我有一个关于将XPath表达式作为属性值传递的快速问题。
所以我有以下文件:
<?xml version="1.0"?>
<personnel>
<person id="EMP01" >
<name><family>Lee</family><given>Ho</given></name>
<email>lee.ho@foo.com</email>
<salary>100000</salary>
<links>
<subordinate>EMP02</subordinate>
<subordinate>EMP03</subordinate>
</links>
</person>
<person id="EMP02">
<name><family>Chan</family><given>Tai Man</given></name>
<email>chan.tai.man@foo.com</email>
<salary>20000</salary>
<links>
<manager>EMP01</manager>
<subordinate>EMP04</subordinate>
<subordinate>EMP05</subordinate>
</links>
</person>
<person id="EMP03">
<name><family>Cheung</family><given>Siu Fan</given></name>
<email>cheung.siu.fan@foo.com</email>
<salary>20200</salary>
<links>
<manager>EMP01</manager>
<subordinate>EMP07</subordinate>
</links>
</person>
<person id="EMP04">
<name><family>Ng</family><given>Ho</given></name>
<email>ng.ho@foo.com</email>
<salary>11000</salary>
<links>
<manager>EMP02</manager>
</links>
</person>
<person id="EMP05">
<name><family>Chow</family><given>Sing Sing</given></name>
<email>chow.sing.sing@foo.com</email>
<salary>20050</salary>
<links>
<manager>EMP02</manager>
<subordinate>EMP06</subordinate>
</links>
</person>
<person id="EMP06">
<name><family>Law</family><given>Lai</given></name>
<email>law.lai@foo.com</email>
<salary>5050</salary>
<links>
<manager>EMP05</manager>
</links>
</person>
<person id="EMP07">
<name><family>Chan</family><given>Siu Ming</given></name>
<email>chan.siu.ming@foo.com</email>
<salary>5000</salary>
<links>
<manager>EMP03</manager>
</links>
</person>
而且我需要一个XPath表达式来获取名称为Law的家庭管理者的人结点,因此在这种情况下为周先生。但是,如何在XPath中获得此信息而无需显式使用名称Chow?
我的尝试是:
personnel/person[@id="personnel/person/name/family='Law'"]
但这实际上不起作用...我可以将XPath作为这样的属性的值传递,还是该怎么做?
最佳答案
如果希望将字符串作为XPath表达式而不是文字字符串读取,请不要使用引号。试试这样的事情:
/personnel/person[@id=/personnel/person[name/family='Law']/links/manager]/name/family
xpathtester demo
输出:
<family>Chow</family>
关于xpath - 如何通过被引用的属性查找元素?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34065531/