Question

下面是我用于验证XSD的XML架构的groovy代码

import java.io.File;
import java.io.IOException;

import javax.xml.XMLConstants;
import javax.xml.transform.stream.StreamSource;
import javax.xml.validation.Schema;
import javax.xml.validation.SchemaFactory;
import javax.xml.validation.Validator;
import javax.xml.transform.sax.SAXSource
import javax.xml.parsers.SAXParserFactory
import org.xml.sax.SAXException 
import org.xml.sax.InputSource
import org.xml.sax.SAXParseException
import org.xml.sax.ErrorHandler


def validateXMLSchema(String xsdPath, String xmlPath) {
    final List < SAXParseException > exceptions = new LinkedList < SAXParseException > ();
     try {
      SchemaFactory factory =
       SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
      Schema schema = factory.newSchema(new File(xsdPath));
      Validator validator = schema.newValidator();  
      validator.setErrorHandler(new ErrorHandler() {
       @Override
       public void warning(SAXParseException exception) throws SAXException {
        exceptions.add(exception);
       }

       @Override
       public void fatalError(SAXParseException exception) throws SAXException {
        exceptions.add(exception);
       }

       @Override
       public void error(SAXParseException exception) throws SAXException {
        exceptions.add(exception);
       }
      });
      def xmlFile = new File(xmlPath);
      validator.validate(new StreamSource(xmlFile));
      exceptions.each {
       println 'lineNumber : ' + it.lineNumber + '; message : ' + it.message
      }
     } catch (IOException | SAXException e) {
      println("Exception: line ${e.lineNumber} " + e.getMessage());
      return false;
     }
     return exceptions.size() == 0;
}

以下是一些验证错误，我可以访问每封邮件的行号，并尝试查找相应的节点名称

lineNumber : 106; message : cvc-datatype-valid.1.2.1: '' is not a valid value for 'date'. 
lineNumber : 248; message : cvc-enumeration-valid: Value 'Associate' is not facet-valid with respect to enumeration '[ADJSTR, ADJSMT]

是否有一种简单的方法可以使用行号找到相应错误消息的节点名称？或者我是否必须阅读该特定行并使用XmlSlurper解析它（如下所示（试图避免这种方法，因为对于生产中较大用户负载的大型XML文件，它会更慢）？

def getNodeName(xmlFile, lineNumber){
  def xmlLine =  xmlFile.readLines().get(lineNumber)  
  def node = new XmlSlurper().parseText(xmlLine.toString())
  node.name()
}

Answer 1

这不是很优雅，但以下getNodeName()应该更快（full example here）：

def getNodeName(xmlFile, lineNumber) {
    def result = "unknown"
    def count = 1
    def NODE_REGEX = /.*?<(.*?)>.*/ 
    def br 

    try {
        br = new BufferedReader(new FileReader(xmlFile)) 
        String line
        def isDone = false
        while ((! isDone) && (line = br.readLine()) != null) {
            if (count == lineNumber) {
                def matcher = (line =~ NODE_REGEX) 
                if (matcher.matches()) {
                    result = matcher[0][1]
                }
                isDone = true
            }
            count++
        }
    } finally {
        // TODO: better exception handling
        br.close()
    }

    return result
}

它只是直到有问题的行读取行，然后使用基本的正则表达式来获取名称。如果愿意，您可以在示例中使用XmlSlurper。关键是文件IO /内存应该少得多。

Java / Groovy：按行号

1 个答案: