使用 xpath 将多个属性与节点值连接起来 [英] Join multiple concatenations of attributes with node values using xpath

问题描述

我有以下 xml 内容:

I have the following xml content:

<ns2:ItemAttributes xml:lang="de-DE">
    <ns2:Creator Role="Role1">Creator One</ns2:Creator>
    <ns2:Creator Role="Role2">Creator Two</ns2:Creator>
    <ns2:Creator Role="Role3">Creator Three</ns2:Creator>
</ns2:ItemAttributes>

我正在尝试使用 xpath 将其格式化并将其合并为一行.类似的东西:

I'm trying to format and combine this into one line using xpath. Something like:

string-join(//ns2:Creator/concat(./text(), @Role), ', ')

我想，我离某个地方很近，因为:

I think, i'm somewhere close, because this:

string-join(//ns2:Creator/@Role , ', ')

工作并给我一个逗号分隔的角色列表:Role1、Role2、Role3

works and gives me a comma-separated list of roles: Role1, Role2, Role3

还有这个

string-join(//ns2:Creator/node(), ', ')

结合创作者的价值观:创作者一、创作者二、创作者三".

combines the values of creators: "Creator One, Creator Two, Creator Three".

我想要

Role1: Creator One, Role2: Creator Two, Role3: Creator Three

你能帮忙吗.

推荐答案

您需要选择 @Role 作为调用 concat() 的第一个参数，并且使用.选择上下文节点的字符串值:

You need to select @Role as the first parameter to your call to concat(), and use . to select the string value of the context node:

string-join($xml//ns2:Creator/concat(@Role, ': ', .), ', ')

代替 . 您还可以使用 string() 或 string(.) 来明确转换，否则会隐式发生:

Instead of . you can also use string() or string(.) which makes explicit the conversion that would otherwise happen implicitly:

string-join($xml//ns2:Creator/concat(@Role, ': ', string(.)), ', ')

返回:

Role1: Creator One, Role2: Creator Two, Role3: Creator Three

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