Bug #16459 Updated: way to get *actual* filename code is in (include() breaks available methods)
| From: | olli at ukgamer dot net | Date: | Sat, 06 Apr 2002 00:17:22 +0000 |
| Subject: | Bug #16459 Updated: way to get *actual* filename code is in (include() breaks available methods) | ||
| References: | 1 | Groups: | php.bugs |
| Request: | Send a blank email to php-bugs+get-4183@lists.php.net to get a copy of this message | ||
ID: 16459
Updated by: olli@ukgamer.net
Reported By: olli@ukgamer.net
-Status: Open
+Status: Bogus
Bug Type: Feature/Change Request
Operating System: Win*/Slackware
PHP Version: 4.1.2
New Comment:
Sorry, i missed the constant "__FILE__".
Previous Comments:
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[2002-04-05 19:07:48] olli@ukgamer.net
There doesn't appear to be a way to get the filename of code that is
currently executing.
Eg. echo $SCRIPT_FILENAME; or echo $PHP_SELF;
works fine in most situations, but as soon as the file containing this
code is included in another, they start to return the filename of the
includING file. (not the includED one).
I understand this generally goes against the purpose of include() but
this functionality would be very useful.
Thanks
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Edit this bug report at http://bugs.php.net/?id=16459&edit=1