[英]XSLT: Can't select attribute
I'm writing simple transformation using XSLT 1.0 and getting strange result.. Looks like I don't understand something, but I found I can't select attribute of a node for some reason. 我正在使用XSLT 1.0编写简单的转换,并得到奇怪的结果。看起来好像我不明白,但由于某种原因我发现无法选择节点的属性。 Here is my input XML:
这是我的输入XML:
<?xml version="1.0"?>
<weekreport>
<employee name="Emp1">
<day date="25.06.2012">
<entry>
<project>Proj1</project>
<time>08:00</time>
<description>Bla-bla-bla</description>
</entry>
</day>
</employee>
<employee name="Emp2">
<day date="25.06.2012">
<entry>
<project>Proj2</project>
<time>08:00</time>
<description></description>
</entry>
</day>
</employee>
</weekreport>
and here is XSLT: 这是XSLT:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="entry" name="entry_t">
<xsl:param name="name"/>
<xsl:param name="date"/>
<Row>
<Cell><xsl:value-of select="$date"/></Cell>
<Cell><xsl:value-of select="$name"/></Cell>
<Cell><xsl:value-of select="time"/></Cell>
<Cell><xsl:value-of select="project"/></Cell>
<Cell><xsl:value-of select="description"/></Cell>
</Row>
</xsl:template>
<xsl:template match="day" name="day_t">
<xsl:param name="name"/>
<xsl:for-each select="entry">
<xsl:call-template name="entry_t">
<xsl:with-param name="name"><xsl:value-of select="$name"/></xsl:with-param>
<xsl:with-param name="date"><xsl:value-of select="@date"/></xsl:with-param>
</xsl:call-template>
</xsl:for-each>
</xsl:template>
<xsl:template match="employee" name="employee_t">
<xsl:for-each select="day">
<xsl:call-template name="day_t">
<xsl:with-param name="name"><xsl:value-of select="@name"/></xsl:with-param>
</xsl:call-template>
</xsl:for-each>
</xsl:template>
<xsl:template match="/weekreport">
<xsl:for-each select="employee">
<xsl:call-template name="employee_t"/>
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
For some reason calling template day_t is done with parameter "name" set to empty value. 由于某种原因,调用模板day_t时会将参数“名称”设置为空值。 Why?..
为什么?..
When you pass the param, the current node isn't an employee element, it's a day element (xsl:for-each changes the current node). 传递参数时,当前节点不是employee元素,而是day元素(xsl:for-each会更改当前节点)。 Therefore, you're attempting to access the name attribute on the day element, not its parent employee element.
因此,您尝试访问day元素上的name属性,而不是其父雇员元素。 Try this instead:
尝试以下方法:
<xsl:call-template name="day_t">
<xsl:with-param name="name"><xsl:value-of select="parent::employee/@name"/></xsl:with-param>
</xsl:call-template>
You are calling template day_t
inside <xsl:for-each select="day">
, and all the XPath expressions inside it, including @name
are evaluated relative to element <day>
. 您正在
<xsl:for-each select="day">
中调用模板day_t
,并且其中的所有XPath表达式(包括@name
都相对于元素<day>
进行求值。 But this element doesn't have attribute name
. 但是这个元素没有属性
name
。
Apart from fix, think of optimization of XSLT
code too. 除了修复之外,还应该考虑
XSLT
代码的优化。
Currently, you are using <xsl:for-each>
for 3 times. 当前,您正在使用
<xsl:for-each>
3次。 Rather than doing this way, you can do using <xsl:apply-templates>
您可以使用
<xsl:apply-templates>
来<xsl:apply-templates>
这种方式
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml"/>
<xsl:template match="entry" mode="entry_t">
<xsl:param name="name"/>
<xsl:param name="date"/>
<Row>
<Cell>
<xsl:value-of select="$date"/>
</Cell>
<Cell>
<xsl:value-of select="$name"/>
</Cell>
<Cell>
<xsl:value-of select="time"/>
</Cell>
<Cell>
<xsl:value-of select="project"/>
</Cell>
<Cell>
<xsl:value-of select="description"/>
</Cell>
</Row>
</xsl:template>
<xsl:template match="day" mode="day_t">
<xsl:param name="name1"/>
<xsl:apply-templates mode="entry_t" select="entry">
<xsl:with-param name="name" select="$name1"/>
<xsl:with-param name="date" select="@date"/>
</xsl:apply-templates>
</xsl:template>
<xsl:template match="employee" mode="employee_t">
<xsl:apply-templates mode="day_t" select="day">
<xsl:with-param name="name1" select="@name"/>
</xsl:apply-templates>
</xsl:template>
<xsl:template match="/weekreport">
<root>
<xsl:apply-templates mode="employee_t" select="employee"/>
</root>
</xsl:template>
</xsl:stylesheet>
<?xml version="1.0" encoding="UTF-8"?>
<root>
<Row>
<Cell>25.06.2012</Cell>
<Cell>Emp1</Cell>
<Cell>08:00</Cell>
<Cell>Proj1</Cell>
<Cell>Bla-bla-bla</Cell>
</Row>
<Row>
<Cell>25.06.2012</Cell>
<Cell>Emp2</Cell>
<Cell>08:00</Cell>
<Cell>Proj2</Cell>
<Cell/>
</Row>
</root>
You need to pass the value for employee name
and day date
. 您需要传递员工
name
和date
的值。
I have done some modification, 我做了一些修改,
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml"/>
<xsl:template match="entry" name="entry_t">
<xsl:param name="name2"/>
<xsl:param name="date1"/>
<Row>
<Cell>
<xsl:value-of select="$date1"/>
</Cell>
<Cell>
<xsl:value-of select="$name2"/>
</Cell>
<Cell>
<xsl:value-of select="time"/>
</Cell>
<Cell>
<xsl:value-of select="project"/>
</Cell>
<Cell>
<xsl:value-of select="description"/>
</Cell>
</Row>
</xsl:template>
<xsl:template match="day" name="day_t">
<xsl:param name="name1"/>
<xsl:param name="date"/>
<xsl:for-each select="entry">
<xsl:call-template name="entry_t">
<xsl:with-param name="name2" select="$name1"/>
<xsl:with-param name="date1" select="$date"/>
</xsl:call-template>
</xsl:for-each>
</xsl:template>
<xsl:template match="employee" name="employee_t">
<xsl:param name="name"/>
<xsl:for-each select="day">
<xsl:call-template name="day_t">
<xsl:with-param name="name1" select="$name"/>
<xsl:with-param name="date" select="@date"/>
</xsl:call-template>
</xsl:for-each>
</xsl:template>
<xsl:template match="/weekreport">
<root>
<xsl:for-each select="employee">
<xsl:call-template name="employee_t">
<xsl:with-param name="name" select="@name"/>
</xsl:call-template>
</xsl:for-each>
</root>
</xsl:template>
</xsl:stylesheet>
<?xml version="1.0" encoding="UTF-8"?>
<root>
<Row>
<Cell>25.06.2012</Cell>
<Cell>Emp1</Cell>
<Cell>08:00</Cell>
<Cell>Proj1</Cell>
<Cell>Bla-bla-bla</Cell>
</Row>
<Row>
<Cell>25.06.2012</Cell>
<Cell>Emp2</Cell>
<Cell>08:00</Cell>
<Cell>Proj2</Cell>
<Cell/>
</Row>
</root>
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.