’ folder, so it is feasible and simple to manage small projects, but when there is a case of large and heavy projects, then there should be a hierarchy and sub-folders needed inside the layout folder.

for e.g.

-- layout_personal
   -- personal_detail.xml
   -- personal_other.xml
  -- address1.xml
  -- address2.xml

Like the same way, we would like to have sub-folders for the large application, so is there any way to do so inside the Android project?

I am able to create layout-personal and layout_address sub-folders inside the layout folder, but when the time comes to access the XML layout file using R.layout._______ , at that time there is no any XML layout pop-up inside the menu.

#12 Best answer 1 of Can the Android layout folder contain subfolders? (Score: 506)

Created: 2014-03-15 Last updated: 2016-11-22

You CAN do this with gradle. I’ve made a demo project showing how.

The trick is to use gradle’s ability to merge multiple resource folders, and set the res folder as well as the nested subfolders in the sourceSets block.

The quirk is that you can’t declare a container resource folder before you declare that folder’s child resource folders.

Below is the sourceSets block from the build.gradle file from the demo. Notice that the subfolders are declared first.

sourceSets {
    main {
        res.srcDirs =

nested resources picture

Also, the direct parent of your actual resource files (pngs, xml layouts, etc..) does still need to correspond with the specification.

#12 Best answer 2 of Can the Android layout folder contain subfolders?(Score: 241)

Created: 2011-02-08 Last updated: 2012-06-21

The answer is no.

I would like to draw your attention towards this book Pro Android 2 that states:

It is also worth noting a few constraints regarding resources. First, Android supports only a linear list of files within the predefined folders under res. For example, it does not support nested folders under the layout folder (or the other folders under res).

Second, there are some similarities between the assets folder and the raw folder under res. Both folders can contain raw files, but the files within raw are considered resources and the files within assets are not.

Note that because the contents of the assets folder are not considered resources, you can put an arbitrary hierarchy of folders and files within it.

See also original question in stackoverflow

#13: How to pretty print XML from the command line? (Score: 587)

Created: 2013-04-18 Last updated: 2017-05-23

Tags: xml, unix, command-line


Is there a (unix) shell script to format XML in human-readable form?

Basically, I want it to transform the following:

<root><foo a="b">lorem</foo><bar value="ipsum" /></root>

… into something like this:

    <foo a="b">lorem</foo>
    <bar value="ipsum" />

#13 Best answer 1 of How to pretty print XML from the command line? (Score: 1010)

Created: 2013-04-18 Last updated: 2019-01-21


This utility comes with libxml2-utils:

echo '<root><foo a="b">lorem</foo><bar value="ipsum" /></root>' |
    xmllint --format -

Perl’s XML::Twig

This command comes with XML::Twig [tag:perl] module, sometimes xml-twig-tools package:

echo '<root><foo a="b">lorem</foo><bar value="ipsum" /></root>' |


This command comes with xmlstarlet:

echo '<root><foo a="b">lorem</foo><bar value="ipsum" /></root>' |
    xmlstarlet format --indent-tab


Check the tidy package:

echo '<root><foo a="b">lorem</foo><bar value="ipsum" /></root>' |
    tidy -xml -i -


Python’s xml.dom.minidom can format XML (both python2 and python3):

echo '<root><foo a="b">lorem</foo><bar value="ipsum" /></root>' |
    python -c 'import sys;import xml.dom.minidom;;print(xml.dom.minidom.parseString(s).toprettyxml())'


You need saxon-lint:

echo '<root><foo a="b">lorem</foo><bar value="ipsum" /></root>' |
    saxon-lint --indent --xpath '/' -


You need saxon-HE:

 echo '<root><foo a="b">lorem</foo><bar value="ipsum" /></root>' |
    java -cp /usr/share/java/saxon/saxon9he.jar net.sf.saxon.Query \
    -s:- -qs:/ '!indent=yes'

#13 Best answer 2 of How to pretty print XML from the command line?(Score: 175)

