java - 使用 Java 进行 XML 动态验证

Question

基础设施:我使用的是 JAVA 1.5，这是强制性的。但我可以加载任何外部库，所以没问题。

问题:

我有一个通过“外部 channel ”接收的 XML 文件，我可以将其用作输入流

如果有人需要得到相同的，可以使用:

InputStream is = new FileInputStream(file);

我需要根据嵌套 XSD 的 XSD 验证 XML

<?xml version="1.0" encoding="UTF-8"?>
<xs:schema xmlns:xs="http://www.w3.org/2001/XMLSchema" 
           xmlns:BODY="urn:CBI:xsd:CBIBdySDDReq.00.00.06" 
           xmlns:HTRT="urn:CBI:xsd:CBIHdrTrt.001.07" 
           xmlns:HE2E="urn:CBI:xsd:CBIHdrSrv.001.07" 
           xmlns:SGNT="urn:CBI:xsd:CBISgnInf.001.04" 
           xmlns:LMSG="urn:CBI:xsd:CBISDDReqLogMsg.00.00.06"
           xmlns="urn:CBI:xsd:CBISDDReqPhyMsg.00.00.06"
           targetNamespace="urn:CBI:xsd:CBISDDReqPhyMsg.00.00.06"
           elementFormDefault="qualified">
    <xs:import namespace="urn:CBI:xsd:CBIHdrTrt.001.07" schemaLocation="CBIHdrTrt.001.07.xsd"/>
    <xs:import namespace="urn:CBI:xsd:CBIHdrSrv.001.07" schemaLocation="CBIHdrSrv.001.07.xsd"/>
    <xs:import namespace="urn:CBI:xsd:CBIBdySDDReq.00.00.06" schemaLocation="CBIBdySDDReq.00.00.06.xsd"/>
    <xs:element name="CBISDDReqPhyMsg" type="CBISDDReqPhyMsg.00.00.06">
        <xs:annotation>
            <xs:documentation>1. - Tag root dell'intero messaggio fisico di richiesta SDD CBI</xs:documentation>
        </xs:annotation>
    </xs:element>
    <xs:complexType name="CBISDDReqPhyMsg.00.00.06">
        <xs:sequence>
            <xs:element name="CBIHdrTrt" type="HTRT:CBIHdrTrt.001.07">
                <xs:annotation>
                    <xs:documentation>1.1. - Header di tratta CBI</xs:documentation>
                </xs:annotation>
            </xs:element>
            <xs:element name="CBIHdrSrv" type="HE2E:CBIHdrSrv.001.07">
                <xs:annotation>
                    <xs:documentation>1.2. - Header di servizio CBI</xs:documentation>
                </xs:annotation>
            </xs:element>
            <xs:element name="CBIBdySDDReq" type="BODY:CBIBdySDDReq.00.00.06">
                <xs:annotation>
                    <xs:documentation>1.3. - Body di servizio CBI</xs:documentation>
                </xs:annotation>
            </xs:element>
        </xs:sequence>
    </xs:complexType>
</xs:schema>

所以我有一个 XSD 文件的早午餐。

block XML 文件是这样的

<?xml version="1.0" encoding="UTF-8"?>
<MSG:CBISDDReqPhyMsg xmlns:MSG="urn:CBI:xsd:CBISDDReqPhyMsg.00.00.06"
    xmlns:HTRT="urn:CBI:xsd:CBIHdrTrt.001.07" xmlns:HE2E="urn:CBI:xsd:CBIHdrSrv.001.07"
    xmlns:BODY="urn:CBI:xsd:CBIBdySDDReq.00.00.06" xmlns:LMSG="urn:CBI:xsd:CBISDDReqLogMsg.00.00.06"
    xmlns:SGNT="urn:CBI:xsd:CBISgnInf.001.04" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
    <MSG:CBIHdrTrt>
        <HTRT:IdCBISndrf>0000636J</HTRT:IdCBISndrf>
        <HTRT:IdCBIRcvrf>0000641V</HTRT:IdCBIRcvrf>
        <HTRT:SrvNm>INC-SDDC</HTRT:SrvNm>
        <HTRT:IdMsgTrt>
            0000636JP12312111154007381042010010000636J000000636J0000641V0
        </HTRT:IdMsgTrt>

