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Java XMLUnit-2.0 xpath不';t忽略XML节点顺序_Java_Xml_Xmlunit_Xmlunit 2 - Fatal编程技术网

Java XMLUnit-2.0 xpath不';t忽略XML节点顺序

Java XMLUnit-2.0 xpath不';t忽略XML节点顺序,java,xml,xmlunit,xmlunit-2,Java,Xml,Xmlunit,Xmlunit 2,我有一个XML,如下所示 <?xml version='1.0' encoding='UTF-8'?> <!DOCTYPE ResourceObject PUBLIC "my_corp.dtd" "my_corp.dtd"> <ResourceObject displayName="TESTNGAD\AggUserFSP test" identity="CN=AggUserFSP test,OU=FSPAggeFrame,OU=unittests,DC=TestNG

我有一个XML,如下所示

<?xml version='1.0' encoding='UTF-8'?>
<!DOCTYPE ResourceObject PUBLIC "my_corp.dtd" "my_corp.dtd">
<ResourceObject displayName="TESTNGAD\AggUserFSP test" identity="CN=AggUserFSP test,OU=FSPAggeFrame,OU=unittests,DC=TestNGAD,DC=local" objectType="account" uuid="{97182a65-61f2-443c-b0fa-477d0821d8c4}">
   <Attributes>
     <Map>
       <entry key="accountFlags">
         <value>
           <List>
             <String>Normal User Account</String>
             <String>Password Cannot Expire</String>
           </List>
         </value>
       </entry>
       <entry key="homePhone" value="6555"/>
       <entry key="l" value="Pune"/>
       <entry key="memberOf">
         <value>
           <List>
             <String>CN=FSPGRP2,OU=ADAggF,OU=unittests2,DC=AUTODOMAIN,DC=LOCAL</String>
             <String>CN=FSPGRP1,OU=ADAggF,OU=unittests2,DC=AUTODOMAIN,DC=LOCAL</String>
             <String>CN=LocalAggFrame,OU=FSPAggeFrame,OU=unittests,DC=TestNGAD,DC=local</String>
           </List>
         </value>
       </entry>
       <entry key="objectClass">
         <value>
           <List>
             <String>top</String>
             <String>person</String>
             <String>organizationalPerson</String>
             <String>user</String>
           </List>
         </value>
       </entry>
       <entry key="sn" value="test"/>
       <entry key="st" value="MH"/>
       <entry key="streetAddress" value="SB ROAD"/>
       <entry key="title" value="QA"/>
       <entry key="userPrincipalName" value="AggUserFSP test@TestNGAD.local"/>
     </Map>
   </Attributes>
 </ResourceObject>

我假设您希望匹配Map元素中的条目元素,并且希望通过属性匹配它们,从而忽略顺序

如果是这样,我建议您这样编写节点匹配器:

.withNodeMatcher(新的默认NodeMatcher(ElementSelectors.conditionalBuilder())
.当要素进入(“条目”)。然后使用(
(e1,e2)->StringUtils.equals(e1.getAttribute(“key”)、e2.getAttribute(“key”))
.elseUse(姓名)
.build()

diff = DiffBuilder
                    .compare(control)
                    .withTest(test)
                    .checkForSimilar().checkForIdentical() //ignore the order of 'elements` but check they are identical
                    .normalizeWhitespace()
                    .ignoreComments()
                    .ignoreWhitespace()
                    //.ignoreElementContentWhitespace()
                    .withNodeMatcher(new DefaultNodeMatcher(ElementSelectors.conditionalBuilder()
                            .whenElementIsNamed("Attributes").thenUse(ElementSelectors.byXPath("./Map/entry[@key]", ElementSelectors.byNameAndText))
                            .elseUse(ElementSelectors.byName)
                            .build()))
                    .build();