小编pan*_*kaj的帖子

我有一个XML文件:

<?xml version="1.0" standalone="yes"?>
<Questionnaire>
  <Temp_GridTypeTable_6>
    <Column2>Select Yes/No</Column2>
  </Temp_GridTypeTable_6>
  <Temp_GridTypeTable_1>
    <Column2>Rank 1,2,3</Column2>
  </Temp_GridTypeTable_1>
  <Temp_GridTypeTable_1>
    <Column1>I needed the income</Column1>
    <Column2>Why did you take a job on this project?</Column2>
  </Temp_GridTypeTable_1>
  <Temp_GridTypeTable_1>
    <Column1>Other</Column1>
    <Column2></Column2>
  </Temp_GridTypeTable_1>
  <Temp_GridTypeTable_2>
    <Column2>Select "Yes/No"</Column2>
  </Temp_GridTypeTable_2>
  <Temp_GridTypeTable_2>
    <Column1>No jobs</Column1>
    <Column2>344</Column2>
  </Temp_GridTypeTable_2>
  <Temp_GridTypeTable_3>
    <Column2>Input</Column2>
  </Temp_GridTypeTable_3>
  <Temp_GridTypeTable_3>
    <Column1>Unit</Column1>
    <Column2>123</Column2>
  </Temp_GridTypeTable_3>
</Questionnaire> 

我想访问

<xsl:for-each select="Questionnaire/concat('Temp_GridTypeTablenode_',"1"))>

但是这个说法不起作用.