{" was not expected.} Deserializing Twitter XML

Question

I'm pulling in the XML from Twitter via OAuth.

I'm doing a request to http://twitter.com/account/verify_credentials.xml, which returns the following XML:

<?xml version="1.0" encoding="UTF-8"?>
<user>
  <id>16434938</id>
  <name>Lloyd Sparkes</name>
  <screen_name>lloydsparkes</screen_name>
  <location>Hockley, Essex, UK</location>
  <description>Student</description>
  <profile_image_url>http://a3.twimg.com/profile_images/351849613/twitterProfilePhoto_normal.jpg</profile_image_url>
  <url>http://www.lloydsparkes.co.uk</url>
  <protected>false</protected>
  <followers_count>115</followers_count>
  <profile_background_color>9fdaf4</profile_background_color>
  <profile_text_color>000000</profile_text_color>
  <profile_link_color>220f7b</profile_link_color>
  <profile_sidebar_fill_color>FFF7CC</profile_sidebar_fill_color>
  <profile_sidebar_border_color>F2E195</profile_sidebar_border_color>
  <friends_count>87</friends_count>
  <created_at>Wed Sep 24 14:26:09 +0000 2008</created_at>
  <favourites_count>0</favourites_count>
  <utc_offset>0</utc_offset>
  <time_zone>London</time_zone>
  <profile_background_image_url>http://s.twimg.com/a/1255366924/images/themes/theme12/bg.gif</profile_background_image_url>
  <profile_background_tile>false</profile_background_tile>
  <statuses_count>1965</statuses_count>
  <notifications>false</notifications>
  <geo_enabled>false</geo_enabled>
  <verified>false</verified>
  <following>false</following>
  <status>
    <created_at>Mon Oct 12 19:23:47 +0000 2009</created_at>
    <id>4815268670</id>
    <text>&#187; @alexmuller your kidding? it should all be &quot;black tie&quot; dress code</text>
    <source>&lt;a href=&quot;http://code.google.com/p/wittytwitter/&quot; rel=&quot;nofollow&quot;&gt;Witty&lt;/a&gt;</source>
    <truncated>false</truncated>
    <in_reply_to_status_id>4815131457</in_reply_to_status_id>
    <in_reply_to_user_id>8645442</in_reply_to_user_id>
    <favorited>false</favorited>
    <in_reply_to_screen_name>alexmuller</in_reply_to_screen_name>
    <geo/>
  </status>
</user>

I am using the following code to deserialize:

    public User VerifyCredentials()
    {
        string url = "http://twitter.com/account/verify_credentials.xml";
        string xml = _oauth.oAuthWebRequestAsString(oAuthTwitter.Method.GET, url, null);

        XmlSerializer xs = new XmlSerializer(typeof(User),"");

        MemoryStream ms = new MemoryStream(Encoding.UTF8.GetBytes(xml));

        return (User)xs.Deserialize(ms);
    }

And I have the following for my User class:

 [System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public partial class User
{

    [XmlElement(ElementName = "id")]       
    public long Id { get; set; }

    [XmlElement(ElementName = "name")] 
    public string Name { get; set; }

    [XmlElement(ElementName = "screen_name")]       
    public string ScreenName { get; set; }

    [XmlElement(ElementName = "location")]       
    public string Location { get; set; }

    [XmlElement(ElementName = "description")]      
    public string Description { get; set; }

    [XmlElement(ElementName = "profile_image_url")]      
    public string ProfileImageUrl { get; set; }

    [XmlElement(ElementName = "url")]       
    public string Url { get; set; }

    [XmlElement(ElementName = "protected")]      
    public bool Protected { get; set; }

    [XmlElement(ElementName = "followers_count")]      
    public int FollowerCount { get; set; }

    [XmlElement(ElementName = "profile_background_color")]       
    public string ProfileBackgroundColor { get; set; }

    [XmlElement(ElementName = "profile_text_color")]       
    public string ProfileTextColor { get; set; }

