The Application serializes a List to xml:
<?xml version="1.0"?>
<Tools xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<Tool>
<Name>test1</Name>
<Path>C:\Program Files\FreePDF_XP\fpmailer.exe</Path>
</Tool>
<Tool>
<Name>test2</Name>
<Path>C:\Program Files\FreePDF_XP\fpassist.exe</Path>
</Tool>
<Tool>
<Name>test3</Name>
<Path>C:\Program Files\FreePDF_XP\ShellMail.exe</Path>
</Tool>
</Tools>
If I'm don't on the wrong way I must create new Objects and add them to a list:
Tool tool = new Tool();
XmlSerializer deserializer = new XmlSerializer(typeof(List<Tool>));
using (var reader = new StreamReader(#Start.userConfigurePath + "\\config.xml"))
{
tool = (Tool)deserializer.Deserialize(reader);
reader.Close();
}
toolList.Add(tool);
}
In the result there isn't a Object in the List. How can I deserialize the serialized objects in the xml in a List? Could it be that there is something wrong at the code for derisalization?
Edit:
Something seems to be wrong with my xml declaration (XML File Error 2,2). That I don't understand because i create the xml document on this way:
XmlDocument toolConfig = new XmlDocument();
XmlNode myRoot;
myRoot = toolConfig.CreateElement("Tool");
toolConfig.AppendChild(myRoot);
It should be
toolList = (List<Tool>)deserializer.Deserialize(reader);
Related
I'm trying to get a single value from an XML using the SelectSingleNode, but it keeps returning null on me.
I've been looking here on SO and it seems it has something to do with the namespace. I tried adding it but I keep getting null.
The XML looks like this:
<?xml version="1.0" encoding="UTF-8"?>
<EML xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xnl="urn:oasis:names:tc:ciq:xnl:4"
xmlns:xal="urn:oasis:names:tc:ciq:xal:4"
xmlns="urn:oasis:names:tc:evs:schema:eml"
xmlns:martine="http://www.martine.be/extensions"
Id="510"
SchemaVersion="7.0"
xsi:schemaLocation="urn:oasis:names:tc:evs:schema:eml schema/510-count-v7-0.xsd
http://www.martine.be/extensions schema/martine-eml-extensions.xsd">
<EMLHeader>
<TransactionId>01</TransactionId>
<ManagingAuthority>
<AuthorityIdentifier>2</AuthorityIdentifier>
<AuthorityName>
<NameElement ElementType="">VLR</NameElement>
</AuthorityName>
<Description>Some Description</Description>
<OrganizationURL>Unknown</OrganizationURL>
<AuthorityAddress/>
</ManagingAuthority>
</EMLHeader>
I'm trying to extract the Description using the code below:
XmlDocument doc = new XmlDocument();
doc.LoadXml(content);
var nsmgr = new XmlNamespaceManager(doc.NameTable);
nsmgr.AddNamespace("ns", "urn:oasis:names:tc:evs:schema:eml");
XmlNode testNode = doc.SelectSingleNode("/ns:EML/ns:EMLHeader/ns:ManagingAuthority/ns:Description", nsmgr);
if (testNode != null)
{
Console.WriteLine(testNode.InnerText);
}
What am doing wrong?
Tested this and you are missing closing </EML> tag. This was the error I got
Unhandled Exception: System.Xml.XmlException: Unexpected end of file has occurred. The following elements are not closed: EML. Line 24, position 17.
TestCodeApp.cs
using System;
using System.Xml;
public class Program
{
public static void Main()
{
XmlDocument doc = new XmlDocument();
doc.Load("input.xml");
var nsmgr = new XmlNamespaceManager(doc.NameTable);
nsmgr.AddNamespace("ns", "urn:oasis:names:tc:evs:schema:eml");
XmlNode testNode = doc.SelectSingleNode("/ns:EML/ns:EMLHeader/ns:ManagingAuthority/ns:Description", nsmgr);
if (testNode != null)
{
Console.WriteLine(testNode.InnerText);
}
}
}
input.xml
<?xml version="1.0" encoding="UTF-8"?>
<EML xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xnl="urn:oasis:names:tc:ciq:xnl:4"
xmlns:xal="urn:oasis:names:tc:ciq:xal:4"
xmlns="urn:oasis:names:tc:evs:schema:eml"
xmlns:martine="http://www.martine.be/extensions"
Id="510"
SchemaVersion="7.0"
xsi:schemaLocation="urn:oasis:names:tc:evs:schema:eml schema/510-count-v7-0.xsd
http://www.martine.be/extensions schema/martine-eml-extensions.xsd">
<EMLHeader>
<TransactionId>01</TransactionId>
<ManagingAuthority>
<AuthorityIdentifier>2</AuthorityIdentifier>
<AuthorityName>
<NameElement ElementType="">VLR</NameElement>
</AuthorityName>
<Description>Some Description</Description>
<OrganizationURL>Unknown</OrganizationURL>
<AuthorityAddress/>
</ManagingAuthority>
</EMLHeader>
</EML>
Your code look OK, except:
you missed close node in the xml file: ""
if your content is contain information of above xml, then you can use doc.LoadXML(content), otherwise you should use doc.Load(fileName).
