我有一个 xml 文档:
<?xml version="1.0" encoding="utf-8"?>
<Root>
<Child name="MyType" compareMode="EQ"></Child>
</Root>
我想在以下 xsd 的帮助下验证此 xml:
<?xml version="1.0" encoding="utf-8"?>
<xs:schema attributeFormDefault="unqualified" elementFormDefault="qualified" xmlns:xs="http://www.w3.org/2001/XMLSchema">
<xs:element name="Root">
<xs:complexType>
<xs:sequence>
<xs:element name="Child">
<xs:complexType>
<xs:attribute name="name" type="xs:string" use="required" />
<xs:attribute name="compareMode" type="xs:string" use="required" />
</xs:complexType>
</xs:element>
</xs:sequence>
</xs:complexType>
</xs:element>
</xs:schema>
当我尝试验证它时,我收到以下错误:
Exception in thread "main" org.xml.sax.SAXException: Validation failed against correct.xml. ErrorMessage:s4s-elt-schema-ns: The namespace of element 'Root' must be from the schema namespace, 'http://www.w3.org/2001/XMLSchema'.
我的问题是为什么Root必须位于模式的命名空间中?难道是我验证xml文档不正确?
public synchronized boolean isValid(String xmlFragment, File xmlSchema) throws SAXException, IOException{
// 1. Lookup a factory for the W3C XML Schema language
SchemaFactory factory = SchemaFactory.newInstance("http://www.w3.org/2001/XMLSchema"); // 2. Compile the schema. Schema schema = factory.newSchema(xmlSchema); // 3. Get a validator from the schema. Validator validator = schema.newValidator(); // 4. Parse the document you want to check. Source source = new StreamSource(new ByteArrayInputStream(xmlFragment.getBytes())); // 5. Check the document validator.validate(source); return true; }
最佳答案
难道 xmlSchema 保存的不是您向我们展示的模式,而是 W3C 发布的“模式文档的模式”?错误消息片段“ErrorMessage:s4s-elt-schema-ns:”似乎暗示了这一点。
关于xml - 元素 'Root' 的命名空间必须来自架构命名空间 'http://www.w3.org/2001/XMLSchema' ?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4922452/