Created: 2013-11-15 Last updated: 2019-01-27

xmllint --format yourxmlfile.xml

xmllint is a command line XML tool and is included in libxml2 (


Note: If you don’t have libxml2 installed you can install it by doing the following:


cd /tmp
tar xzf libxml2-2.8.0.tar.gz
cd libxml2-2.8.0/
sudo make install


sudo apt-get install libxml2-utils


apt-cyg install libxml2


To install this on MacOS with Homebrew just do: brew install libxml2


Also available on Git if you want the code: git clone git://

See also original question in stackoverflow

#14: How to make layout with rounded corners..? (Score: 567)

Created: 2013-04-23 Last updated: 2021-03-24

Tags: android, xml, image, layout, android-shapedrawable

How can I make a layout with rounded corners? I want to apply rounded corners to my LinearLayout.

#14 Best answer 1 of How to make layout with rounded corners..? (Score: 1151)

Created: 2013-04-23 Last updated: 2017-10-05

1: Define layout_bg.xml in drawables:

<?xml version="1.0" encoding="UTF-8"?>
<shape xmlns:android="">
    <solid android:color="#FFFFFF"/>
    <stroke android:width="3dp" android:color="#B1BCBE" />
    <corners android:radius="10dp"/>
    <padding android:left="0dp" android:top="0dp" android:right="0dp" android:bottom="0dp" />

2: Add layout_bg.xml as background to your layout


#14 Best answer 2 of How to make layout with rounded corners..?(Score: 154)

Created: 2015-06-07 Last updated: 2021-01-21

For API 21+, Use Clip Views

Rounded outline clipping was added to the View class in API 21. See this training doc or this reference for more info.

This in-built feature makes rounded corners very easy to implement. It works on any view or layout and supports proper clipping.

Here’s What To Do:

  • Create a rounded shape drawable and set it as your view’s background: android:background="@drawable/round_outline"
  • Clip to outline in code: setClipToOutline(true)

The documentation used to say that you can set android:clipToOutline="true" the XML, but this bug is now finally resolved and the documentation now correctly states that you can only do this in code.

What It Looks Like:

examples with and without clipToOutline

Special Note About ImageViews

setClipToOutline() only works when the View’s background is set to a shape drawable. If this background shape exists, View treats the background’s outline as the borders for clipping and shadowing purposes.

This means that if you want to round the corners on an ImageView with setClipToOutline(), your image must come from android:src instead of android:background (since background is used for the rounded shape). If you MUST use background to set your image instead of src, you can use this nested views workaround:

  • Create an outer layout with its background set to your shape drawable
  • Wrap that layout around your ImageView (with no padding)
  • The ImageView (including anything else in the layout) will now be clipped to the outer layout’s rounded shape.

See also original question in stackoverflow

#15: How do I escape ampersands in XML so they are rendered as entities in HTML? (Score: 548)

Created: 2009-08-25 Last updated: 2019-11-06

Tags: html, xml, escaping, ampersand

I have some XML text that I wish to render in an HTML page. This text contains an ampersand, which I want to render in its entity representation: &amp;.

How do I escape this ampersand in the source XML? I tried &amp;, but this is decoded as the actual ampersand character (&), which is invalid in HTML.

So I want to escape it in such a way that it will be rendered as &amp; in the web page that uses the XML output.

#15 Best answer 1 of How do I escape ampersands in XML so they are rendered as entities in HTML? (Score: 441)

Created: 2009-08-25 Last updated: 2016-08-30

When your XML contains &amp;amp;, this will result in the text &amp;.

When you use that in HTML, that will be rendered as &.

#15 Best answer 2 of How do I escape ampersands in XML so they are rendered as entities in HTML?(Score: 202)

Created: 2009-08-25 Last updated: 2017-10-04

As per §2.4 of the XML 1.0 spec, you should be able to use &amp;.

I tried &amp; but this isn’t allowed.

Are you sure it isn’t a different issue? XML explicitly defines this as the way to escape ampersands.

