xsl格式化xml后能输出为XML格式吗?不要输出html格式,该如何解决

xsl格式化xml后能输出为XML格式吗?不要输出html格式
我想用xsl实现模糊查找功能,实现方法是用xsl进行过滤。
但是过滤后输出是html形式的,我想要过滤后还是xml格式,跟过滤之前的xml格式一样的,只显示我要的xml数据。

比如:我的xml格式是这样的:
<?xml   version= "1.0 "   encoding= "gb2312 "?>
<company>
<department   name= "Research   &amp;   Development ">
    <employee   name= 'JohnDoe '   job= 'Software   Analyst '   salary= '2000 '/>
    <employee   name= 'Jane   Fletcher '   job= 'Designer '   salary= '2500 '/>
    <employee   name= 'Chris   Benton '   job= 'Programmer '   salary= '3100 '/>
</department>
</company>
-------我只想取name属性中包含a的那一行,我想用xsl过滤后数据是下面的格式,而且是xml格式,可以用xmlDom.loadXml()加载的,而不是以html的格式显示在网页上。
<?xml   version= "1.0 "   encoding= "gb2312 "?>
<company>
<department   name= "Research   &amp;   Development ">
    <employee   name= 'Jane   Fletcher '   job= 'Designer '   salary= '2500 '/>
</department>
</company>

请帮忙想想办法。谢谢

附上代码:

-------test.xml
<?xml   version= "1.0 "   encoding= "iso-8859-1 "?>
<?xml-stylesheet   type= "text/xsl "   href= "test.xsl "?>
<company>
<department   name= "Research   &amp;   Development ">
    <employee   name= 'JohnDoe '   job= 'Software   Analyst '   salary= '2000 '/>
    <employee   name= 'Jane   Fletcher '   job= 'Designer '   salary= '2500 '/>
    <employee   name= 'Chris   Benton '   job= 'Programmer '   salary= '3100 '/>
</department>
</company>
-------------test.xsl
<?xml   version= "1.0 "   encoding= "gb2312 "?>
<xsl:stylesheet   version= "1.0 "   xmlns:xsl= "http://www.w3.org/1999/XSL/Transform ">
<xsl:template   match= "/ ">
&lt;company&gt;
<br   />
<xsl:for-each   select= "company/department/employee ">
<xsl:if       test= "contains(@name, 'a ') ">      
&lt;employee   name= ' <xsl:value-of   select= "@name "/> '   job= ' <xsl:value-of   select= "@job "/> '   salary= ' <xsl:value-of       select= "@salary "/> '/&gt;
<br   />
</xsl:if>
</xsl:for-each>
&lt;company/&gt;
</xsl:template>
</xsl:stylesheet>
---结果是以html格式显示的,不是XML格式:
<company>  
<employee   name= 'Jane   Fletcher '   job= 'Designer '   salary= '2500 '/>  
<employee   name= 'Jane   Fletcher '   job= 'Designer '   salary= '2500 '/>  
<company/>  




------解决方案--------------------
<?xml version= "1.0 "?>
<xsl:stylesheet version= "1.0 " xmlns:xsl= "http://www.w3.org/1999/XSL/Transform ">
<xsl:output indent= "yes " method= "xml " standalone= "yes "/>