    [XmlElement(ElementName = "profile_link_color")]       
    public string ProfileLinkColor { get; set; }

    [XmlElement(ElementName = "profile_sidebar_fill_color")]       
    public string ProfileSidebarFillColor { get; set; }

    [XmlElement(ElementName = "profile_sidebar_border_color")]      
    public string ProfileSidebarBorderColor { get; set; }

    [XmlElement(ElementName = "friends_count")]     
    public int FriendsCount { get; set; }

    [XmlElement(ElementName = "created_at")]     
    public string CreatedAt { get; set; }

    [XmlElement(ElementName = "favourties_count")]      
    public int FavouritesCount { get; set; }

    [XmlElement(ElementName = "utc_offset")]      
    public int UtcOffset { get; set; }

    [XmlElement(ElementName = "time_zone")]       
    public string Timezone { get; set; }

    [XmlElement(ElementName = "profile_background_image_url")]        
    public string ProfileBackgroundImageUrl { get; set; }

    [XmlElement(ElementName = "profile_background_tile")]        
    public bool ProfileBackgroundTile { get; set; }

    [XmlElement(ElementName = "statuese_count")]        
    public int StatusesCount { get; set; }

    [XmlElement(ElementName = "notifications")]       
    public string Notifications { get; set; }

    [XmlElement(ElementName = "geo_enabled")]       
    public bool GeoEnabled { get; set; }

    [XmlElement(ElementName = "Verified")]        
    public bool Verified { get; set; }

    [XmlElement(ElementName = "following")]
    public string Following { get; set; }

    [XmlElement(ElementName = "status", IsNullable=true)]
    public Status CurrentStatus { get; set; }

}

But when it's deserializing the above XML the application throws the following:

$exception {"There is an error in XML document (2, 2)."} System.Exception {System.InvalidOperationException}

InnerException {"<user xmlns=''> was not expected."} System.Exception {System.InvalidOperationException}

Now I have searched around and the best solution I can find is to add blank namespaces to the serializer when you serialize the content, but i'm not serializing it so I can't.

I also have some code for receiving Statuses, which works fine.

So can someone explain to me why the error is happening? As well as a possible solution?

In my case, it was because of wrong declaration of `XmlSerializer`. So check that also. — Mangesh, Jun 02 '17 at 07:00
I had to add field with **XmlAttribute** into class. See [link](https://stackoverflow.com/a/59142448/4331722) — Stefan Varga, Dec 02 '19 at 15:59

score 334 · Accepted Answer · edited Aug 12 '13 at 15:53

334

Either decorate your root entity with the XmlRoot attribute which will be used at compile time.

[XmlRoot(Namespace = "www.contoso.com", ElementName = "MyGroupName", DataType = "string", IsNullable=true)]

Or specify the root attribute when de serializing at runtime.

XmlRootAttribute xRoot = new XmlRootAttribute();
xRoot.ElementName = "user";
// xRoot.Namespace = "http://www.cpandl.com";
xRoot.IsNullable = true;

XmlSerializer xs = new XmlSerializer(typeof(User),xRoot);

edited Aug 12 '13 at 15:53

Bronumski

14,009
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answered Oct 12 '09 at 21:51

david valentine

4,625
2
21
15

This works and fixes the issue, so thank you. What i dont understand is why do i need to do this? I do deserializing all the time, and this is the first time i have seen this error. – lloydsparkes Oct 12 '09 at 22:08
6

You could add an XmlRoot attribute to the class http://msdn.microsoft.com/en-us/library/83y7df3e(VS.71).aspx [XmlRoot(Namespace = "www.contoso.com", ElementName = "MyGroupName", DataType = "string", IsNullable=true)] – david valentine Oct 12 '09 at 23:55
50

XML is case-sensitive. "User" and "user" are different names. – John Saunders Oct 14 '09 at 01:20
4

