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

ID: 6817 Updated by: sniper Reported By: carl.schreiber@web.de Status: Feedback Bug Type: Filesystem function related Assigned To: Comments: I tried this (with latest CVS) and it works. Please try latest CVS or snapshot from http://snaps.php.net/ --Jani Previous Comments: --------------------------------------------------------------------------- [2000-09-20 18:43:01] carl.schreiber@web.de 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 --------------------------------------------------------------------------- Full Bug description available at: http://bugs.php.net/?id=6817