输入:
<Root><output>
<queries>
<query name="Test">
<parameters>
<parameter>298674,298673,298675,298676</parameter>
</parameters>
<queryResults>
<record id="1">
<column name="Order">272334</column>
<column name="Task">272093</column>
<column name="FirstAction">2709305</column>
</record>
<record id="2">
<column name="Order">272334</column>
<column name="Task">272093</column>
<column name="FirstAction">2709301</column>
</record>
<record id="3">
<column name="Order">272334</column>
<column name="Task">272093</column>
<column name="FirstAction">2709306</column>
</record>
<record id="4">
<column name="Order">268997</column>
<column name="Task">268756</column>
<column name="FirstAction">2709307</column>
</record>
<record id="5">
<column name="Order"/>
<column name="Task"/>
<column name="FirstAction">2709307</column>
</record>
</queryResults>
</query>
</queries></output>
如何删除clumn Order中没有值的记录?
期望的输出:
<Root><output>
<queries>
<query name="Test">
<parameters>
<parameter>298674,298673,298675,298676</parameter>
</parameters>
<queryResults>
<record id="1">
<column name="Order">272334</column>
<column name="Task">272093</column>
<column name="FirstAction">2709305</column>
</record>
<record id="2">
<column name="Order">272334</column>
<column name="Task">272093</column>
<column name="FirstAction">2709301</column>
</record>
<record id="3">
<column name="Order">272334</column>
<column name="Task">272093</column>
<column name="FirstAction">2709306</column>
</record>
<record id="4">
<column name="Order">268997</column>
<column name="Task">268756</column>
<column name="FirstAction">2709307</column>
</record>
</queryResults>
</query>
</queries></output>
我当前的XSL只删除空节点
<xsl:strip-space elements="*"/>
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="*[not(child::node())]"/>
我似乎无法掌握模板匹配以删除整个“记录”标记... 非常感谢你!
答案 0 :(得分:1)
添加一个新模板到删除clumn中没有值的记录
<xsl:template match="record[column[ @name='Order']= '']"/>
此匹配记录在名称为Order的列中没有值,并忽略它们。