I downvoted because I feel that the XmlRoot information should be defined on the class itself, not where it is being deserialized. For this reaon, Junto's solution is, in my opinion, superior. – GuiSim May 02 '12 at 15:26
Anyone looking for the correct solution to a namespace or root element error, look at @davidvalentine's comment and Junto's answer. – Sam Sussman Sep 18 '12 at 14:28
6

@GuiSim You are assuming that the OP has control over the original entity. Both answers are valid. In my case I cannot add XmlRoot to the entity as it is used as part of a MessageContract. Using the above method works for my scenario. – Bronumski Aug 12 '13 at 15:50
@Bronumski From the OP's post: _And i have the following for my User Class_ . I think it is fair to assume that the OP has control over the original entity. – GuiSim Aug 12 '13 at 21:09
5

@GuiSim I don't disagree with what the OP said. I have complete control of my root entity but cannot use the rootattribute because it conflicts with the MessageContract attribute. Both answers are valid and the alternative was presented in the comments. The point is you are down voting a valid answer not a wrong answer. If you don't agree that it is the best just don't vote. – Bronumski Aug 13 '13 at 12:40
@Bronumski I removed my downvote. I still think that Junto's answer is superior and should be the accepted solution. – GuiSim Aug 14 '13 at 14:08
You should probably mention that this solution is quite slow and will cause memory leaks of not careful. Se https://msdn.microsoft.com/en-us/library/system.xml.serialization.xmlserializer%28v=vs.110%29.aspx Dynamically Generated Assemblies – oddbear Jun 03 '15 at 10:07
In my case adding the XmlRoot attribute did not solve the problem but adding it to the serializer at runtime did. Not sure why but it validates putting both solutions here. – Steve Hiner Sep 01 '16 at 23:35
The data annotation somehow didn't work for me but the second method worked. – tedi Sep 28 '20 at 11:56
Thanks. Add "Namespace" in [XmlRoot] attribute helped my for Deserialize kml file. – bob Jun 09 '21 at 08:34

score 154 · Answer 2 · answered Jan 21 '11 at 17:41

154

Even easier is just to add the following annotations to the top of your class:

[Serializable, XmlRoot("user")]
public partial class User
{
}

answered Jan 21 '11 at 17:41

Rebecca

13,914
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136

Simple and Straightforward – mayowa ogundele Feb 10 '17 at 10:38
Ahh!! You Just saved my 6 manual convertions from XML to class obj! Works as expected.. Thanks – Thameem Apr 14 '21 at 14:34

Ranadheer Reddy · Answer 3 · 2013-03-11T06:12:43.030

32

XmlSerializer xs = new XmlSerializer(typeof(User), new XmlRootAttribute("yourRootName"));

edited Mar 11 '13 at 06:12

answered Mar 07 '13 at 06:52

Ranadheer Reddy

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3

This is fantastic for cases where otherwise you would have to rely on conditionals for attributes. – Den Sep 24 '15 at 18:00
2

Very simple and fixed my case completely. Thanks! – A.W Dec 05 '17 at 03:48

Jeremy Thompson · Answer 4 · 2020-02-11T06:22:05.263

15

The error message is so vague, for me I had this code:

var streamReader = new StreamReader(response.GetResponseStream());
var xmlSerializer = new XmlSerializer(typeof(aResponse));
theResponse = (bResponse) xmlSerializer.Deserialize(streamReader);

Notice xmlSerializer is instantiated with aResponse but on deserializing I accidentally casted it to bResonse.

edited Feb 11 '20 at 06:22

answered Apr 04 '12 at 03:48

Jeremy Thompson

61,933
36
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2

Just had a similar problem. xmlserializer was initialized to typeof(T) and I was casting to List – nurettin Aug 29 '12 at 05:18
1

I declared `XmlRoot(..)` on child class `ClassB` of parent class `ClassA`. I used `new XmlSerializer(typeof(ClassA)` instead of on `ClassB` and also casted object to it. Thanks for pointing out! – Shishir Gupta Apr 25 '17 at 14:21

score 8 · Answer 5 · edited Sep 10 '18 at 19:46

8

The simplest and best solution is just to use XMLRoot attribute in your class, in which you wish to deserialize.

