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Bug 61720 - <xsl:if ... or does not evaluate correctly
<xsl:if ... or does not evaluate correctly
Status: VERIFIED NOTABUG
Product: libxslt
Classification: Platform
Component: general
unspecified
Other Linux
: Normal normal
: ---
Assigned To: Daniel Veillard
Daniel Veillard
Depends on:
Blocks:
 
 
Reported: 2001-10-04 11:48 UTC by astra
Modified: 2009-08-15 18:40 UTC
See Also:
GNOME target: ---
GNOME version: ---

Attachments
the xsl generates lots of html. I placed some lines like ERROR! at the appropriate place. Also there is a line with "---" to show the place. (4.29 KB, application/octet-stream)
2001-10-04 11:51 UTC, astra 		Details

Description astra 2001-10-04 11:48:38 UTC 
I have an expression with or:
/Object/user/id=/Object/owner/object/id or owner/object/id=/Object/user/id
which does not work. I am attaching a testsuite.
I am not sure about the usage of or, but this works with sablotron.
Comment 1 astra 2001-10-04 11:51:32 UTC 
Created attachment 5757 [details]
the xsl generates lots of html. I placed some lines like ERROR! at the appropriate place. Also there is a line with "---" to show the place.
Comment 2 Daniel Veillard 2001-10-04 13:16:47 UTC 
The error seems to be in your stylesheet.
The fact that the two node sets are equal is not defined
as the string values of the nodes are equal.
A priori the two node sets don't share any nodes
and hence they are not equal. The fact that their string
values are is not relevant.
If you want to compare the string values not the node sets then
use string(xxx) = string(yyy) !
Please read the section of the XPath spec on node-sets equality

Daniel
Comment 3 Daniel Veillard 2001-10-30 18:55:47 UTC 
Seems there is nothing new to add,

Daniel