根據XSD驗證XML

Question

我需要驗證具有給定XSD文件的XML文件。我只需要該方法返回true，否則驗證正常或錯誤。

Answer 1

返回簡單對或錯（您也不需要任何外部庫）：

static boolean validateAgainstXSD(InputStream xml, InputStream xsd) 
{ 
    try 
    { 
     SchemaFactory factory = 
      SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI); 
     Schema schema = factory.newSchema(new StreamSource(xsd)); 
     Validator validator = schema.newValidator(); 
     validator.validate(new StreamSource(xml)); 
     return true; 
    } 
    catch(Exception ex) 
    { 
     return false; 
    } 
} 

Answer 2

public boolean validate() { 
    DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance(); 

    factory.setValidating(true); 

    factory.setAttribute(
     "http://java.sun.com/xml/jaxp/properties/schemaLanguage", 
     "http://www.w3.org/2001/XMLSchema"); 
    factory.setAttribute(
     "http://java.sun.com/xml/jaxp/properties/schemaSource", 
     "http://domain.com/mynamespace/mySchema.xsd"); 
    Document doc = null; 
    try { 
    DocumentBuilder parser = factory.newDocumentBuilder(); 
    doc = parser.parse("data.xml"); 
    return true; 
    } catch (Exception e) { 
    return false; 
    } 
} 

Answer 3

這可能取決於你所使用的庫，但谷歌搜索周圍的「如何驗證XML文件中的Java」給我的這些結果，你可能會發現你的答案：

first interesting result

second interesting result

Answer 4

XMLUnit有一些不錯的班級要做到這一點，有一個例子，他們的README文件：

Validator v = Validator.forLanguage(Languages.W3C_XML_SCHEMA_NS_URI); 
v.setSchemaSources(Input.fromFile("local.xsd").build()); 
ValidationResult result = v.validateInstance(new StreamSource(new File("local.xml"))); 
return result.isValid();