所以我需要根据 CBISDDReqPhyMsg.00.00.06 验证 XML。所以我只知道在运行时针对 xml 来使用。例如另一个文件可以加载 CBISDDReqPhyMsg.00.00.05

我有两个主要问题

1)我需要从 XML 获取 xsd 文件名，XML 可能很大 1/2GB(stax/sax 是一个很好的解决方案)

2) 我需要从 jar 加载 xsd，因为整个应用程序无法访问文件系统。

对于验证，我更喜欢像 http://www.edankert.com/validate.html 这样的系统

我读到了有关包含在此答案中的内容 Problem validating an XML file using Java with an XSD having an include

但是使用加载器我无法获取有关文件的信息。

有什么想法吗？

Answer 1

I have two major problem

1) I need to obtain the file name from XML and XML could be BIG 1/2GB

2) I need to load the xsd from a jar because the whole app cannot have access to the file system.

1.) 文件名不在 XML 内...您应该通过其他方式(用户输入、硬编码等)知道 XML 的文件名是什么。如果您的目录仅包含 XML 文件，则可以对这些文件运行循环。一些伪代码:

for(getNextFile)
{
    if(xmlValidationWithXSD(nextFile))
        //passed validation
    else
        //failed validation
}

我通常运行脚本将所有 XML 文件从一个位置移动到纯 XML 文件夹。

2.) 我相信 StreamSource 允许您做到这一点。这是我最近所做的:

            public static boolean xmlValidationWithXSD(String xmlLocation){
                Source xmlFile = new StreamSource(new File(xmlLocation));

                 //IMPORTANT:  Here is what you need.  Multiple XSDs (that relate to each other) AND getResource will access form .jar files
                //Treat XSD as resource found in the class path, assume that the full package name is passed in xsdLocation
                Source schemaFileSource1 = new StreamSource(BenefitEnrollmentRequestFileUtil.class.getResource(NEW_XSD_FILE_RESOURCE1).toString());
                Source schemaFileSource2 = new StreamSource(BenefitEnrollmentRequestFileUtil.class.getResource(NEW_XSD_FILE_RESOURCE2).toString());
                Source[] schemaFileSources = {schemaFileSource1, schemaFileSource2};  


                SchemaFactory schemaFactory = SchemaFactory.newInstance(XMLConstants.W3C_XML_SCHEMA_NS_URI);

                try {
                    Schema schema = schemaFactory.newSchema(schemaFileSources);
                    Validator validator = schema.newValidator();
                    validator.validate(xmlFile);
                    return true;
                } catch (SAXException e) {
                    LOGGER.debug(xmlFile.getSystemId() + " is NOT valid", e);
                    LOGGER.debug("Reason: " + e.getLocalizedMessage());
                } catch (IOException e) {
                    LOGGER.debug(xmlFile.getSystemId() + " is NOT valid", e);
                    LOGGER.debug("Reason: " + e.getLocalizedMessage());                 
                }
                return false;
            }

来源:

Referencing a XSD schema within jar file

Problem validating an XML file using Java with an XSD having an include

编辑:

the relation between file is the list contained in the first tag i have the list of xsd to use and the first xsd is the one that has the same namesace as the first tag

<MSG:CBISDDReqPhyMsg xmlns:MSG="urn:CBI:xsd:CBISDDReqPhyMsg.00.00.06" xmlns:HTRT="urn:CBI:xsd:CBIHdrTrt.001.07" xmlns:HE2E="urn:CBI:xsd:CBIHdrSrv.001.07" xmlns:BODY="urn:CBI:xsd:CBIBdySDDReq.00.00.06" xmlns:LMSG="urn:CBI:xsd:CBISDDReqLogMsg.00.00.06" xmlns:SGNT="urn:CBI:xsd:CBISgnInf.001.04" xmlns:xsi="w3.org/2001/XMLSchema-instance">;

你不能读入第一行，然后提取数据吗？使用 FileReader/DOM(如果您只读取一行，它应该不会太慢)只是为了检索这一行。然后解析字符串。创建一个 ArrayList 并添加您提取的每个单独的部分。

看看效率如何:

https://stackoverflow.com/a/2134533/2498729

https://stackoverflow.com/a/12273296/2498729