Question

在以下示例中，要求是通过移动<userSearchRecords>子元素来合并<userAccount/login>元素（由<otherAttributes>元素标识），如果需要，这意味着：如果有2 <userSearchRecords> xml中的元素，通过将<otherAttributes>从第二个<userSearchRecords>移动到第一个<userSearchRecords>来连接它们，并删除第二个<SearchReply_OutputVariable> <part name="parameters"> <searchResponse> <userSearchRecords> <userAccount> <login>A</login> <otherAttributes> <name>X</name> <value>1</value> </otherAttributes> </userAccount> </userSearchRecords> <userSearchRecords> <userAccount> <login>B</login> <otherAttributes> <name>X</name> <value>2</value> </otherAttributes> </userAccount> </userSearchRecords> <userSearchRecords> <userAccount> <login>C</login> <otherAttributes> <name>X</name> <value>1</value> </otherAttributes> </userAccount> </userSearchRecords> <userSearchRecords> <userAccount> <login>D</login> <otherAttributes> <name>X</name> <value>1</value> </otherAttributes> </userAccount> </userSearchRecords> <userSearchRecords> <userAccount> <login>G</login> <otherAttributes> <name>Y</name> <value>5</value> </otherAttributes> </userAccount> </userSearchRecords> <userSearchRecords> <userAccount> <login>C</login> <otherAttributes> <name>Y</name> <value>6</value> </otherAttributes> </userAccount> </userSearchRecords> <userSearchRecords> <userAccount> <login>H</login> <otherAttributes> <name>Y</name> <value>7</value> </otherAttributes> </userAccount> </userSearchRecords> <userSearchRecords> <userAccount> <login>A</login> <otherAttributes> <name>Y</name> <value>7</value> </otherAttributes> </userAccount> </userSearchRecords> </searchResponse> </part> </SearchReply_OutputVariable>元素;否则保持原样。

示例输入XML：

<SearchReply_OutputVariable>
  <part name="parameters">
    <searchResponse>
      <userSearchRecords>
        <userAccount>
          <login>A</login>
          <otherAttributes>
            <name>X</name>
            <value>1</value>
          </otherAttributes>
          <otherAttributes>
            <name>Y</name>
            <value>7</value>
          </otherAttributes>
        </userAccount>
      </userSearchRecords>
      <userSearchRecords>
        <userAccount>
          <login>B</login>
          <otherAttributes>
            <name>X</name>
            <value>2</value>
          </otherAttributes>
        </userAccount>
      </userSearchRecords>
      <userSearchRecords>
        <userAccount>
          <login>C</login>
          <otherAttributes>
            <name>X</name>
            <value>1</value>
          </otherAttributes>
          <otherAttributes>
            <name>Y</name>
            <value>6</value>
          </otherAttributes>
        </userAccount>
      </userSearchRecords>
      <userSearchRecords>
        <userAccount>
          <login>D</login>
          <otherAttributes>
            <name>X</name>
            <value>1</value>
          </otherAttributes>
        </userAccount>
      </userSearchRecords>
      <userSearchRecords>
        <userAccount>
          <login>G</login>
          <otherAttributes>
            <name>Y</name>
            <value>5</value>
          </otherAttributes>
        </userAccount>
      </userSearchRecords>
      <userSearchRecords>
        <userAccount>
          <login>H</login>
          <otherAttributes>
            <name>Y</name>
            <value>7</value>
          </otherAttributes>
        </userAccount>
      </userSearchRecords>
    </searchResponse>
  </part>
</SearchReply_OutputVariable>

预期输出XML：

<userSearchRecords>

在输出XML中，现在有6个<userSearchRecords>，而在输入XML中有8个。在输入XML中有2个<login>，<userSearchRecords>值为A，而2 { {1}} <login>值C.在输出XML中，这些元素被合并（<otherAttributes>子项从匹配的登录对中移出，<userSearchRecords>元素从匹配的对中移除）。

有人能指出我正确的方向，如何解决这个问题？

谢谢！

Answer 1

感谢Martin，我查找了与xsl相关的“分组”，这就是我提出的解决方案：

{{1}}

XSLT合并具有相同标识符元素的节点

1 个答案: