在 xQuery 中使用属性名称

发布于 2024-09-06 22:58:34 字数 749 浏览 5 评论 0原文

我有这个 Xml：

<Item key = "id">
   <SubItem id = "1" a = "2"/>
   <SubItem id = "1" b = "3"/>
   <SubItem x = "1"/>
   <SubItem y = "1"/>
   <SubItem z = "1"/>
</Item>

我希望选择具有名称等于 (.) 键的值的属性的子项

我想要一个返回此值的查询：

<SubItem id = "1" a = "2"/>
<SubItem id = "1" b = "3"/>

所以我尝试了：

let $x := ... my xml here...
let $key = $x/@id

return $x/*/@*[name(.) = $key]/..

这有效......

但我想要更简洁的东西，例如：

let $x := ... my xml here...
let $key = $x/@id

return $x/*/@$key

问题是，该代码无法编译，因为它不允许我在 @ 之后使用非文字（或通配符）。

难道真的不可能吗？提前致谢！！

原文

I have this Xml:

<Item key = "id">
   <SubItem id = "1" a = "2"/>
   <SubItem id = "1" b = "3"/>
   <SubItem x = "1"/>
   <SubItem y = "1"/>
   <SubItem z = "1"/>
</Item>

I wish to select the SubItems that have an attribute whose name equals the value of (.) key

I want a query that returns this:

<SubItem id = "1" a = "2"/>
<SubItem id = "1" b = "3"/>

So I tried:

let $x := ... my xml here...
let $key = $x/@id

return $x/*/@*[name(.) = $key]/..

this works...

but I wanted something more succinct like:

let $x := ... my xml here...
let $key = $x/@id

return $x/*/@$key

problem is, that code does not compile because it won't let me use a non literal (or wildcard) after the @.

Is it really impossible? Thanks in advance!!

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挽你眉间 2024-09-13 22:58:34

您没有提及您正在使用的系统，但是有一种 MarkLogic 特定的方法可以做到这一点

let $data :=
  <Item key = "id">
   <SubItem id = "1" a = "2"/>
   <SubItem id = "1" b = "3"/>
   <SubItem x = "1"/>
   <SubItem y = "1"/>
   <SubItem z = "1"/>
  </Item>
let $path := fn:concat("$data/SubItem[exists(./@", fn:data($data/@key), ")]")
return xdmp:value($path)

：据我所知，标准 XQuery 没有比您所做的更简单的方法。其他系统可能有类似的扩展。

You didn't mention what system you're using, but there is a MarkLogic-specific way to do this:

let $data :=
  <Item key = "id">
   <SubItem id = "1" a = "2"/>
   <SubItem id = "1" b = "3"/>
   <SubItem x = "1"/>
   <SubItem y = "1"/>
   <SubItem z = "1"/>
  </Item>
let $path := fn:concat("$data/SubItem[exists(./@", fn:data($data/@key), ")]")
return xdmp:value($path)

As far as I know, standard XQuery doesn't have a simpler way than what you did. Other systems may have similar extensions.

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