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[2002-10-03 04:28 UTC] k.schroeder@php.net
[2002-10-03 19:50 UTC] tony at marston-home dot demon dot co dot uk
[2002-10-04 02:26 UTC] derick@php.net
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Last updated: Tue Oct 28 10:00:01 2025 UTC |
I am using Windows XP, PHP 4.2.3, Apache 1.3.26 I am using XSLT to produce HTML output. The code I use is as follows: $arg_buffer = array("/xml" => $xml_string, "/xslt" => $xsl_string); $result = xslt_process($xp, "arg:/xml", "arg:/xslt", NULL, $arg_buffer, $params))) My xsl file contains as 'include' statement which points to a file which exists in the same folder as the PHP script. I initially tried it as <xsl:include href="pagination.xsl"/> but it produced this error: 'arg:/pagination.xsl' not found (error code: 65) I then tried it with <xsl:include href="file://pagination.xsl"/> and got this error: cannot open file '/pagination.xsl' (error code: 4) It only works if I enter the full path, as in <xsl:include href="file://c:/blah/blah/blah/pagination.xsl"/> It is very inconvenient to have to specify the full pathname in every <xsl:include> statement. Should it not first look in the current directory? This may be more of an enhancement request than a bug.