我的最终XML应该采用以下格式：
router.post('/some-irl', function (req, res) {
//some code
return {some JSON}
});
我目前正在合并两个XML，它为我提供了以下XML：
<?xml version="1.0" encoding="UTF-8"?>
<Rowsets>
<Rowset>
<Columns></Columns>
<Row></Row>
<Row></Row>
<Row></Row>
</Rowset>
</Rowsets>
现在我只需要一个<?xml version="1.0" encoding="UTF-8"?><Rowsets>
<Rowset>
<Columns>
<Column Description="" MaxRange="1" MinRange="0" Name="Material" SQLDataType="1" SourceColumn="Material"/>
<Column Description="" MaxRange="1" MinRange="0" Name="DS_MATERIAL" SQLDataType="1" SourceColumn="DS_MATERIAL"/>
<Column Description="" MaxRange="1" MinRange="0" Name="Press" SQLDataType="1" SourceColumn="Press"/>
<Column Description="" MaxRange="1" MinRange="0" Name="Mold" SQLDataType="1" SourceColumn="Mold"/>
</Columns>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>111</DS_MATERIAL>
<Press>14</Press>
<Mold>3864F</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>111</DS_MATERIAL>
<Press>25</Press>
<Mold>4306K-1</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>111</DS_MATERIAL>
<Press>25</Press>
<Mold>4306K-1</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>111</DS_MATERIAL>
<Press>Summary</Press>
<Mold>---</Mold>
</Row>
<Row>
<Material>300-6953-1</Material>
<DS_MATERIAL>111</DS_MATERIAL>
<Press>23</Press>
<Mold>4305P-1</Mold>
</Row>
<Row>
<Material>300-6953-1</Material>
<DS_MATERIAL>111</DS_MATERIAL>
<Press>23</Press>
<Mold>4305P-1</Mold>
</Row>
<Row>
<Material>300-6953-1</Material>
<DS_MATERIAL>---</DS_MATERIAL>
<Press>Summary</Press>
<Mold>---</Mold>
</Row>
</Rowset>
<Rowset>
<Columns>
<Column Description="" MaxRange="1" MinRange="0" Name="Material" SQLDataType="1" SourceColumn="Material"/>
<Column Description="" MaxRange="1" MinRange="0" Name="DS_MATERIAL" SQLDataType="1" SourceColumn="DS_MATERIAL"/>
<Column Description="" MaxRange="1" MinRange="0" Name="Press" SQLDataType="1" SourceColumn="Press"/>
<Column Description="" MaxRange="1" MinRange="0" Name="Mold" SQLDataType="1" SourceColumn="Mold"/>
</Columns>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>14</Press>
<Mold>3864F</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>25</Press>
<Mold>4306K-1</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>25</Press>
<Mold>4306K-1</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>Summary</Press>
<Mold>---</Mold>
</Row>
<Row>
<Material>300-6953-1</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>23</Press>
<Mold>4305P-1</Mold>
</Row>
<Row>
<Material>300-6953-1</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>23</Press>
<Mold>4305P-1</Mold>
</Row>
<Row>
<Material>300-6953-1</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>Summary</Press>
<Mold>---</Mold>
</Row>
</Rowset>
</Rowsets>
以及<Rowset>内的所有<Row>，我正在将以下XSLT应用于上面的XML：
<Rowset>这将在<?xml version="1.0" encoding="UTF-8"?><xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:java="http://xml.apache.org/xslt/java" exclude-result-prefixes="java" version="1.0">
<xsl:output media-type="text/xml" method="xml"/>
<!-- Merges multiple rowsets -->
<xsl:param name="SD"/>
<xsl:param name="ED"/>
<xsl:param name="RowCount"/>
<xsl:template match="/">
<Rowsets>
<Rowset>
<xsl:copy-of select="/Rowsets/Rowset/Columns"/>
<xsl:for-each select="Rowsets/Rowset">
<xsl:copy-of select="Row"/>
</xsl:for-each>
</Rowset>
</Rowsets>
</xsl:template>
</xsl:stylesheet>
内创建所有<Row>，但会创建两个<Rowset>个节点。我只需要一个<Columns>节点。
以上XSLT的结果XML如下：
<Columns>我的XSLT需要进行哪些修改才能删除这个额外的<?xml version="1.0" encoding="UTF-8"?><Rowsets>
<Rowset>
<Columns>
<Column Description="" MaxRange="1" MinRange="0" Name="Material" SQLDataType="1" SourceColumn="Material"/>
<Column Description="" MaxRange="1" MinRange="0" Name="DS_MATERIAL" SQLDataType="1" SourceColumn="DS_MATERIAL"/>
<Column Description="" MaxRange="1" MinRange="0" Name="Press" SQLDataType="1" SourceColumn="Press"/>
<Column Description="" MaxRange="1" MinRange="0" Name="Mold" SQLDataType="1" SourceColumn="Mold"/>
</Columns>
<Columns>
<Column Description="" MaxRange="1" MinRange="0" Name="Material" SQLDataType="1" SourceColumn="Material"/>
<Column Description="" MaxRange="1" MinRange="0" Name="DS_MATERIAL" SQLDataType="1" SourceColumn="DS_MATERIAL"/>
<Column Description="" MaxRange="1" MinRange="0" Name="Press" SQLDataType="1" SourceColumn="Press"/>
<Column Description="" MaxRange="1" MinRange="0" Name="Mold" SQLDataType="1" SourceColumn="Mold"/>
</Columns>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>111</DS_MATERIAL>
<Press>14</Press>
<Mold>3864F</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>111</DS_MATERIAL>
<Press>25</Press>
<Mold>4306K-1</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>111</DS_MATERIAL>
<Press>25</Press>
<Mold>4306K-1</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>111</DS_MATERIAL>
<Press>Summary</Press>
<Mold>---</Mold>
</Row>
<Row>
<Material>300-6953-1</Material>
<DS_MATERIAL>111</DS_MATERIAL>
<Press>23</Press>
<Mold>4305P-1</Mold>
</Row>
<Row>
<Material>300-6953-1</Material>
<DS_MATERIAL>111</DS_MATERIAL>
<Press>23</Press>
<Mold>4305P-1</Mold>
</Row>
<Row>
<Material>300-6953-1</Material>
<DS_MATERIAL>---</DS_MATERIAL>
<Press>Summary</Press>
<Mold>---</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>14</Press>
<Mold>3864F</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>25</Press>
<Mold>4306K-1</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>25</Press>
<Mold>4306K-1</Mold>
</Row>
<Row>
<Material>300-6953</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>Summary</Press>
<Mold>---</Mold>
</Row>
<Row>
<Material>300-6953-1</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>23</Press>
<Mold>4305P-1</Mold>
</Row>
<Row>
<Material>300-6953-1</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>23</Press>
<Mold>4305P-1</Mold>
</Row>
<Row>
<Material>300-6953-1</Material>
<DS_MATERIAL>222</DS_MATERIAL>
<Press>Summary</Press>
<Mold>---</Mold>
</Row>
</Rowset>
</Rowsets>
节点？
我需要的XML如下：
<Columns>请帮忙。
答案 0 :(得分：1)
以这种方式试试吗？
XSLT 1.0
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/Rowsets">
<Rowsets>
<Rowset>
<xsl:copy-of select="Rowset[1]/Columns"/>
<xsl:copy-of select="Rowset/Row"/>
</Rowset>
</Rowsets>
</xsl:template>
</xsl:stylesheet>
我目前正在合并两个XML，它为我提供了以下XML：
您可以在合并XSLT中修复此问题，并节省自己需要进行额外转换。