Like:

[XmlRoot(ElementName = "YourPreferableNameHere")]
public class MyClass{
...
}

Also, use the following Assembly :

using System.Xml.Serialization;

edited Sep 10 '18 at 19:46

wp78de

18,207
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answered Sep 28 '16 at 07:42

Khawaja Asim

1,327
18
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8

Anyone want to give an explanation? *Why* is `XmlRoot()` attribute needed to fix this problem? There are 5 answers here that say, "just add this code" and not one explanation. People are answering 7 years later and it's still just, "Add this XmlRoot code". Of all the answers I picked the newest one to ask for an explanation. – Quantic Oct 13 '16 at 15:44
Thankyou @Quantic, XmlRoot allow you to create a new xml value rather than using default root you can customize that. It just kind of decoration you will add to your xml at compile time. Will eliminate some times compatibility issues. Hope you get my point. – Khawaja Asim Oct 14 '16 at 07:06

score 7 · Answer 6 · answered Jun 20 '19 at 17:39

In my case, my xml had multiple namespaces and attributes. So I used this site to generate the objects - https://xmltocsharp.azurewebsites.net/

And used the below code to deserialize

 XmlDocument doc =  new XmlDocument();
        doc.Load("PathTo.xml");
        User obj;
        using (TextReader textReader = new StringReader(doc.OuterXml))
        {
            using (XmlTextReader reader = new XmlTextReader(textReader))
            {
                XmlSerializer serializer = new XmlSerializer(typeof(User));
                obj = (User)serializer.Deserialize(reader);
            }
        }

The link provided a huge help to deserialize xml to class. – smr5 Oct 16 '19 at 18:58 — smr5, Oct 16 '19 at 18:58
The provided URL was huge for me too. – thargenediad Jul 09 '20 at 17:17 — thargenediad, Jul 09 '20 at 17:17

score 6 · Answer 7 · answered Mar 16 '10 at 13:39

6

As John Saunders says, check if the class/property names matches the capital casing of your XML. If this isn't the case, the problem will also occur.

answered Mar 16 '10 at 13:39

Luuk

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score 3 · Answer 8 · answered Jun 08 '16 at 23:03

The only thing that worked in my case was by using david valentine code. Using Root Attr. in the Person class did not help.

I have this simple Xml:

<?xml version="1.0"?>
<personList>
 <Person>
  <FirstName>AAAA</FirstName>
  <LastName>BBB</LastName>
 </Person>
 <Person>
  <FirstName>CCC</FirstName>
  <LastName>DDD</LastName>
 </Person>
</personList>

C# class:

public class Person
{
    public string FirstName { get; set; }
    public string LastName { get; set; }
}

De-Serialization C# code from a Main method:

XmlRootAttribute xRoot = new XmlRootAttribute();
xRoot.ElementName = "personList";
xRoot.IsNullable = true;
using (StreamReader reader = new StreamReader(xmlFilePath))
{
List<Person> result = (List<Person>)(new XmlSerializer(typeof(List<Person>), xRoot)).Deserialize(reader);
 int numOfPersons = result.Count;
}

This one worked for me but I had to have the same lower-upper case name of "Person" in this example as the class name. — tedi, Sep 28 '20 at 12:07
Or you can use `[XmlType(TypeName = "Person")]` annotation above your item class. — tedi, Sep 28 '20 at 12:14

score 3 · Answer 9 · answered Oct 20 '16 at 19:44

3

My issue was that the root element actually has a xmlns="abc123"

So had to make XmlRoot("elementname",NameSpace="abc123")

answered Oct 20 '16 at 19:44

Nick

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Definitely helped me... You must add the namespace if your XML has one. – rios0rios0 Jul 22 '21 at 20:02

Adel Mourad · Answer 10 · 2020-05-01T10:11:29.500

Nothing worked for me for these errors EXCEPT

... was not expected, 
... there is an error in XML document (1,2)
... System.FormatException Input String was not in correct format ...

