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java - 如何在根据 XML 模式验证 XML 文件时获取错误的行号

转载 作者:太空狗 更新时间:2023-10-29 22:46:12 25 4
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我正在尝试根据 W3C XML 模式验证 XML。

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。

有没有简单的方法获取行号?

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();
}
}
}

最佳答案

我找到了这个

http://www.herongyang.com/XML-Schema/Xerces2-XSD-Validation-with-XMLReader.html

似乎提供了以下详细信息(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.

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());
}
}
}

关于java - 如何在根据 XML 模式验证 XML 文件时获取错误的行号,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4348285/

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