Re: [PHP] Get filename in php command line
$PHP_SELF returns the filename with the full path. from there, you can use split to split the string wherever a / occurs, then call count, to get the total number of values in the array, decrememnt the count by 1 (Arrays begin at 0) then, set a variable to store just the filename, like so: $fullpath = split('/',$PHP_SELF); $c = count($fullpath); $c--; $filename = $fullpath[$c]; There may be a better way to do it, but this works for me, and it's only 4 lines. :) JB - Original Message - From: Reuben D Budiardja [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Tuesday, July 17, 2001 7:48 AM Subject: [PHP] Get filename in php command line Hello, How do I get the filename or script name if I run it in php comamnd line? For example, if I run this bash$ php test.php I want to get something inside test.php that tells me the filename is test.php, but *without* the actual path. Thanks Reuben D. Budiardja -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] Get filename in php command line
Seems that $PHP_SELF is not defined when it's run from the command line. Reuben D. Budiardja On Tuesday 17 July 2001 10:29 am, Jason Bell wrote: $PHP_SELF returns the filename with the full path. from there, you can use split to split the string wherever a / occurs, then call count, to get the total number of values in the array, decrememnt the count by 1 (Arrays begin at 0) then, set a variable to store just the filename, like so: $fullpath = split('/',$PHP_SELF); $c = count($fullpath); $c--; $filename = $fullpath[$c]; There may be a better way to do it, but this works for me, and it's only 4 lines. :) JB - Original Message - From: Reuben D Budiardja [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Tuesday, July 17, 2001 7:48 AM Subject: [PHP] Get filename in php command line Hello, How do I get the filename or script name if I run it in php comamnd line? For example, if I run this bash$ php test.php I want to get something inside test.php that tells me the filename is test.php, but *without* the actual path. Thanks Reuben D. Budiardja -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP] Get filename in php command line
Seems that $PHP_SELF is not defined when it's run from the command line. Try: __FILE__ Chris
Re: [PHP] Get filename in php command line
I had that before. OK, seems that my problem is not that simple. The thing is , I have something like this in my file test.php: require('my_api.inc'); echo This is test.php3; Now, in my_api.inc, I want to know that I run test.php3, when I do bash$ php test.php3 and the filename is test.php3 . __FILE__ will give my my_api.inc instead. I need it because my_api.inc do all sort of authenticating, session, etc. Any solutions? Thanks. Reuben D. Budiardja On Tuesday 17 July 2001 10:52 am, Boget, Chris wrote: Seems that $PHP_SELF is not defined when it's run from the command line. Try: __FILE__ Chris -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP] Get filename in php command line
I had that before. OK, seems that my problem is not that simple. The thing is , I have something like this in my file test.php: require('my_api.inc'); echo This is test.php3; Now, in my_api.inc, I want to know that I run test.php3, when I do bash$ php test.php3 and the filename is test.php3 . __FILE__ will give my my_api.inc instead. I need it because my_api.inc do all sort of authenticating, session, etc. __FILE__ will give you the name of the file that uses that constant. So, based on the above, I believe that you are using __FILE__ only in my_api.inc (which is pretty much what you said, I'm just looking for verification). What you can do instead is this: in my_api.inc, have the following code: echo Currently Running file is: ; echo ( $CurrentRunningFile ) ? $CurrentRunningFile : __FILE__; in test.php ? $CurrentRunningFile = __FILE__; include( my_api.inc ); echo This is test.php3br\n; ? Define $CurrentRunningFile on line one of all your files. Or something along those lines... Chris
Re: [PHP] Get filename in php command line
On Tuesday 17 July 2001 11:14 am, Boget, Chris wrote: in my_api.inc, have the following code: echo Currently Running file is: ; echo ( $CurrentRunningFile ) ? $CurrentRunningFile : __FILE__; in test.php ? $CurrentRunningFile = __FILE__; include( my_api.inc ); echo This is test.php3br\n; ? Define $CurrentRunningFile on line one of all your files. Or something along those lines... Chris Well, I thought about that too. The thing is, this is part of the bigger project to make our apps can be run from command line, and we already have tons of apps that currently running from web. Doing that for every single apps would be a pain. But apparently since there is no other way, at least at this point, it will have do for some apps. Thanks. Reuben D. Budiardja -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]