CODEWITHSUNDEEP
c#xmlxsdxml-namespaces
Richard Barraclough
Richard Barraclough

Reputation: 2964

Using XSDs with includes

Here is an XSD:

<?xml version="1.0"?>
<xsd:schema 
elementFormDefault='unqualified' 
attributeFormDefault='unqualified' 
xmlns:xsd='http://www.w3.org/2001/XMLSchema' 
>

  <xsd:simpleType name='TheSimpleType'>
    <xsd:restriction base='xsd:string' />
  </xsd:simpleType>
</xsd:schema>

Here is a second XSD that includes the one above:

<?xml version="1.0" encoding="UTF-8" ?>
<xsd:schema 
elementFormDefault='unqualified' 
attributeFormDefault='unqualified' 
xmlns:xsd='http://www.w3.org/2001/XMLSchema' 
targetNamespace='a'
xmlns='a'
>

  <xsd:include schemaLocation='Include.xsd' />

  <xsd:element name = "TheElement" >
  <xsd:complexType>
  <xsd:attribute name="Code" type="TheSimpleType" use="required"/>
  </xsd:complexType>
  </xsd:element>
</xsd:schema>

I need to read the (second) XSD into C# and:

  1. check that it is a valid XSD, and
  2. validate documents against it.

Here is some C# to read in the schemata:

    XmlSchemaSet schemaSet = new XmlSchemaSet();
    foreach (string sd in Schemas)
    {
        using (XmlReader r = XmlReader.Create(new FileStream(sd, FileMode.Open)))
        {
            schemaSet.Add(XmlSchema.Read(r, null));
        }
    }
    schemaSet.CompilationSettings = new XmlSchemaCompilationSettings();
    schemaSet.Compile();

The .Compile() fails because "Type 'a:TheSimpleType' is not declared, or is not a simple type."

However, it works if either:

The question is: how do I get C# to accept it without editing the schemata?

I suspect the problem is that although I have put both schemata into the XmlSchemaSet, I still need to tell C# that one is included into the other, i.e., it hasn't worked it out for itself. Indeed, if I only tell the XmlSchemaSet about the main XSD (and not the include) (both without (or with) namespaces) then "Type 'TheSimpleType' is not declared, or is not a simple type."

Thus this seems to be a question about resolving includes: how?!

Upvotes: 19

Views: 12099

Answers (6)

Raviraj Bhalerao
Raviraj Bhalerao

Reputation: 59

With .net6 the code goes as follows -


XmlReaderSettings settings = new XmlReaderSettings();
settings.Schemas.XmlResolver = new XmlUrlResolver();     // Need this for resolving include and import
settings.ValidationType = ValidationType.Schema; // This might not be needed, I am using same settings to validate the input xml
settings.ValidationFlags |= XmlSchemaValidationFlags.ProcessInlineSchema;
settings.ValidationFlags |= XmlSchemaValidationFlags.ProcessSchemaLocation;
settings.Schemas.Add(null, "yourpath\\yourxsd.xsd");

settings.Schemas.Compile();

string xmlFilePath = "yourpath\\your.xml";
ValidationEventHandler eventHandler = new ValidationEventHandler(ValidationEventHandler);

// Create an XmlReader for the XML file
using (XmlReader reader = XmlReader.Create(xmlFilePath, settings))
{
}

Wanted to highlight that with net6, we have to set settings.Schemas.XmlResolver = new XmlUrlResolver() instead of settings.XmlResolver = new XmlUrlResolver()

Upvotes: 0

Michel
Michel

Reputation: 51

The default behaviour of the XmlSchemaSet is to not try to resolve any XSD included schemas. To do this, the XmlResolver property must be initialised.

XmlSchemaSet schemas = new XmlSchemaSet
{
    XmlResolver = new XmlUrlResolver()
};

Additionally, you have to set baseUri for XmlReader as per @Richard Barraclough's answer.

Upvotes: 2

CountZero
CountZero

Reputation: 6379

Here is the method I wrote to handle xsd validation. Hope this helps some one. 

