在 XQuery 中組合兩個 for 循環和聯合結果

Question

假設我們有這個文件夾

not_my_files
    collections
        collection1.xml
        collection2.xml
        collection3.xml
        etc...
my_files
    my_documents
        mydoc1.xml
        mydoc2.xml
        mydoc3.xml
        etc...

有xml文件的結構

collection1.xml（collection2.xml、collection3.xml 等的結構相同...）

<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（mydoc2.xml、mydoc3.xml 等的結構相同...）

<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>

所以：1)不同文件夾中的 xml 文件具有不同的結構和 2) ollection1.xml 可以包含許多標題，而 mydoc1.xml 只能包含 1 個時間標題。

我想從 collections/collection1.xml（等）和 my_documents/mydoc1.xml（等）中獲取所有標題。 這是一個想要的結果：

<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>

我當前的 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>

我目前的結果：

<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**

所以，我在 XQuery 中尋找某種 SQL“UNION”替代方案......我有這種感覺，就像我有一些基本的愚​​蠢問題，但我是 XQuery 的新手，所以請原諒我:)

Answer 1

它可能與

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>
  }
)