Xquery如何检查具有属性的元素的值

Question

I am really getting mad because I just don't know how to solve this X-Path/ X-Query problem我真的很生气，因为我只是不知道如何解决这个 X-Path/X-Query 问题

I have a Lido-File.我有一个丽都档案。 Let assume this:让我们假设：

    <lido:actorInRole>
                     <lido:actor lido:type="Person">
                        <lido:actorID lido:type="d-nb.info" lido:source="http://d-nb.info/gnd/116001313">116001313</lido:actorID>
                        <lido:actorID lido:type="uuid">2ca86edd-6113-4133-a86e-dd61135133ae</lido:actorID>
                        <lido:nameActorSet>
                           <lido:appellationValue lido:pref="preferred">Abel, Alfred</lido:appellationValue>
                        </lido:nameActorSet>
                        <lido:vitalDatesActor>
                           <lido:earliestDate>1879</lido:earliestDate>
                           <lido:latestDate>1937</lido:latestDate>
                        </lido:vitalDatesActor>
                        <lido:genderActor>männlich</lido:genderActor>
                     </lido:actor>
                       <lido:actor lido:type="Person">
                        <lido:actorID lido:type="d-nb.info" lido:source="http://d-nb.info/gnd/118501402">118501402</lido:actorID>
                        <lido:actorID lido:type="uuid">2e00b761-d766-47c5-80b7-61d76607c58d</lido:actorID>
                        <lido:nameActorSet>
                           <lido:appellationValue lido:pref="preferred">Albers, Hans</lido:appellationValue>
                        </lido:nameActorSet>
                        <lido:vitalDatesActor>
                           <lido:earliestDate>1891</lido:earliestDate>
                           <lido:latestDate>1960</lido:latestDate>
                        </lido:vitalDatesActor>
                        <lido:genderActor>männlich</lido:genderActor>
                     </lido:actor>
   </lido:actorInRole>

I am using a index.qxl where I am asking for the value of //lido:actorID[@lido:type="uuid"]我正在使用 index.qxl 我要求//lido:actorID[@lido:type="uuid"]的值

So i want to iterate through the these elements and check if the number of the number of //lido:actorID[@lido:type="uuid"] is 2ca86edd-6113-4133-a86e-dd61135133ae or something else.所以我想遍历这些元素并检查//lido:actorID[@lido:type="uuid"]的数量是否为2ca86edd-6113-4133-a86e-dd61135133ae或其他。

like "is the content of the object lido:actor withe the attribut lido:type containing the value 'uuid' 2ca86edd-6113-4133-a86e-dd61135133ae or not?. if yes, then give me the ancestor of this object. if not, do nothing"像“object lido:actor 的内容是否包含属性lido:type 包含值'uuid' 2ca86edd-6113-4133-a86e-dd61135133ae？，没做什么”

How can I achieve this?!我怎样才能做到这一点？！

Again.再次。 I DONT WANT THE VALUE OF lido:type..!!我不想要lido的价值：type..!! I want the content of the whole element.我想要整个元素的内容。 The ID Number!!身份证号码！！ And let this checked via an if-else which is nested in a for loop.并通过嵌套在 for 循环中的 if-else 进行检查。

Please I really need help!!!请问我真的需要帮助！！！

Answer 1

Assuming that you want to select the node假设你要 select 节点

<lido:actor lido:type="Person">...

you can use the following XPath-1.0 expression您可以使用以下 XPath-1.0 表达式

//lido:actor[lido:actorID[@lido:type='uuid']='2ca86edd-6113-4133-a86e-dd61135133ae']

It should give you the lido:actor node which has a child <actorID> with a lido:type attribute with the desired value .它应该为您提供具有子<actorID>的lido lido:actor节点，该子节点具有具有所需值的lido lido:type属性。

You can use this XPath-1.0 expression in an <xsl:if test='...'> or a template's match rule.您可以在<xsl:if test='...'>或模板的匹配规则中使用此 XPath-1.0 表达式。 In both cases the content of the matched node can be retrieved with <xsl:copy-of select="." />在这两种情况下，都可以使用<xsl:copy-of select="." />检索匹配节点的内容。 <xsl:copy-of select="." /> . <xsl:copy-of select="." /> .

Answer 2

thanks so far with the answer!到目前为止，感谢您的回答！

It's still not working.它仍然无法正常工作。

I have a new file calles "personeninfo.xql"我有一个名为“personeninfo.xql”的新文件

here the code:这里的代码：

    xquery version "3.1";

declare namespace lido = "http://www.lido-schema.org";

declare namespace output = "http://www.w3.org/2010/xslt-xquery-serialization";
declare option output:method "html5";
declare option output:media-type "text/html";




let $data := collection("/db/apps/eli/data")
let $id := request:get-parameter("e",())
let $actorID: = $data//lido:actor[@lido:type="Person"]//lido:actorID[@lido:type="uuid"]/text()



return 

<html>
<head>
    <title>Übersicht Weber-Korpus</title>
    <link rel="stylesheet" href="resources/bootstrap.min.css"/> 
</head>
<body>
        <div class="container" id="main-container">
            <div>
                <h1>Steckbrief </h1>
            </div>
          
               
               
                
                 { for $a in $actorID
                 
                 where  //lido:actor[lido:actorID[@lido:type='uuid']/text()=$id]    
                 return <li>{$a/ancestor::lido:actorInRole//lido:appellationValue[@lido:pref="preferred"]}</li>
                 
                 }
                
           
        </div>
</body>
</html>

I don't kno why it is not working.我不知道为什么它不起作用。 The e-parameter is correctly been given with from the index.xql from which I link to the personeninfo.xql从我链接到 personeninfo.xql 的 index.xql 中正确给出了 e 参数

Here the URL:这里是 URL：

http://localhost:8080/exist/apps/eli/personeninfo.xql?e=2ca86edd-6113-4133-a86e-dd61135133ae

You can see, that the ID is in e.可以看到，ID 在 e 中。

But why is not working?: I really need help :( It's driving me mad!但是为什么不工作？：我真的需要帮助:(这让我发疯！

Xquery如何检查具有属性的元素的值

问题描述

2 个解决方案

解决方案1
0 2021-01-29 01:11:48

解决方案2
0 2021-01-29 08:48:43

Xquery如何检查具有属性的元素的值

问题描述

2 个解决方案

解决方案1 0 2021-01-29 01:11:48

解决方案2 0 2021-01-29 08:48:43

解决方案1
0 2021-01-29 01:11:48

解决方案2
0 2021-01-29 08:48:43