is it possible to rewrite XML file according to other schema XSD using c# ?
this is XML file
this is the current schema XSD file
and this is the new schema same everything but changed node names
so how to get a new XML from the old one based on the new XSD using c#?
Using xml linq :
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;
namespace ConsoleApplication13
{
class Program
{
const string FILENAME = #"c:\temp\test.xml";
static void Main(string[] args)
{
XDocument doc = XDocument.Load(FILENAME);
List<XElement> shipTo = doc.Descendants("shipto").ToList();
foreach (XElement ship in shipTo)
{
ship.Element("name").ReplaceWith(new XElement("FullName", (string)ship.Element("name")));
ship.Element("address").ReplaceWith(new XElement("FirstAddress", (string)ship.Element("address")));
ship.Element("city").ReplaceWith(new XElement("homeTown", (string)ship.Element("city")));
ship.Element("country").ReplaceWith(new XElement("HomeLand", (string)ship.Element("country")));
}
}
}
}
Related
Say I have an XSLT file like below:
`<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:math="http://www.w3.org/2005/xpath-functions/math" exclude-result-
prefixes="xs math"
version="3.0">`
....... and so on.
I need the output as 3.0 because the above file has version="3.0". I want to use C# to get this given the XSLT is in string format
Using xml linq :
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;
namespace ConsoleApplication1
{
class Program
{
const string FILENAME = #"c:\temp\test.xml";
static void Main(string[] args)
{
XDocument doc = XDocument.Load(FILENAME);
string version = (string)doc.Root.Attribute("version");
}
}
}
Use XElement.Parse(yourString).Attribute("version").Value, where you add using System.Xml.Linq;. See https://learn.microsoft.com/en-us/dotnet/csharp/programming-guide/concepts/linq/linq-to-xml-overview for details of the used API.
I generated a XML file through API call then I tried to read the file using XML source component in ssis but it is read only data sets except all data contains in file .
Here my file
<?XML version 1.0 >
<ABC>
<a>info<a/>
<ABC/>
But I want file like below then only I can easily read file using component
We can manipulate the file manually for single file but not for thousand files
<?XML Version 1.0>
<X>
<ABC>
<a>info <a/>
<ABC/>
</X>
How to add that 'X' node to the existing file .
I am not having much exposure on .Net technology .
Kindly help me at the earliest of time .
Thank You
KiranKumar
Using xml linq
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
string xml =
"<?xml version=\"1.0\" encoding=\"utf-8\" ?>" +
"<ABC>" +
"<a>info</a>" +
"</ABC>";
XDocument doc = XDocument.Parse(xml);
XElement root = doc.Root;
root.ReplaceWith(new XElement("X", root));
}
}
}
Try streaming API.
using (var reader = XmlReader.Create("test.xml"))
using (var writer = XmlWriter.Create("test2.xml"))
{
writer.WriteStartElement("X");
reader.MoveToContent();
writer.WriteNode(reader.ReadSubtree(), true);
writer.WriteEndElement();
}
This approach handles xml without excessive memory consumption.
Also, this method allows to modify xml on the fly, getting it from the input API stream and writing to output stream.
using (var reader = XmlReader.Create(inputStream))
using (var writer = XmlWriter.Create(outputStream))
I want to read XML document from a property which is created in edit mode of Episerver.
I have made one property of type 'URL to Document'.
When I try to fetch it from code behind, it gives only file path. I am not able to read the content of XML file which is uploaded in property.
string XMLContent = Currentpage.Getproperty<string>("XMLFile");
Can anyone help out on this?
You need to load the file as well. Something like this:
var path = CurrentPage["XMLFile"] as string;
if (HostingEnvironment.VirtualPathProvider.FileExists(path))
{
var file = HostingEnvironment.VirtualPathProvider.GetFile(path) as UnifiedFile;
if (file != null)
{
using (var stream = file.Open())
{
// Here is your XML document
var xml = XDocument.Load(stream);
}
}
}
You can also load the file content by using the local path on disk, file.LocalPath.
try this
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
string XMLContent = "";
//using XML
XmlDocument doc1 = new XmlDocument();
doc1.LoadXml(XMLContent);
//using xml linq
XDocument doc2 = XDocument.Parse(XMLContent);
}
}
}
My issue is, when selecting a node from xml file (packet in my case) is decoding an url that is the inner text of the node within the xml. Example
<url>"http://my.xml.org/?id=AAA%2DDDD%3dNNNLKLKJLKL%2"</url>
using an instance of XmlDocument:
xmlDoc.SelectSingleNode("url").InnerText;
The string that is returned is: "http://my.xml.org/?id=AAA/DDDD/NNNLKLKJLKL/"
How do I prevent this from happening? Notice that the %2 and %3's have been changed to forward slashes.
The code below gets correct results. What are you doing differently?
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
string xml =
"<?xml version=\"1.0\" encoding=\"utf-8\" ?>" +
"<url>\"http://my.xml.org/?id=AAA%2DDDD%3dNNNLKLKJLKL%2\"</url>";
XmlDocument xmlDoc = new XmlDocument();
xmlDoc.LoadXml(xml);
string results = xmlDoc.SelectSingleNode("url").InnerText;
}
}
}
How i can open an xml or xsl file inside the same assembly from source code in C#?
Any ideas?
Google found this
using System;
using System.IO;
using System.Reflection;
using System.Xml;
class Application
{
static void Main(string[] args)
{
Stream s =
Assembly.GetExecutingAssembly().GetManifestResourceStream("File1.xml");
XmlDocument xdoc = new XmlDocument();
using (StreamReader reader = new StreamReader(s))
xdoc.LoadXml(reader.ReadToEnd());
}
}