sql-server - XQuery - 迭代每个属性

Question

我正在尝试使用这种结构迭代 XML:

<Query id="Y120-M2-D1-H8-M42-S12-MS276-6676" name="Transactions_ModelGenerator_v_1.0.0.104">
  <Description />
  <Columns>
    <Column caption="Client Code" sorting="None" sortIndex="-1">
      <Expr class="ENTATTR" id="c.Number" />
    </Column>
    <Column caption="Client Name" sorting="None" sortIndex="-1">
      <Expr class="ENTATTR" id="c.Name" />
    </Column>
    <Column caption="Market Code" sorting="None" sortIndex="-1">
      <Expr class="ENTATTR" id="r.Number" />
    </Column>
    <Column caption="Market Name" sorting="None" sortIndex="-1">
      <Expr class="ENTATTR" id="r.Name" />
    </Column>
  </Columns>
  <JustSortedColumns />
  <Conditions linking="All">
    <Condition class="SMPL" enabled="True" readOnly="False">
      <Operator id="Equal" />
      <Expressions>
        <Expr class="ENTATTR" id="c.Number" />
        <Expr class="CONST" type="String" kind="Scalar" value="A1B" text="A1B" />
      </Expressions>
    </Condition>
    <Condition class="SMPL" enabled="True" readOnly="False">
      <Operator id="Equal" />
      <Expressions>
        <Expr class="ENTATTR" id="ConversionCurrency" />
        <Expr class="CONST" type="String" kind="Scalar" value="EUR" text="EUR" />
      </Expressions>
    </Condition>
  </Conditions>
</Query>

我想遍历列，并替换某些关键字，特别是标题属性。例如，我想用“Cedent”替换关键字“Client”。

我试过写这样的东西:

update #data
set data.modify('
    for $col in /Query/Columns[1]/Column/caption
    replace value of $col with fn:replace($col, "Client", "Cedent")
    where fn:contains($col, "Client")
')

但当然我收到错误我不知道如何解决:

XQuery [#data.data.modify()]: Syntax error near 'replace', expected 'where', '(stable) order by' or 'return'.

我对 XQuery 不是很熟悉。有人可以帮我吗？

Answer 1

   for $col in /Query/Columns[1]/Column/@caption[fn:contains(., "Client")]
   let $x := fn:replace($col, "Client", "Cedent")
   return replace value of node $col with $x