Except this way

1- You need to inspect the xml response as string (the response that you are trying to de-serialize to an object)

2- Use online tools for string unescape and xml prettify/formatter

3- MAKE SURE that the C# Class (main class) you are trying to map/deserialize the xml string to HAS AN XmlRootAttribute that matches the root element of the response.

Exmaple:

My XML Response was staring like :

<ShipmentCreationResponse xmlns="http://ws.aramex.net/ShippingAPI/v1/">
       <Transaction i:nil="true" xmlns:i="http://www.w3.org/2001/XMLSchema-instance"/>
           ....

And the C# class definition + attributes was like :

[System.Diagnostics.DebuggerStepThroughAttribute()]
[System.CodeDom.Compiler.GeneratedCodeAttribute("System.ServiceModel", "4.0.0.0")]
[System.ServiceModel.MessageContractAttribute(WrapperName="ShipmentCreationResponse", WrapperNamespace="http://ws.aramex.net/ShippingAPI/v1/", IsWrapped=true)]
public partial class ShipmentCreationResponse {
  .........
}

Note that the class definition does not have "XmlRootAttribute"

[System.Xml.Serialization.XmlRootAttribute(Namespace = "http://ws.aramex.net/ShippingAPI/v1/", IsNullable = false)]

And when i try to de serialize using a generic method :

public static T Deserialize<T>(string input) where T : class
{
    System.Xml.Serialization.XmlSerializer ser = new System.Xml.Serialization.XmlSerializer(typeof(T));

    using (System.IO.StringReader sr = new System.IO.StringReader(input))
    {
        return (T)ser.Deserialize(sr);
    }
}





var _Response = GeneralHelper.XMLSerializer.Deserialize<ASRv2.ShipmentCreationResponse>(xml);

I was getting the errors above

... was not expected, ... there is an error in XML document (1,2) ...

Now by just adding the "XmlRootAttribute" that fixed the issue for ever and for every other requests/responses i had a similar issue with:

[System.Xml.Serialization.XmlRootAttribute(Namespace = "http://ws.aramex.net/ShippingAPI/v1/", IsNullable = false)]

..

[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true, Namespace = "http://ws.aramex.net/ShippingAPI/v1/")]
[System.Xml.Serialization.XmlRootAttribute(Namespace = "http://ws.aramex.net/ShippingAPI/v1/", IsNullable = false)]
public partial class ShipmentCreationResponse
{
    ........
}

score 2 · Answer 11 · answered Feb 09 '16 at 17:05

My problem was one of my elements had the xmlns attribute:

<?xml version="1.0" encoding="utf-8"?>
<RETS ReplyCode="0">
    <RETS-RESPONSE xmlns="blahblah">
        ...
    </RETS-RESPONSE>
</RETS>

No matter what I tried the xmlns attribute seemed to be breaking the serializer, so I removed any trace of xmlns="..." from the xml file:

<?xml version="1.0" encoding="utf-8"?>
<RETS ReplyCode="0">
    <RETS-RESPONSE>
        ...
    </RETS-RESPONSE>
</RETS>

and voila! Everything worked.

I now parse the xml file to remove this attribute before deserializing. Not sure why this works, maybe my case is different since the element containing the xmlns attribute is not the root element.

Your file specified that RETS-RESPONSE was in the "blahblah" namespace. If that didn't match your schema, the element would not have been found. Also, adding a default namespace causes all sorts of other reference problems, too. — Suncat2000, Mar 10 '20 at 21:26

score 2 · Answer 12 · answered Oct 28 '18 at 06:41

2

All above not worked for me, but this was: Check that the name of Root element of class is exactly like the one from XML case sensitive.

answered Oct 28 '18 at 06:41

shdr

868
10
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{" was not expected.} Deserializing Twitter XML

12 Answers12

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