XSLT1.0跨多个节点的不同属性列表
XSLT1.0跨多个节点的不同属性列表
提问于 2018-06-22 06:01:56
我被困在没有扩展的XSLT1.0上。
我需要获取属性不同的节点列表。根据下面的XML，第一个post_press_version (25)的结果应该是1,2,3,5，第二个(26)的结果应该是1,2,3,6
<order>
<post_press_version>
<post_press_version_id>25</post_press_version_id>
<hopper_allocations>
<hopper_sequence>
<sequence_id>2</sequence_id>
<sequence_part_produced_id>37</sequence_part_produced_id>
<sequence_part_produced_name>96854-12WS10</sequence_part_produced_name>
<sequence_qty>2</sequence_qty>
<hopper number="0" id="" is_insert="False"/>
<hopper number="1" id="23" is_insert="False">96854-6FL</hopper>
<hopper number="2" id="13" is_insert="False">96854-8FL003</hopper>
<hopper number="3" id="15" is_insert="False">96854-10FL004</hopper>
<hopper number="4" id="" is_insert="False"/>
<hopper number="5" id="" is_insert="False"/>
<hopper number="6" id="" is_insert="False"/>
</hopper_sequence>
<hopper_sequence>
<sequence_id>3</sequence_id>
<sequence_part_produced_id>38</sequence_part_produced_id>
<sequence_part_produced_name>96854-12WS20</sequence_part_produced_name>
<sequence_qty>250</sequence_qty>
<hopper number="0" id="" is_insert="False"/>
<hopper number="1" id="24" is_insert="False">96854-6FL10</hopper>
<hopper number="2" id="29" is_insert="False">96854-8FL</hopper>
<hopper number="3" id="" is_insert="False"/>
<hopper number="4" id="" is_insert="False"/>
<hopper number="5" id="31" is_insert="False">96854-10FL</hopper>
<hopper number="6" id="" is_insert="False"/>
</hopper_sequence>
</hopper_allocations>
</post_press_version>
<post_press_version>
<post_press_version_id>26</post_press_version_id>
<hopper_allocations>
<hopper_sequence>
<sequence_id>2</sequence_id>
<sequence_part_produced_id>37</sequence_part_produced_id>
<sequence_part_produced_name>96854-12WS10</sequence_part_produced_name>
<sequence_qty>2</sequence_qty>
<hopper number="0" id="" is_insert="False"/>
<hopper number="1" id="23" is_insert="False">96854-6FL</hopper>
<hopper number="2" id="13" is_insert="False">96854-8FL003</hopper>
<hopper number="3" id="15" is_insert="False">96854-10FL004</hopper>
<hopper number="4" id="" is_insert="False"/>
<hopper number="5" id="" is_insert="False"/>
<hopper number="6" id="" is_insert="False"/>
</hopper_sequence>
<hopper_sequence>
<sequence_id>3</sequence_id>
<sequence_part_produced_id>38</sequence_part_produced_id>
<sequence_part_produced_name>96854-12WS20</sequence_part_produced_name>
<sequence_qty>250</sequence_qty>
<hopper number="0" id="" is_insert="False"/>
<hopper number="1" id="24" is_insert="False">96854-6FL10</hopper>
<hopper number="2" id="29" is_insert="False">96854-8FL</hopper>
<hopper number="3" id="" is_insert="False"/>
<hopper number="4" id="" is_insert="False"/>
<hopper number="5" id="31" is_insert="False"/>
<hopper number="6" id="" is_insert="False">96854-10FL</hopper>
</hopper_sequence>
</hopper_allocations>
</post_press_version>
我尝试了下面的XSL，但它只返回1,2,3:第一个hopper_sequence中第一个hopper节点的结果
<xsl:template match="order">
<html>
<body>
<xsl:for-each
select="/order/post_press_version/hopper_allocations/hopper_sequence/hopper[node() and not(@number = preceding::hopper/@number)]]">
<p>
<xsl:value-of select="../../../post_press_version_id"/>-<xsl:value-of select="@number"/>
</p>
</xsl:for-each>
</body>
</html>
</xsl:template>我在努力找出我哪里错了。我用这个密钥尝试了Meunchen方法，但得到了类似的结果。
<xsl:key name="unique_post_press_version_hoppers" match="/order/post_press_version/hopper_allocations/hopper_sequence/hopper" use="concat(../../../post_press_version_id, @number)"/>有什么建议吗？
谢谢
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回答 1
Stack Overflow用户
发布于 2018-06-23 00:51:26
我更新并使用了您的XML，并根据您的需要找到了适当的输出。请再看一遍，它是正确的。只有零个人会来。
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原文链接：
https://stackoverflow.com/questions/50978029
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