Your xml file should be:
<?xml version="1.0" encoding="UTF-8"?>
<EML xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xnl="urn:oasis:names:tc:ciq:xnl:4"
xmlns:xal="urn:oasis:names:tc:ciq:xal:4"
xmlns="urn:oasis:names:tc:evs:schema:eml"
xmlns:martine="http://www.martine.be/extensions"
Id="510"
SchemaVersion="7.0"
xsi:schemaLocation="urn:oasis:names:tc:evs:schema:eml schema/510-count-v7-0.xsd
http://www.martine.be/extensions schema/martine-eml-extensions.xsd">
<EMLHeader>
<TransactionId>01</TransactionId>
<ManagingAuthority>
<AuthorityIdentifier>2</AuthorityIdentifier>
<AuthorityName>
<NameElement ElementType="">VLR</NameElement>
</AuthorityName>
<Description>Some Description</Description>
<OrganizationURL>Unknown</OrganizationURL>
<AuthorityAddress/>
</ManagingAuthority>
</EMLHeader>
</EML>
And then you can read it:
XmlDocument doc = new XmlDocument();
doc.Load(fileName);
var nsmgr = new XmlNamespaceManager(doc.NameTable);
nsmgr.AddNamespace("ns", "urn:oasis:names:tc:evs:schema:eml");
XmlNode testNode = doc.SelectSingleNode("/ns:EML/ns:EMLHeader/ns:ManagingAuthority/ns:Description", nsmgr);
if (testNode != null)
{
Console.WriteLine(testNode.InnerText);
}
I build an XML Document which needs to be validated against a xsd file. Thus I need a reference to the xsd file in the root element of the xml. So far I use this C# Code:
var ser = new XmlSerializer(typeof(myspecialtype));
XmlSerializerNamespaces MainNamespace = new XmlSerializerNamespaces();
MainNamespace.Add("xlink", "http://www.w3.org/1999/xlink");
MainNamespace.Add("xsi", "http://www.w3.org/2001/XMLSchema-instance");
using (XmlWriter w = XmlWriter.Create(#"C:\myxmlfile.xml"))
{
w.WriteProcessingInstruction("xml-stylesheet", "type=\"text/xsl\" href=\"utils/somexsl.xsl\"");
ser.Serialize(w, LeBigObject, HauptNs);
}
The resulting Xml begins like this:
<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="utils/somexsl.xsl"?>
<caddy-xml xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xlink="http://www.w3.org/1999/xlink" xmlVersion="03.07.00">
but I need this:
<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="utils/somexsl.xsl"?>
<caddy-xml xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xlink="http://www.w3.org/1999/xlink" xmlVersion="03.07.00" xsi:noNamespaceSchemaLocation="utils/theveryimportant.xsd">
I came across "CreateAttribute" here: Add Namespace to an xml root node c# but I can't put it together with the Serializer. Thank you!
I was pointed to the solution here:
https://social.msdn.microsoft.com/Forums/en-US/e43585c6-181b-4449-8806-b07f82681a2a/how-to-include-xsinonamespaceschemalocation-in-the-xml?forum=asmxandxml
I added this to my class:
[XmlAttribute("noNamespaceSchemaLocation", Namespace = XmlSchema.InstanceNamespace)]
public string attr = "utils/theveryimportant.xsd";
and it works.
I have an XML file with some data on it..for example
<?xml version="1.0" encoding="utf-8"?>
<CreateAndSendMessageRequest xmlns:xsi="http://www.w3.org/2001/XMLSchema-
instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema"
xmlns="http://schemas.communisis.com/lv">
<CompositionRequest>
<Metadata xmlns="http://lv.com/gi/si/common/CommonTypes">
<PolicyReference>250028766505DN</PolicyReference>
<AccountReference>Test1234</AccountReference>
<QuoteReference>Test3214</QuoteReference>
<OutboundTransactionID>string</OutboundTransactionID>
</Metadata>
Now i want to replace POLICY REFERENCE value with some dummy data. I am able to do that but now i have to save it as a new file with a different file name..
How can i achieve that..
For reference i am giving my code.
XmlDocument doc = new XmlDocument();
doc.Load(filepath);
XmlNodeList node = doc.GetElementsByTagName("PolicyReference");
var item =node.Item(0);
string value = item.FirstChild.Value;
string nevalue = value.Replace(value, "Test123");
doc.DocumentElement.;
doc.Save(#"C:\Test\file.xml");
The xml wasn't valid.