        /// <summary>
        /// Ensure all xsd imported xsd documented are in same folder as master xsd
        /// </summary>
        public XsdXmlValidatorResult Validate(string xmlPath, string xsdPath, string xsdNameSpace)
        {
            var result = new XsdXmlValidatorResult();
            var readerSettings = new XmlReaderSettings {ValidationType = ValidationType.Schema};
            readerSettings.ValidationFlags |= XmlSchemaValidationFlags.ProcessInlineSchema; 
            readerSettings.ValidationFlags |= XmlSchemaValidationFlags.ProcessSchemaLocation;
            readerSettings.Schemas.Add(null, xsdPath);

            readerSettings.ValidationEventHandler += (sender, args) =>
                {
                    switch (args.Severity)
                    {
                        case XmlSeverityType.Warning:
                            result.Warnings.Add(args.Message);
                            break;
                        case XmlSeverityType.Error:
                            result.IsValid = false;
                            result.Warnings.Add(args.Message);
                            break;
                    }
                };

            var reader = XmlReader.Create(xmlPath, readerSettings);

            while (reader.Read()) { }

            return result;
        }

Upvotes: 0

Richard Barraclough
Richard Barraclough

Reputation: 2964

The problem is with the way the schema is opened for reading on the line:

XmlReader.Create(new FileStream(sd, FileMode.Open)

I had to write my own XmlResolver before I could see how the paths to the include files were being resolved: it was from the directory of the executable and not from the directory of the parent schema. The problem is that the parent schema was not getting its BaseURI set. Here's how the schema must be opened:

XmlReader.Create(new FileStream(pathname, FileMode.Open, FileAccess.Read),null, pathname)

Upvotes: 29

Richard Barraclough
Richard Barraclough

Reputation: 2964

Try this :D

public static XmlSchema LoadSchema(string pathname)
{
    XmlSchema s = null;
    XmlValidationHandler h = new XmlValidationHandler();
    using (XmlReader r = XmlReader.Create(new FileStream(pathname, FileMode.Open)))
    {
        s = XmlSchema.Read(r, new ValidationEventHandler(h.HandleValidationEvent));
    }

    if (h.Errors.Count > 0)
    {
        throw new Exception(string.Format("There were {1} errors reading the XSD at {0}. The first is: {2}.", pathname, h.Errors.Count, h.Errors[0]));
    }

    return s;
}

public static XmlSchema LoadSchemaAndResolveIncludes(string pathname)
{
    FileInfo f = new FileInfo(pathname);
    XmlSchema s = LoadSchema(f.FullName);

    foreach(XmlSchemaInclude i in s.Includes)
    {
        XmlSchema si = LoadSchema(f.Directory.FullName + @"\" + i.SchemaLocation);
        si.TargetNamespace = s.TargetNamespace;
        i.Schema = si;
    }

    return s;
}

public static List<ValidationEventArgs> Validate(string pathnameDocument, string pathnameSchema)
{
    XmlSchema s = LoadSchemaAndResolveIncludes(pathnameSchema);

    XmlValidationHandler h = new XmlValidationHandler();

    XmlDocument x = new XmlDocument();
    x.Load(pathnameDocument);
    x.Schemas.Add(s);
    s.Compile(new ValidationEventHandler(h.HandleValidationEvent));
    x.Validate(new ValidationEventHandler(h.HandleValidationEvent));
    return h.Errors;
}

Note in particular the si.TargetNamespace = s.TargetNamespace;.

Obviously, this assumes that the includes are specified as file paths relative to the schema into which they are included.

Upvotes: 1

Jord&#227;o
Jord&#227;o

Reputation: 56457

You can use the XmlSchema.Includes to link them together. Then, you just need to add the main schema to the schema set:

var includeSchema = XmlSchema.Read(XmlReader.Create(...), null);
var mainSchema = XmlSchema.Read(XmlReader.Create(...), null);

var include = new XmlSchemaInclude();
include.Schema = includeSchema;
mainSchema.Includes.Add(include);

var schemaSet = new XmlSchemaSet();
schemaSet.Add(mainSchema);
schemaSet.Compile();

Upvotes: 6

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