4

我在三个表上使用内部联接，并显示结果表。我为此使用 Xml。

<SQLInformation>
      <Table>tblechecklistprogramworkpackagexref prwpxref</Table>
      <TableJoins>
        INNER JOIN tblechecklistprogram pr ON pr.ixProgram=prwpxref.ixProgram
        INNER JOIN tblechecklistworkpackage wp ON wp.ixWorkPackage=prwpxref.ixWorkPackage
        INNER JOIN tblechecklistworkpackageactivityxref wpaxref ON wpaxref.ixWorkPackage=wp.ixWorkPackage
        INNER JOIN tblechecklistactivity act ON act.ixActivity=wpaxref.ixActivity
      </TableJoins>
      <WhereClause>
      </WhereClause>
      <GroupBy>
      </GroupBy>
      <OrderBy></OrderBy>
      <HeaderInformation></HeaderInformation>
      <FooterInformation></FooterInformation>
    </SQLInformation>
    <Columns>
      <Column>
        <Id>ProgramName</Id>
        <Title>Program Name</Title>
        <Table>pr</Table>
        <Field>sName</Field>
        <Alias>sName</Alias>
        <Sortable>true</Sortable>
        <GroupSort>true</GroupSort>
        <ColumnWidth>100</ColumnWidth>
        <JavascriptFormatter>
          <![CDATA[
            DesignDataFileFormatter = function(span,row,columns,data,item)
            {              
              $(span).createAppend('a', {href: '../EcheckList/TemplateUsage.aspx?IxTemplate=' + 2, title: 'Go to Corresponding Schedule Item.', innerHTML:sName });

              return true;
            }
          ]]>
        </JavascriptFormatter>
        <Display>true</Display>
        <Filter type="Database">
          <FilterContainer name="Program" DisplayMember="sName" ValueMember="ixProgram" UseHavingClause="false" >
            <FilterQuery>SELECT sName,ixProgram FROM tblechecklistprogram</FilterQuery>
          </FilterContainer>
        </Filter>
      </Column>

我想使用输出表的列值之一创建一个超链接。每当我单击链接时，它都会给出格式异常。

4

1 回答 1

2

我发现问题出在虚拟路径上的解决方案..

$(span).createAppend('a', { href : 'Customer.aspx?ixProgram=' + row[columns.ixProgram] }).text(data); 返回真；} ]]>

于 2012-05-03T11:58:27.880 回答