Anyway, this is how to make it work:
Xml file:
<?xml version="1.0" encoding="UTF-8"?>
<CreateAndSendMessageRequest>
<CompositionRequest>
<Metadata xmlns="http://lv.com/gi/si/common/CommonTypes">
<PolicyReference>250028766505DN</PolicyReference>
<AccountReference>Test1234</AccountReference>
<QuoteReference>Test3214</QuoteReference>
<OutboundTransactionID>string</OutboundTransactionID>
</Metadata>
</CompositionRequest>
</CreateAndSendMessageRequest>
Code:
XmlDocument doc = new XmlDocument();
doc.Load("bho.xml");
XmlNodeList node = doc.GetElementsByTagName("PolicyReference");
node[0].InnerText = "Test123";
doc.Save("file.xml");
Result xml:
<?xml version="1.0" encoding="UTF-8"?>
<CreateAndSendMessageRequest>
<CompositionRequest>
<Metadata xmlns="http://lv.com/gi/si/common/CommonTypes">
<PolicyReference>Test123</PolicyReference>
<AccountReference>Test1234</AccountReference>
<QuoteReference>Test3214</QuoteReference>
<OutboundTransactionID>string</OutboundTransactionID>
</Metadata>
</CompositionRequest>
</CreateAndSendMessageRequest>
Hope it helps you ^^
I have problem when I'm trying to deserialize an XML to object. My XML look like:
<?xml version="1.0" encoding="utf-16"?>
<Products
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<AllProducts>
<Product>
<ID>8</ID>
<GID>51280</GID>
<Kod>RNIKAKC1.6</Kod>
<Name>SB-800</Name>
<Ean>0018208048014</Ean>
<CommodityGroup>
<ID>86</ID>
<Name>AKCESORIA FOTO</Name>
<Path>
<Category>
<ID>60798</ID>
<Name>ARCHIWALNE</Name>
</Category>
</Path>
</CommodityGroup>
</Product>
....
Next products
...
My method code:
var MemoryStream = APIAccess.DownloadFileToStream("example.xml", "exampleContainer");
using (MemoryStream)
{
MemoryStream.Position = 0;
using (StreamReader StreamReader = new StreamReader(MemoryStream))
{
XmlSerializer serializer = new XmlSerializer(typeof(CommodityGroup));
var products = serializer.Deserialize(StreamReader);
}
}
Method DownloadFileToStream is working good, because it is useful in other classes.
I'm geting error:
InvalidOperationException: Products xmlns='' was not expected.
I want to create object of a Node CommodityGroup. I've created class selecting this node, coping it and pasting in the new class like Paste Special -> XML
Attributes of this class looks like:
[Serializable()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[XmlTypeAttribute(AnonymousType = true)]
[XmlRootAttribute(Namespace = "CommodityGroup", IsNullable = false)]
I don't know to fix it. When I'm adding into XML Serializer param new XmlRootAttribute("Products"), I'm getting "0" values.
Do you have any suggestions?
If you want to deserialize only part of an xml document, you should skip unnecessary nodes.
Do it using XmlReader.
using (StreamReader StreamReader = new StreamReader(MemoryStream))
using (var xmlReader = XmlReader.Create(StreamReader))
{
xmlReader.ReadToFollowing("CommodityGroup");
XmlSerializer serializer = new XmlSerializer(typeof(CommodityGroup));
var commodityGroup = (CommodityGroup)serializer.Deserialize(xmlReader);
}
I created a .NET class for XML serialization
[Serializable()]
[XmlRoot("documents")]
public class BdfXmlData
{
[XmlElement("document")]
public List<XmlElement> Documents { get; set; }
public BdfXmlData()
{
Documents = new List<XmlElement>();
}
}
When I try to serialize an object, I get with this tree:
<?xml version="1.0" encoding="utf-8"?>
<ArrayOfXmlElement xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<XmlElement>
<documents>
<document>
...
<document>
<documents>
</XmlElement>
</ArrayOfXmlElement>
How can I have the following three?
<?xml version="1.0" encoding="utf-8"?>
<documents>
<document>
...
<document>
<documents>
Thanks in advance.
The code to serialize my class is the following:
public static string GetSerializedObject<T>(T t)
{
using (MemoryStream stream = new MemoryStream())
{
XmlSerializer serializer = new XmlSerializer(typeof(T));
serializer.Serialize(stream, t);
stream.Position = 0;
using (StreamReader reader = new StreamReader(stream))
return reader.ReadToEnd();
}
}
I think you can at least simplify your xml-code:
<ArrayOfdocuments xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<documents>
<document>
...
<document>
<documents>
</ArrayOfdocuments>
The "XmlElement" element can be left out.
Surely you are serialising an array of BdfXmlData
in pseudo code
BdfXmlData[] dataInArray = new [] {new BdfXmlData()};
GetSerializedObject(dataInArray)
would produce what you have whereas,
in pseudo code..
BdfXmlData data = new BdfXmlData();
GetSerializedObject(data)
should give you what you want!
Exact, I serialise the Documents collection :
SerializationHelper.GetSerializedObject(xmlRootData.Documents);
In fact, I receive some XmlElement with this tree :
<document>
...
<document>
and I add these elements to my List "Documents".
xmlRootData.Documents.Add(xmlData);
When I serialise my final object (BdfXmlData), the xml contains 2 documents tags and 2 document tags
string xmlData = SerializationHelper.GetSerializedObject(xmlRootData);
<documents xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<document>
<documents>
<document>
</document>
<documents>
</document>
</documents>