Question

我尝试创建删除功能时遇到了问题。我目前的代码是：

的Xquery：

declare variable $d as xs:string;
declare variable $p as xs:string;

let $xp := saxon:evaluate(concat("doc('",$d,"')",$p))

return document {for $n in doc($d)/* return qsx10p8:delete($n, $xp)}

declare function qsx10p8:delete
($n as node(), $xp as node()*) 
as node() { 
 if  ($n[self::element()])
 then element
   {fn:local-name($n)}
  {for $c in $n/(*|@*)
      return qsx10p8:delete($c, $xp),  
     if (some $x in $xp satisfies ($n is $x)) 
    then ()
   else ($n/text())}

 else $n   
};

如果输入为：$d = C:/supplier.xml and $p= /Suppliers/Supplier/* 结果是：

<Suppliers><Supplier><address /><Phone /></Supplier></Suppliers>

但我希望结果为<Suppliers><Supplier></Supplier></Suppliers>。有没有办法编辑我的功能代码以删除那些必要的标签？

Answer 1

您可以尝试使用以下递归函数作为删除所需元素的方法。

declare function local:transform ($x as node())
{   
  typeswitch ($x)
      case element(Supplier) return element {"Supplier"} {}
      case text() return $x
      default 
        return element { fn:node-name($x) }
        {
                $x/attribute::*,
                for $z in $x/node() return local:transform($z)
        }
};


let $d := <Suppliers>
<Supplier><address /><Phone /></Supplier>
<Supplier><address /><Phone /></Supplier>
<Supplier><address /><Phone /></Supplier>
</Suppliers> 
return local:transform($d)

Answer 2

这个XQuery：

declare variable $pPath as xs:string external;
declare variable $vPath := tokenize($pPath,'\|');
declare function local:copy-match($x as element()) {
   element
      {node-name($x)}
      {for $child in $x/node()
       return
          if ($child instance of element())
          then
             local:match($child)
          else
             $child
      }
};
declare function local:match($x as element()) {
   let $element-path := string-join(
                           $x/ancestor-or-self::node()/name(),
                           '/'
                        )
   where
      not(
         some $path in $vPath
         satisfies
            ends-with($element-path,$path)
      )
   return
      local:copy-match($x)
};
local:match(/*)

将此xs:string作为$pPath参数：'Supplier/address|Supplier/Phone'

输出：

<Suppliers>
    <Supplier></Supplier>
</Suppliers>

如何在XQuery中创建删除功能

2 个答案: