xml - XQuery - 连接查询

Question

我有两个 XML 文件，想合并它们以获得所需的输出。

XML 文件 -1

<Countries>
<Country>
   <No C="1"/>
   <Name>India</Name>
</Country>
<Country>
   <No C="2"/>
   <Name>USA</Name>
</Country>
<Country>
   <No C="3"/>
   <Name>Srilanka</Name>
</Country>
<Country>
   <No C="4"/>
   <Name>Thailand</Name>
</Country>
</Countries>

XML 文件 - 2

<Capitals>
  <Capital>
      <No C="1"/>
      <Name>New Delhi</Name>
  </Capital>
  <Capital>
      <No C="2"/>
      <Name>Washington DC</Name>
  </Capital>
  <Capital>
      <No C="3"/>
      <Name>Colombo</Name>
  </Capital>
  <Capital>
      <No C="4"/>
      <Name>Bangkok</Name>
  </Capital>
</Capitals>

所需的输出 -

India - New Delhi
USA - Washington DC
Srilanka - Colombo
Thailand - Bangkok

我试过是这样的，但是无法得到想要的输出:(

for $x in doc('Country')/Countries
  return 
  <a>
 {$x/Country/Name}
 {
   for $y in doc('Capital')/Capitals
   where $y/Capital/No/@C = $x/Country/No/@C
    return $y/Capital/Name
  }
  </a>

Answer 1

我得到了答案，它是——

谓词中的双向连接 -

for $x in doc('country')//Country
  return 
  string-join( ($x/Name/text(),  
     for $y in doc('capital')//Capital
       where $y/No/@C = $x/No/@C
       return $y/Name/text()
                ), " - ")