用xsl排序出的？
我有XML文件，
<?xml version= "1.0 " encoding= "UTF-8 "?>
<section name= "CallDetailCDRTypeOrder ">
<param name= "CSMOC " seq= "1 "/>
<param name= "CSMOT " seq= "9 "/>
<param name= "CSMTC " seq= "2 "/>
<param name= "CSCFW " seq= "3 " />
<param name= "WAP01 " seq= "4 " />
<param name= "DAOMT " seq= "5 " />
<param name= "SMSMO " seq= "6 " />
<param name= "GPRSS " seq= "7 " />
<param name= "GPRSB " seq= "8 " />
</section>
1.我想用XSLseq按升序徘序，我的xsl如下：
<?xml version= "1.0 " encoding= "UTF-8 "?>
<xsl:stylesheet version= "1.0 " xmlns:xsl= "http://www.w3.org/1999/XSL/Transform " xmlns:fo= "http://www.w3.org/1999/XSL/Format ">
<xsl:template match= "/ ">
<xsl:for-each select= "param ">
<xsl:sort select= "seq " data-type= "number() " order= "ascending "> </xsl:sort>
<xsl:value-of select= "param "> </xsl:value-of>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet> ，用XML SPY後出，大家我看？
2.如果我想把seq放在name前面如 <param seq= "1 " name= "CSMOC "/> 那在排序xsl怎么，是？
<xsl:attribute name= "seq "> <xsl:value-of select= "seq "/> </xsl:attribute>


[解决办法]
<?xml version= "1.0 " encoding= "UTF-8 "?>
<xsl:stylesheet version= "1.0 " xmlns:xsl= "http://www.w3.org/1999/XSL/Transform ">
<xsl:output indent= "yes " method= "xml "/>
<xsl:template match= "/ ">
<section name= "CallDetailCDRTypeOrder ">
<xsl:for-each select= "//param ">
<xsl:sort select= "@seq " data-type= "number " order= "ascending "/>
<xsl:element name= "param ">
<xsl:attribute name= "seq "> <xsl:value-of select= "@seq "/> </xsl:attribute>
<xsl:attribute name= "name "> <xsl:value-of select= "@name "/> </xsl:attribute>
</xsl:element>
</xsl:for-each>
</section>
</xsl:template>
</xsl:stylesheet>