PHP 4.0 Bug #6817: fopen doesn't work with $var ?

From: carl.schreiber@web.de Operating system: Linux Suse 6.4 PHP version: 4.0.2 PHP Bug Type: Filesystem function related Bug description: fopen doesn't work with $var ? Hello, I'm lost with a problem that I don't really understand, may it's a bug? (working on Suse Linux 6.4 & php4) This way the file is created written and closed: $fp = fopen("/tmp/zytest.tmp","w"); $i = fputs($fp, "TEST"); $j = fclose($fp); All results are ok: $fp =..#1, $i=4, $j=1; This way all the results (echo...) seems to be ok, but no file is created, no file is in that folder hereafter? $dfn="/tmp/ztest.tmp"; $fp = fopen($dfn, "w"); $i = fputs($fp, "TEST"); $j = fclose($fp); The echo.. results are the same: $fp =..#1, $i=4, $j=1; So if I hardcode folder and filename fopen() works, if I assign the folder and file to a variable, nothing happens? Does anybody know what to do? Thanks a lot in advance, Carl