问题描述
比方说我们有这个文件夹
not_my_files
collections
collection1.xml
collection2.xml
collection3.xml
etc...
my_files
my_documents
mydoc1.xml
mydoc2.xml
mydoc3.xml
etc...
有 xml文件的结构
collection1.xml(与collection2.xml,collection3.xml等相同的结构)
<collection xml:id="name_of_collection_1">
<ref id="id_of_ref_1">
<title>This is title 1 of first document in this collection</title>
</ref>
<ref id="id_of_ref_2">
<title>This is title 2 of second document in this collection</title>
</ref>
</collection>
mydoc1.xml(与mydoc2.xml,mydoc3.xml等相同的结构)
<mydoc id="my_doc_id_1">
<tag1>
<tag2>
<reference_tag>
<my_title>This is title 1 of my documents</my_title>
</reference_tag>
</tag2>
</tag1>
</mydoc>
因此:1)不同文件夹中的 xml文件具有不同的结构,并且2)c ollection1.xml可以包含许多标题,而mydoc1.xml只能包含1个标题 >。
我要从collections / collection1.xml(等)和my_documents / mydoc1.xml(等)中获取所有标题。 这是期望的结果:
<doc>
<folder>Not my files</folder>
<title>This is title 1 of first document in this collection</title>
</doc>
<doc>
<folder>Not my files</folder>
<title>This is title 2 of second document in this collection</title>
</doc>
<doc>
<folder>My files</folder>
<title>This is title 1 of my documents</title>
</doc>
我的当前XQuery:
xquery version "3.1";
for $doc_not_my_files in collection("/not_my_files/collections")
let $folder_not_my_files := "Not my files"
for $ref in $doc_not_my_files//ref
let $title_not_my_files := $ref/title/text()
for $doc_my_files in collection("/my_files/my_documents")
let $folder_my_files := "My files"
let $title_my_files := $doc_my_files//reference_tag/my_title/text()
return
if ($folder_my_files="My files")
then
<doc>
<folder>{$folder_my_files}</folder>
<title>{$title_my_files}</title>
</doc>
else
<doc>
<folder>{$folder_not_my_files}</folder>
<title>{$title_not_my_files}</title>
</doc>
我的当前结果:
<doc>
<folder>My files</folder>
<title>This is title 1 of my documents</title>
</doc>
<doc>
<folder>My files</folder>
<title>This is title 1 of my documents</title>
</doc>
<doc>
<folder>My files</folder>
<title>This is title 1 of my documents</title>
</doc>
<doc>
<folder>My files</folder>
<title>This is title 1 of my documents</title>
</doc>
**etc... 1000 times**
<doc>
<folder>Not my files</folder>
<title>This is title 1 of first document in this collection</title>
</doc>
<doc>
<folder>Not my files</folder>
<title>This is title 1 of first document in this collection</title>
</doc>
<doc>
<folder>Not my files</folder>
<title>This is title 1 of first document in this collection</title>
</doc>
**etc... another 1000 times**
因此,我在XQuery中寻找某种sql“ UNION”替代方案... 我有种基本的愚蠢问题的感觉,但是我是XQuery的新手,请原谅:)
解决方法
它可能与
一起使用for-each-pair(
collection("/my_files/my_documents"),collection("/not_my_files/collections"),function($doc,$col) {
$doc//reference_tag/my_title/text() ! <doc>
<folder>My files</folder>
<title>{.}</title>
</doc>,$col//ref/title/text() ! <doc>
<folder>Not my files</folder>
<title>{.}</title>
</doc>
}
)