Question

我目前正在尝试使用XSL转换以下XML结构

<table name="PERSONS" schema="INO_PERSONS_MGMT" syntheticPK="ID">
<columns>
  <column type="NUMBER" primary="true" unique="true">key</column>
  <column type="varchar">name</column>
  <column type="varchar">comment</column>
  <column type="varchar">email</column>
</columns>
<records>
    <record key="1" name="Test1" email="test1@test.com"/>
    <record key="2" name="Test2" comment="something" mail="test2@test.com"/>
</records>

到以下xml输出：

<table:table table:name="INO_PERSONS_MGMT.PERSONS">
            <table:table-row>
                <table:table-cell>
                    <text:p>key</text:p>
                </table:table-cell>
                <table:table-cell>
                    <text:p>name</text:p>
                </table:table-cell>
                <table:table-cell>
                    <text:p>comment</text:p>
                </table:table-cell>
                <table:table-cell>
                    <text:p>email</text:p>
                </table:table-cell>
            </table:table-row>
            <table:table-row>
                <table:table-cell>
                    <text:p>NUMBER</text:p>
                </table:table-cell>
                <table:table-cell>
                    <text:p>varchar</text:p>
                </table:table-cell>
                <table:table-cell>
                    <text:p>varchar</text:p>
                </table:table-cell>
                <table:table-cell>
                    <text:p>varchar</text:p>
                </table:table-cell>
            </table:table-row>
            <table:table-row>
                <table:table-cell>
                    <text:p>1</text:p>
                </table:table-cell>
                <table:table-cell>
                    <text:p>Test1</text:p>
                </table:table-cell>
                <table:table-cell>
                    <text:p></text:p>
                </table:table-cell>
                <table:table-cell>
                    <text:p>test1@test.com</text:p>
                </table:table-cell>
            </table:table-row>
            <table:table-row>
                <table:table-cell>
                    <text:p>2</text:p>
                </table:table-cell>
                <table:table-cell>
                    <text:p>Test2</text:p>
                </table:table-cell>
                <table:table-cell>
                    <text:p>something</text:p>
                </table:table-cell>
                <table:table-cell>
                    <text:p>test2@test.com</text:p>
                </table:table-cell>
            </table:table-row>
        </table:table>

因此，数据表行（第2行和第3行）的结构应该由列的结构定义。因此，它应迭代给定的列并生成表格单元格，即使它们未在记录中定义。

Answer 1

您显示的输出不是格式良好的XML，因为table:前缀未绑定到命名空间。

要最小化实际问题的示例，请考虑以下样式表：

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:template match="/table">
    <xsl:variable name="cols" select="columns/column" />
    <table name="{@name}">
        <!-- header rows -->
        <row>
            <xsl:for-each select="$cols">
                <cell>
                    <xsl:value-of select="."/>
                </cell>
            </xsl:for-each>
        </row>
        <row>
            <xsl:for-each select="$cols">
                <cell>
                    <xsl:value-of select="@type"/>
                </cell>
            </xsl:for-each>
        </row>
        <!-- data rows -->
        <xsl:for-each select="records/record">
            <xsl:variable name="curr-row" select="." />
            <row>
                <xsl:for-each select="$cols">
                    <cell>
                        <xsl:value-of select="$curr-row/@*[name()=current()]"/>
                    </cell>
                </xsl:for-each>
            </row>
        </xsl:for-each>
    </table>
</xsl:template>

</xsl:stylesheet>

应用于您的输入示例，结果将是：

<?xml version="1.0" encoding="UTF-8"?>
<table name="PERSONS">
   <row>
      <cell>key</cell>
      <cell>name</cell>
      <cell>comment</cell>
      <cell>email</cell>
   </row>
   <row>
      <cell>NUMBER</cell>
      <cell>varchar</cell>
      <cell>varchar</cell>
      <cell>varchar</cell>
   </row>
   <row>
      <cell>1</cell>
      <cell>Test1</cell>
      <cell/>
      <cell>test1@test.com</cell>
   </row>
   <row>
      <cell>2</cell>
      <cell>Test2</cell>
      <cell>something</cell>
      <cell/>
   </row>
</table>

这应该很容易让你适应你的情况。

请注意，"email"与"mail"不同 - 这就是第二行的最后一个单元格为空的原因。

属性值取决于XSL中的其他结构

1 个答案: