XSLT 截取唯一的属性值问题
有如下 XML 文件：
- XML code
<?xml version="1.0" encoding="UTF-8"?><score> <record n="20050001"> <total>342</total> </record> <record n="20050002"> <total>263</total> </record> <record n="20050003"> <total>210</total> </record> <record n="20060009"> <total>120</total> </record> <record n="20060010"> <total>285</total> </record> <record n="20060011"> <total>215</total> </record> <record n="20070001"> <total>170</total> </record> <record n="20070002"> <total>88</total> </record> <record n="20070003"> <total>400</total> </record> <record n="20090006"> <total>330</total> </record> <record n="20090007"> <total>315</total> </record></score>
我想输出：
<select>
<option>2005</option>
<option>2006</option>
<option>2007</option>
<option>2009</option>
</select>
XSLT 应该怎样写？
关键在于提取 record 的 n 的前四位年份。数据量有近万条记录。
如果先提取所有记录的年份后再过滤，运算量很大，速度很慢，有什么好办法吗？有没有直接的办法提取唯一的年份？
[解决办法]
- XML code
<?xml version="1.0" encoding="UTF-8"?><xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:output method="html"/> <xsl:template match="/"><select> <xsl:for-each select="score/record[substring(@n,0,5)!=substring(following::record/@n,0,5)]"> <option><xsl:value-of select="substring(@n,0,5)"/></option> </xsl:for-each></select> </xsl:template></xsl:stylesheet>