相关疑难解决方法(0)

如何在针对XML模式验证XML文件时获取错误的行号

我正在尝试针对W3C XML Schema验证XML.

以下代码执行作业并报告发生错误的时间.但是我无法得到错误的行号.它总是返回-1.

有没有简单的方法来获得行号?

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 …
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java xml xsd sax

16
推荐指数
2
解决办法
2万
查看次数

如何使用SAX XML Schema Validator的验证消息进行内部化?

我正在使用此代码来针对XSD验证XML:

SchemaFactory factory = SchemaFactory.newInstance("http://www.w3.org/2001/XMLSchema");
Schema schema = factory.newSchema(xmlSchema);
Validator validator = schema.newValidator();
Source source = new StreamSource(myXmlFile);

try {
    validator.validate(source);
    return null;
}catch (SAXException ex) {
    String validationMessage = ex.getMessage();
    return validationMessage;
}
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但是当XML无效时,消息总是如下所示:

cvc-minLength-valid: Value '-' with length = '1' is not facet-valid with respect to minLength '2' for type '#AnonType_xLgrTEndereco'.
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有没有办法以我的语言返回用户友好的消息,而无需以编程方式翻译消息?

更新 这是XSD中发生消息的字段的一段代码:

<xs:element name="xLgr">
    <xs:annotation>
        <xs:documentation>Logradouro</xs:documentation>
    </xs:annotation>
    <xs:simpleType>
        <xs:restriction base="TString">
            <xs:maxLength value="60"/>
            <xs:minLength value="2"/>
        </xs:restriction>
    </xs:simpleType>
</xs:element>
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你可以看到它甚至有一个描述来返回字段的"名称"(Logradouro),但模式验证器似乎不承认它.

java xml validation

6
推荐指数
1
解决办法
662
查看次数

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