See also original question in stackoverflow

#16: XDocument or XmlDocument (Score: 541)

Created: 2009-10-09 Last updated: 2015-08-19

Tags: c#, xml, xmldocument, linq-to-xml

I am now learning XmlDocument but I’ve just ran into XDocument and when I try to search the difference or benefits of them I can’t find something useful, could you please tell me why you would use one over another ?

#16 Best answer 1 of XDocument or XmlDocument (Score: 521)

Created: 2009-10-09 Last updated: 2009-10-09

If you’re using .NET version 3.0 or lower, you have to use XmlDocument aka the classic DOM API. Likewise you’ll find there are some other APIs which will expect this.

If you get the choice, however, I would thoroughly recommend using XDocument aka LINQ to XML. It’s much simpler to create documents and process them. For example, it’s the difference between:

XmlDocument doc = new XmlDocument();
XmlElement root = doc.CreateElement("root");
root.SetAttribute("name", "value");
XmlElement child = doc.CreateElement("child");
child.InnerText = "text node";


XDocument doc = new XDocument(
    new XElement("root",
                 new XAttribute("name", "value"),
                 new XElement("child", "text node")));

Namespaces are pretty easy to work with in LINQ to XML, unlike any other XML API I’ve ever seen:

XNamespace ns = "";
XElement element = new XElement(ns + "elementName");
// etc

LINQ to XML also works really well with LINQ - its construction model allows you to build elements with sequences of sub-elements really easily:

// Customers is a List<Customer>
XElement customersElement = new XElement("customers",
    customers.Select(c => new XElement("customer",
        new XAttribute("name", c.Name),
        new XAttribute("lastSeen", c.LastOrder)
        new XElement("address",
            new XAttribute("town", c.Town),
            new XAttribute("firstline", c.Address1),
            // etc

It’s all a lot more declarative, which fits in with the general LINQ style.

Now as Brannon mentioned, these are in-memory APIs rather than streaming ones (although XStreamingElement supports lazy output). XmlReader and XmlWriter are the normal ways of streaming XML in .NET, but you can mix all the APIs to some extent. For example, you can stream a large document but use LINQ to XML by positioning an XmlReader at the start of an element, reading an XElement from it and processing it, then moving on to the next element etc. There are various blog posts about this technique, here’s one I found with a quick search.

#16 Best answer 2 of XDocument or XmlDocument(Score: 61)

Created: 2012-08-29 Last updated: 2015-01-16

I am surprised none of the answers so far mentions the fact that XmlDocument provides no line information, while XDocument does (through the IXmlLineInfo interface).

This can be a critical feature in some cases (for example if you want to report errors in an XML, or keep track of where elements are defined in general) and you better be aware of this before you happily start to implement using XmlDocument, to later discover you have to change it all.

See also original question in stackoverflow

#17: What's the difference between text/xml vs application/xml for webservice response (Score: 530)

Created: 2011-01-28

Tags: xml, rest, jersey

This is more of a general question about the difference between text/xml and application/xml. I am fairly new to writing webservices (REST - Jersey). I have been producing application/xml since it is what shows up in most tutorials / code examples that I have been using to learn, but I recently found out about text/xml and was wondering what is different about it and when would you use it over application/xml?

#17 Best answer 1 of What's the difference between text/xml vs application/xml for webservice response (Score: 450)

Created: 2011-01-28 Last updated: 2019-01-26

From the RFC (3023), under section 3, XML Media Types:

If an XML document – that is, the unprocessed, source XML document – is readable by casual users, text/xml is preferable to application/xml. MIME user agents (and web user agents) that do not have explicit support for text/xml will treat it as text/plain, for example, by displaying the XML MIME entity as plain text. Application/xml is preferable when the XML MIME entity is unreadable by casual users.

(emphasis mine)

#17 Best answer 2 of What's the difference between text/xml vs application/xml for webservice response(Score: 132)

Created: 2015-04-16 Last updated: 2019-01-26

This is an old question, but one that is frequently visited and clear recommendations are now available from RFC 7303 which obsoletes RFC3023. In a nutshell (section 9.2):

The registration information for text/xml is in all respects the same
as that given for application/xml above (Section 9.1), except that
the "Type name" is "text".

