Question

我用完了如何避免这种“重复”节点的情况。

问题： 考虑一下，我们遵循xml：

<root>
 <cities>
  <city id = "1">
    <name>ABC</name>
    <value>1</value>
  </city>
  <city id = "2">
    <name>DEF</name>
    <value>2</value>
  </city>
  <city id = "1">
    <name>ABC</name>
    <value>1</value>
  </city>
  <city id = "1">
    <name>ABC</name>
    <value>1</value>
  </city>
 <cities>
</root>

我需要存档的内容是避免重复。所以在输出上我想看到并输出2个节点：

 <city id = "1">
    <name>ABC</name>
    <value>1</value>
  </city>
  <city id = "2">
    <name>DEF</name>
    <value>2</value>
  </city>

我在发布之前尝试了很多解决方案（如//*:city[@name='ABC']/@name）但它返回包含所有重复项的列表。（在本例中为3）

任何想法如何解决？

非常感谢

Answer 1

使用示例HTML | XML测试以下XPaths here...

<cities>
<city>
  <name>ABC</name>
  <value>2</value>
</city>
<city>
  <name>ABC</name>
  <value>1</value>
</city>
<city>
  <name>DEF</name>
  <value>2</value>
</city>
<city>
  <name>AB</name>
  <value>1</value>
</city>
<city>
  <name>ABC</name>
  <value>1</value>
</city>
<cities>

来自子元素列表的不同数据：

 //cities/descendant::city[not(.=preceding-sibling::city)]

不同节点Values形成父City兄弟姐妹。

 //cities/descendant::city[not(./value=preceding-sibling::city/value)]/value

来自父Names兄弟姐妹的不同节点City。

 //cities/descendant::city[not(./name=preceding-sibling::city/name)]/name

nodes of same level包含不同的数据。

<root>
    <item type='test'>A</item>
    <item type='test'>B</item>
    <item type='test'>C</item>
    <item type='test'>A</item>
    <item type='other'>A</item>
    <item type='test'>B</item>
    <item type='other'>D</item>
</root>

//preceding::item/preceding::item[not(.=preceding-sibling::item)]/text()

使用节点属性值更新了答案：

<root>
 <cities>
  <city id = "1">
    <name>ABC</name>
    <value>1</value>
  </city>
  <city id = "3">
    <name>DEF</name>
    <value>2</value>
  </city>
  <city id = "1">
    <name>ABC</name>
    <value>1</value>
  </city>
  <city id = "4">
    <name>ABC</name>
    <value>1</value>
  </city>
   <city id = "2">
    <name>DE</name>
    <value>2</value>
  </city>
 <cities>
</root>

的XPath：

 //cities/city/following::city[not(./@id=following::city/@id)]

 //cities/city/following::city[not(./@id=following::city/@id)]/name

 //cities/city/following::city[not(./@id=following::city/@id)]/value

获取基于父节点id的子节点列表，没有任何重复

1 个答案: