没有命名空间的 xml 的 Java xsd 验证
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Java xsd validation of xml without namespace
提问by Alec Holmes
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>.
我想根据 XSD 架构验证 XML 文件。XML 文件根元素没有任何命名空间或 xsi 详细信息。它没有属性所以只是<root>.
I have tried the following code from http://www.ibm.com/developerworks/xml/library/x-javaxmlvalidapi.htmlwith no luck as I receive
cvc-elt.1: Cannot find the declaration of element 'root'
我已经尝试了http://www.ibm.com/developerworks/xml/library/x-javaxmlvalidapi.html 中的以下代码，但我收到时没有运气
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);
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?
使用包含的命名空间标头等（通过 xmlspy 添加），xml 可以很好地验证，但我原以为可以声明 xml 命名空间而无需手动编辑源文件？
Edit and Solution:
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);
}
}
采纳答案by xcut
You have two separate concerns you need to take care of:
您需要处理两个不同的问题：
- Declaring the namespace that your document uses.
- Putting an
xsi:schemaLocationattribute in the file to give a hint (!)where the schema is.
- 声明您的文档使用的命名空间。
xsi:schemaLocation在文件中放置一个属性以给出模式所在的提示（！）。
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:
您可以安全地跳过第二部分，因为位置实际上只是一个提示。您不能跳过第一部分。XML 文件中声明的命名空间与架构匹配。重要的是，这个：
<xml> ... </xml>
Is not the sameas this:
是不一样的，因为这：
<xml xmlns="urn:foo"> ... </xml>
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.
所以你需要在 XML 文档中声明你的命名空间，否则它将与你的架构不对应，你会得到这个错误。