See also original question in stackoverflow

#18: How to Deserialize XML document (Score: 500)

Created: 2008-12-12 Last updated: 2012-08-09

Tags: c#,, xml, serialization, xml-deserialization

How do I Deserialize this XML document:

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

I have this:

public class Car
    public string StockNumber{ get; set; }

    public string Make{ get; set; }

    public string Model{ get; set; }


[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
    public Car[] Car { get; set; }



public class CarSerializer
    public Cars Deserialize()
        Cars[] cars = null;
        string path = HttpContext.Current.ApplicationInstance.Server.MapPath("~/App_Data/") + "cars.xml";

        XmlSerializer serializer = new XmlSerializer(typeof(Cars[]));

        StreamReader reader = new StreamReader(path);
        cars = (Cars[])serializer.Deserialize(reader);

        return cars;

that don’t seem to work :-(

#18 Best answer 1 of How to Deserialize XML document (Score: 459)

Created: 2008-12-12 Last updated: 2018-11-30

How about you just save the xml to a file, and use xsd to generate C# classes?

  1. Write the file to disk (I named it foo.xml)
  2. Generate the xsd: xsd foo.xml
  3. Generate the C#: xsd foo.xsd /classes

Et voila - and C# code file that should be able to read the data via XmlSerializer:

    XmlSerializer ser = new XmlSerializer(typeof(Cars));
    Cars cars;
    using (XmlReader reader = XmlReader.Create(path))
        cars = (Cars) ser.Deserialize(reader);

(include the generated foo.cs in the project)

#18 Best answer 2 of How to Deserialize XML document(Score: 373)

Created: 2008-12-12 Last updated: 2020-04-09

Here’s a working version. I changed the XmlElementAttribute labels to XmlElement because in the xml the StockNumber, Make and Model values are elements, not attributes. Also I removed the reader.ReadToEnd(); (that function reads the whole stream and returns a string, so the Deserialize() function couldn’t use the reader anymore…the position was at the end of the stream). I also took a few liberties with the naming :).

Here are the classes:

public class Car
    public string StockNumber { get; set; }

    public string Make { get; set; }

    public string Model { get; set; }

public class CarCollection
    [XmlArrayItem("Car", typeof(Car))]
    public Car[] Car { get; set; }

The Deserialize function:

CarCollection cars = null;
string path = "cars.xml";

XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));

StreamReader reader = new StreamReader(path);
cars = (CarCollection)serializer.Deserialize(reader);

And the slightly tweaked xml (I needed to add a new element to wrap <Cars>…Net is picky about deserializing arrays):

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

See also original question in stackoverflow

#19: How does one parse XML files? (Score: 492)

Created: 2008-09-11 Last updated: 2015-06-21

Tags: c#, xml

Is there a simple method of parsing XML files in C#? If so, what?

#19 Best answer 1 of How does one parse XML files? (Score: 319)

Created: 2008-09-11 Last updated: 2013-06-15

It’s very simple. I know these are standard methods, but you can create your own library to deal with that much better.

Here are some examples:

XmlDocument xmlDoc= new XmlDocument(); // Create an XML document object
xmlDoc.Load("yourXMLFile.xml"); // Load the XML document from the specified file

// Get elements
XmlNodeList girlAddress = xmlDoc.GetElementsByTagName("gAddress");
XmlNodeList girlAge = xmlDoc.GetElementsByTagName("gAge"); 
XmlNodeList girlCellPhoneNumber = xmlDoc.GetElementsByTagName("gPhone");

// Display the results
Console.WriteLine("Address: " + girlAddress[0].InnerText);
Console.WriteLine("Age: " + girlAge[0].InnerText);
Console.WriteLine("Phone Number: " + girlCellPhoneNumber[0].InnerText);

Also, there are some other methods to work with. For example, here. And I think there is no one best method to do this; you always need to choose it by yourself, what is most suitable for you.

