Does anyone know how I can pass variables into a script being executed
from the command line?
I have a script, named testscript. So far, I have been getting all my
variables from the system env command, not a big deal. But I want to
input a username into the script, like testscript jmchenry.
In
$argv, possibly $_SERVER['argv'].
Cheers,
Rob.
On Thu, 2003-12-11 at 00:31, Jake McHenry wrote:
Does anyone know how I can pass variables into a script being executed
from the command line?
I have a script, named testscript. So far, I have been getting all my
variables from the system env
On Thursday 11 December 2003 12:31 am, Jake McHenry wrote:
Does anyone know how I can pass variables into a script being executed
from the command line?
I have a script, named testscript. So far, I have been getting all my
variables from the system env command, not a big deal. But I want to
-Original Message-
From: Evan Nemerson [mailto:[EMAIL PROTECTED]
Sent: Wednesday, December 10, 2003 9:32 PM
To: Jake McHenry; [EMAIL PROTECTED]
Subject: Re: [PHP] Php shell scripting
On Thursday 11 December 2003 12:31 am, Jake McHenry wrote:
Does anyone know how I can pass
I'm trying to do shell scripting in PHP. I have PHP installed in
/usr/bin/php. I have the following script:
#!/usr/bin/php -q
?php
print Success!\n;
?
It's saved as test.php and CHMODed to 777. When I type test.php at
the command line, it says:
bash: test.php: command not found
When I type
Ever answered your own question five seconds after asking it?
Apparently, I have to type /path/to/test.php even though I'm already
in the right directory.
Leif K-Brooks wrote:
I'm trying to do shell scripting in PHP. I have PHP installed in
/usr/bin/php. I have the following script:
In Unix you need to specify the working folder when you are launching an
executable, otherwise Bash will try to look into its search path, which
does not include the current folder. Try:
./test.php
That should work. BTW--running test by itself only *looks* like
hitting enter--it's really a valid
You don't have ./ in your $PATH so you must type ./test.php to execute it
Leif K-Brooks wrote:
I'm trying to do shell scripting in PHP. I have PHP installed in
/usr/bin/php. I have the following script:
#!/usr/bin/php -q
?php
print Success!\n;
?
It's saved as test.php and CHMODed to 777.
I am setting up a shell script to provide a menu for some tasks that our server admin
would like
to do via script rather then vi'ing the file, now I am almost finished but I would
like to get rid
of this ugly part of the output.
X-Powered-By: PHP/4.0.4pl1
Content-type: text/html
Is there away
Start php with -q, eg: php -q script.php
-Stathis.
Dan McCullough wrote:
I am setting up a shell script to provide a menu for some tasks that our server
admin would like
to do via script rather then vi'ing the file, now I am almost finished but I would
like to get rid
of this ugly part
I have some questions about using PHP as a shell scripting language...
mainly how you pass arguments to a script on the command line. Say I'm
using:
--
#!/usr/local/bin/php -q
?
print I am: $name\n;
?
--
How do I define test from the command
: Ryan Christensen [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 22, 2001 8:19 PM
To: [EMAIL PROTECTED]
Subject: [PHP] PHP shell scripting..
I have some questions about using PHP as a shell scripting language...
mainly how you pass arguments to a script on the command line. Say I'm
using
:30 PM
To: 'Kees Hoekzema'
Subject: RE: [PHP] PHP shell scripting..
I'm slightly confused as to how I would run this from a shell though..
what would I enter in the command line then?
-
Ryan Christensen
(mail) [EMAIL PROTECTED]
(cell) 360-808-1506
-Original
-Original Message-
From: Kees Hoekzema [mailto:[EMAIL PROTECTED]]
Sent: Sunday, July 22, 2001 11:41 AM
To: Ryan Christensen
Cc: [EMAIL PROTECTED]
Subject: RE: [PHP] PHP shell scripting..
Hello Ryan,
Let me give you an example :)
Lets use your script
In article 001c01c112da$b8c0d420$0d6fa8c0@local, Ryan Christensen wrote:
I have some questions about using PHP as a shell scripting language...
mainly how you pass arguments to a script on the command line. Say I'm
using:
--
#!/usr/local/bin/php -q
?
print I am:
I think it's the same as perl, using the backtick operator, ie:
$result = `uptime`;
That's ` NOT '
-Original Message-
From: Johan Vikerskog (ECS) [mailto:[EMAIL PROTECTED]]
Sent: 03 July 2001 09:41
To: [EMAIL PROTECTED]
Subject: [PHP] Shell scripting and PHP.
Hi all!
If i want
Hi all!
If i want to use some shell commands and use the result it displays to set a variable.
How is this done?
Is there a way of doing it like you can do with Perl?
Anyone knows?
Thanks for all the tips you can give me, or better yet, if you know of a tutorial
somewere that covers this.
On Tue, 3 Jul 2001 17:46, Ray Hilton wrote:
I think it's the same as perl, using the backtick operator, ie:
$result = `uptime`;
That's ` NOT '
Look at passthru() exec() and system() as well.
Cheers,
Brad
--
Brad Hubbard
Congo Systems
12 Northgate Drive,
Thomastown, Victoria, Australia
Hello Justin,
(JY == "Justin Yu") [EMAIL PROTECTED] writes:
JY When I tried connecting to Postgres through a shell script that I
JY wrote, I get the following error:
JY "Fatal error: Call to undefined function: pg_connect()"
Because you don't have pgsql (PostgreSQL) support built into your
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