java 如何在根据 XML 模式验证 XML 文件时获取错误的行号
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How to get error's line number while validating a XML file against a XML schema
提问by pablosaraiva
I'm trying to validade a XML against a W3C XML Schema.
我正在尝试针对 W3C XML 架构验证 XML。
The following code does the job and reports when error occurs. But I'm unable to get line number of the error. It always returns -1.
以下代码完成工作并在发生错误时报告。但我无法获得错误的行号。它总是返回-1。
Is there a easy way to get the line number?
有没有简单的方法来获取行号？
import java.io.File;
import javax.xml.XMLConstants;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.transform.Source;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamSource;
import javax.xml.validation.Schema;
import javax.xml.validation.SchemaFactory;
import javax.xml.validation.Validator;
import org.w3c.dom.Document;
import org.xml.sax.SAXParseException;
public class XMLValidation {
public static void main(String[] args) {
try {
DocumentBuilder parser = DocumentBuilderFactory.newInstance().newDocumentBuilder();
Document document = parser.parse(new File("myxml.xml"));
SchemaFactory factory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);
Source schemaFile = new StreamSource(new File("myschema.xsd"));
Schema schema = factory.newSchema(schemaFile);
Validator validator = schema.newValidator();
validator.validate(new DOMSource(document));
} catch (SAXParseException e) {
System.out.println(e.getLineNumber());
e.printStackTrace();
} catch (Exception e) {
e.printStackTrace();
}
}
}
回答by Mads Hansen
I found this
我找到了这个
http://www.herongyang.com/XML-Schema/Xerces2-XSD-Validation-with-XMLReader.html
http://www.herongyang.com/XML-Schema/Xerces2-XSD-Validation-with-XMLReader.html
that appears to provide the following details(to include line numbers)
似乎提供以下详细信息（包括行号）
Error:
Public ID: null
System ID: file:///D:/herong/dictionary_invalid_xsd.xml
Line number: 7
Column number: 22
Message: cvc-datatype-valid.1.2.1: 'yes' is not a valid 'boolean'
value.
using this code:
使用此代码：
/**
* XMLReaderValidator.java
* Copyright (c) 2002 by Dr. Herong Yang. All rights reserved.
*/
import java.io.IOException;
import org.xml.sax.XMLReader;
import org.xml.sax.helpers.DefaultHandler;
import org.xml.sax.helpers.XMLReaderFactory;
import org.xml.sax.SAXException;
import org.xml.sax.SAXParseException;
class XMLReaderValidator {
public static void main(String[] args) {
String parserClass = "org.apache.xerces.parsers.SAXParser";
String validationFeature
= "http://xml.org/sax/features/validation";
String schemaFeature
= "http://apache.org/xml/features/validation/schema";
try {
String x = args[0];
XMLReader r = XMLReaderFactory.createXMLReader(parserClass);
r.setFeature(validationFeature,true);
r.setFeature(schemaFeature,true);
r.setErrorHandler(new MyErrorHandler());
r.parse(x);
} catch (SAXException e) {
System.out.println(e.toString());
} catch (IOException e) {
System.out.println(e.toString());
}
}
private static class MyErrorHandler extends DefaultHandler {
public void warning(SAXParseException e) throws SAXException {
System.out.println("Warning: ");
printInfo(e);
}
public void error(SAXParseException e) throws SAXException {
System.out.println("Error: ");
printInfo(e);
}
public void fatalError(SAXParseException e) throws SAXException {
System.out.println("Fattal error: ");
printInfo(e);
}
private void printInfo(SAXParseException e) {
System.out.println(" Public ID: "+e.getPublicId());
System.out.println(" System ID: "+e.getSystemId());
System.out.println(" Line number: "+e.getLineNumber());
System.out.println(" Column number: "+e.getColumnNumber());
System.out.println(" Message: "+e.getMessage());
}
}
}
回答by tiboo
Replace this line:
替换这一行：
validator.validate(new DOMSource(document));
by
经过
validator.validate(new StreamSource(new File("myxml.xml")));
will let the SAXParseException contain line number & column number
将让 SAXParseException 包含行号和列号
回答by peter.murray.rust
Try using a SAXLocator http://download.oracle.com/javase/1.5.0/docs/api/org/xml/sax/Locator.htmlParsers are not required to supply one, but if they do it should report line numbers
尝试使用 SAXLocator http://download.oracle.com/javase/1.5.0/docs/api/org/xml/sax/Locator.html解析器不需要提供一个，但如果他们这样做应该报告行号
I think your code should include:
我认为你的代码应该包括：
// this will be called when XML-parser starts reading
// XML-data; here we save reference to current position in XML:
public void setDocumentLocator(Locator locator) {
this.locator = locator;
}
The parser will give you a locator which you can then use to get the line number. It's probably worth printing/debugging when this happens to see if you have a valid locator
解析器会给你一个定位器，然后你可以用它来获取行号。发生这种情况时可能值得打印/调试以查看您是否有有效的定位器