#19 Best answer 2 of How does one parse XML files?(Score: 249)

Created: 2008-09-11 Last updated: 2018-02-23

I’d use LINQ to XML if you’re in .NET 3.5 or higher.

See also original question in stackoverflow

#20: Declaring a custom android UI element using XML (Score: 478)

Created: 2010-04-23

Tags: xml, android, user-interface

How do I declare an Android UI element using XML?

#20 Best answer 1 of Declaring a custom android UI element using XML (Score: 845)

Created: 2010-04-23 Last updated: 2017-05-23

The Android Developer Guide has a section called Building Custom Components. Unfortunately, the discussion of XML attributes only covers declaring the control inside the layout file and not actually handling the values inside the class initialisation. The steps are as follows:

1. Declare attributes in values\attrs.xml

<?xml version="1.0" encoding="utf-8"?>
    <declare-styleable name="MyCustomView">
        <attr name="android:text"/>
        <attr name="android:textColor"/>    		
        <attr name="extraInformation" format="string" />

Notice the use of an unqualified name in the declare-styleable tag. Non-standard android attributes like extraInformation need to have their type declared. Tags declared in the superclass will be available in subclasses without having to be redeclared.

2. Create constructors

Since there are two constructors that use an AttributeSet for initialisation, it is convenient to create a separate initialisation method for the constructors to call.

private void init(AttributeSet attrs) {	
    TypedArray a=getContext().obtainStyledAttributes(

    //Use a
         R.styleable.MyCustomView_android_textColor, Color.BLACK));

    //Don't forget this

R.styleable.MyCustomView is an autogenerated int[] resource where each element is the ID of an attribute. Attributes are generated for each property in the XML by appending the attribute name to the element name. For example, R.styleable.MyCustomView_android_text contains the android_text attribute for MyCustomView. Attributes can then be retrieved from the TypedArray using various get functions. If the attribute is not defined in the defined in the XML, then null is returned. Except, of course, if the return type is a primitive, in which case the second argument is returned.

If you don’t want to retrieve all of the attributes, it is possible to create this array manually.The ID for standard android attributes are included in android.R.attr, while attributes for this project are in R.attr.

int attrsWanted[]=new int[]{android.R.attr.text, R.attr.textColor};

Please note that you should not use anything in android.R.styleable, as per this thread it may change in the future. It is still in the documentation as being to view all these constants in the one place is useful.

3. Use it in a layout files such as layout\main.xml

Include the namespace declaration xmlns:app="" in the top level xml element. Namespaces provide a method to avoid the conflicts that sometimes occur when different schemas use the same element names (see this article for more info). The URL is simply a manner of uniquely identifying schemas - nothing actually needs to be hosted at that URL. If this doesn’t appear to be doing anything, it is because you don’t actually need to add the namespace prefix unless you need to resolve a conflict.

	android:text="Test text"
    app:extraInformation="My extra information"

Reference the custom view using the fully qualified name.

Android LabelView Sample

If you want a complete example, look at the android label view sample.

 TypedArray a=context.obtainStyledAttributes(attrs, R.styleable.LabelView);


<declare-styleable name="LabelView">
	<attr name="text"format="string"/>
	<attr name="textColor"format="color"/>
	<attr name="textSize"format="dimension"/>


	app:text="Blue" app:textSize="20dp"/>

This is contained in a LinearLayout with a namespace attribute: xmlns:app=""

#20 Best answer 2 of Declaring a custom android UI element using XML(Score: 91)

Created: 2011-11-28

Great reference. Thanks! An addition to it:

If you happen to have a library project included which has declared custom attributes for a custom view, you have to declare your project namespace, not the library one’s. Eg:

Given that the library has the package “com.example.library.customview” and the working project has the package “com.example.customview”, then:

Will not work (shows the error " error: No resource identifier found for attribute ‘newAttr’ in package ‘com.example.library.customview’" ):

        app:newAttr="value" />

Will work:

        app:newAttr="value" />

See also original question in stackoverflow

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