Question

  <xsl:output method="text" omit-xml-declaration="yes" />
<xsl:template match="/Update/Object/Insert">
  Insert into <xsl:value-of select="@Table" />(
  <xsl:for-each select="Field">
    <xsl:value-of select="@Name" />,      
  </xsl:for-each>)
  Values(
    <xsl:for-each select="Field">
      <xsl:value-of select="@Value" />,
    </xsl:for-each>)
</xsl:template>

Output : Insert into Good( GoodCode, ...
,
) Values( 1, ... ,
)

既然你能够看到还有额外的“,”问题是如何纠正?

    <?xml version="1.0" encoding="UTF-8"?>
<?xml-stylesheet type= text/xsl  href= book.xslt ?>
<Update>
  <Object Name="Good">
    <Insert Table="Good">
      <Field Name="GoodCode" Value="1" Type="Integer" />
      <Field Name="GoodUserCode" Value="" Type="String" />
</insert>
</object>

Answer 1

采取类似行动:

 <xsl:output method="text" omit-xml-declaration="yes" />
   <xsl:template match="/Update/Object/Insert">
      Insert into <xsl:value-of select="@Table" />(
   <xsl:for-each select="Field">
      <xsl:value-of select="@Name" /><xsl:if test = "position() != last()">, </xsl:if>      
   </xsl:for-each>)
   Values(
   <xsl:for-each select="Field">
      <xsl:value-of select="@Value" /><xsl:if test = "position() != last()">, </xsl:if>
   </xsl:for-each>)
 </xsl:template>

http://www.w3schools.com/xsl/el_if.asp

友情链接