没有命名空间的xml的Java xsd验证 [英] Java xsd validation of xml without namespace

问题描述

我想针对XSD架构验证XML文件。 XML文件根元素没有任何名称空间或xsi详细信息。它没有属性所以只有< root> 。

I want to validate an XML file against an XSD schema. The XML files root element does not have any namespace or xsi details. It has no attributes so just <root>.

我尝试了以下代码 http://www.ibm.com/developerworks/xml/library/x-javaxmlvalidapi。 html 没有运气，因为我收到
cvc-elt.1：找不到元素'root'的声明

I have tried the following code from http://www.ibm.com/developerworks/xml/library/x-javaxmlvalidapi.html with no luck as I receive cvc-elt.1: Cannot find the declaration of element 'root'

SchemaFactory factory = SchemaFactory.newInstance("http://www.w3.org/2001/XMLSchema");

File schemaFile = new File("schema.xsd");

Schema xsdScheme = factory.newSchema(schemaFile);

Validator validator = xsdScheme.newValidator();

Source source = new StreamSource(xmlfile);

validator.validate(source);

xml使用包含的命名空间标题（通过xmlspy添加）验证正常，但我会想到可以在不必手动编辑源文件的情况下声明xml命名空间吗？

The xml validates fine with the namespace headers included etc (added via xmlspy), but I would have thought the xml namespace could be declared without having to manually edit the source file?

编辑和解决方案：

public static void validateAgainstXSD(File file) {

    try {
        SchemaFactory factory = SchemaFactory.newInstance("http://www.w3.org/2001/XMLSchema");

        File schemaFile = new File("path/to/xsd");

        Schema xsdScheme = factory.newSchema(schemaFile);

        Validator validator = xsdScheme.newValidator();

        SAXSource source = new SAXSource(
                new NamespaceFilter(XMLReaderFactory.createXMLReader()),
                new InputSource(new FileInputStream(file)));

        validator.validate(source,null);

    } catch (Exception e) {
        e.printStackTrace();
    }

}

protected static class NamespaceFilter extends XMLFilterImpl {

    String requiredNamespace = "namespace";

    public NamespaceFilter(XMLReader parent) {
        super(parent);
    }

    @Override
    public void startElement(String arg0, String arg1, String arg2, Attributes arg3) throws SAXException {
        if(!arg0.equals(requiredNamespace)) 
            arg0 = requiredNamespace;
        super.startElement(arg0, arg1, arg2, arg3);
    }       
}

推荐答案

你您需要注意两个单独的问题：

You have two separate concerns you need to take care of:

1. 声明文档使用的命名空间。


2. 在文件中放置 xsi：schemaLocation 属性，以提示（！）架构所在的位置。

1. Declaring the namespace that your document uses.
2. Putting an xsi:schemaLocation attribute in the file to give a hint (!) where the schema is.

您可以安全地跳过第二部分，因为该位置实际上只是一个提示。你不能跳过第一部分。 XML文件中声明的名称空间与模式匹配。重要的是，

You can safely skip the second part, as the location is really only a hint. You cannot skip the first part. The namespace declared in the XML file is matched against the schema. Important, this:

<xml> ... </xml>

与此不一样：

<xml xmlns="urn:foo"> ... </xml>

所以你需要在XML文档中声明你的命名空间，否则它将不符合你的模式和你会收到这个错误。

So you need to declare your namespace in the XML document, otherwise it will not correspond to your schema and you will get this error.

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