XQuery：简单查询示例

我的任务是创建一个xml文件并使用XQuery查询它。该文件大致如下所示：

<library>
    <book>
        <title>Title1</title>
        <author>Author1</author>
        <author>Author2</author>
        <subject>Subject1</subject>
    </book>
    <book>
        <title>Title2</title>
        <author>Author3</author>
        <subject>Subject1</subject>
    </book>
</library>

let $a:=doc("/home/user/library.xml")/library/book
for $x in $a/title, $y in $a/author
return $x

我得到的结果是： 标题1 标题2 标题1 标题2 标题1 标题2

我发现问题在于，对于每个作者，我都会返回每个书名，但我不确定如何让它只返回与特定作者对应的书名。有什么建议吗

谢谢

使用：

   for $author in distinct-values(/*/*/author)
     return
       /*/book[$author = author]/title/string()

<library>
    <book>
        <title>Title1</title>
        <author>Author1</author>
        <author>Author2</author>
        <subject>Subject1</subject>
    </book>
    <book>
        <title>Title2</title>
        <author>Author3</author>
        <subject>Subject1</subject>
    </book>
</library>

根据提供的XML文档计算此XPath表达式（也就是XQuery，因为XPath是XQuery的子集）时：

   for $author in distinct-values(/*/*/author)
     return
       /*/book[$author = author]/title/string()

<library>
    <book>
        <title>Title1</title>
        <author>Author1</author>
        <author>Author2</author>
        <subject>Subject1</subject>
    </book>
    <book>
        <title>Title2</title>
        <author>Author3</author>
        <subject>Subject1</subject>
    </book>
</library>

---

谢谢，这真的很有帮助。我忘了如何正确使用XPath导航了。@coffeeNjava，不客气。我建议从一本关于XPath的好书开始。