简体   繁体   中英

XSLT - How to select XML Attribute by Attribute?

this is the structure of my source xml:

<root>
<DataSet Value="A">
<Data Value1="1" Value2="anythingA1" />
<Data Value1="2" Value2="anythingA2" />
<Data Value1="3" Value2="anythingA3" />
<Data Value1="4" Value2="anythingA4" />
<Data Value1="5" Value2="anythingA5" />
</DataSet>
</root>

from which I like to create some variables eg from all with Value1="2" and all with Value1="5" should result myVar1 with anythingA2 and myVar2 with anythingA5

My approch looks like this

<xsl:variable name="myVarA" select="/DataSet/Data/[@Value1='2']/@Value2" />

but of course is not working since Value2 is no child of Value1.

thanks for any hints in advance!

只需在Data之后删除斜杠并在前面添加根:

<xsl:variable name="myVarA" select="/root/DataSet/Data[@Value1='2']/@Value2"/>

There are two problems with your xpath - first you need to remove the child selector from after Data like phihag mentioned. Also you forgot to include root in your xpath. Here is what you want to do:

select="/root/DataSet/Data[@Value1='2']/@Value2"

Try this

xsl:variable name="myVarA" select="//DataSet/Data[@Value1='2']/@Value2" />

The '//' will search for DataSet at any depth

Note: using // at the beginning of the xpath is a bit CPU intensitve -- it will search every node for a match. Using a more specific path, such as /root/DataSet will create a faster query.

I would do it by creating a variable that points to the nodes that have the proper value in Value1 then referring to t

<xsl:variable name="myVarANode" select="root//DataSet/Data[@Value1='2']" />
<xsl:value-of select="$myVarANode/@Value2"/>

Everyone else's answers are right too - more right in fact since I didn't notice the extra slash in your XPATH that would mess things up. Still, this will also work , and might work for different things, so keep this method in your toolbox.

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM