Re: Exec command not working through php
| From: | Mike Morton | Date: | Tue, 15 Jul 2003 22:51:18 +0000 |
| Subject: | Re: Exec command not working through php | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-155330@lists.php.net to get a copy of this message | ||
David:
Permissions were the first guess - I chmoded 777 everything - still no go.
On the server, there should be a second file created identical to the first
with the results of the sed find and replace in it - the only thing that is
created is a blank file.
So, it is not permissions, nothing is returned in the results array, and the
exact same query run on the server command line works properly - the only
difference between it executing through php and command line is, as far as I
can tell, the user being apache for the php exec and a different user
command line, but with mode 777 on the files and directories (ya ya, I know
that is not secure but it is just for testing purposes ;)) the user should
not make a difference.....
On 7/15/03 6:23 PM, "David Nicholson" <me@djnicholson.com> wrote:
> Hello,
>
> This is a reply to an e-mail that you wrote on Tue, 15 Jul 2003 at 23:17,
> lines prefixed by '>' were originally written by you.
>> "if I print out the exec line with the
>> dynamically generated paths, etc, then execute that through the
>> terminal, no
>> problem. "
>>> From the first email meaning that I did just like you suggested
>> already and
>> it works in the terminal and contains exactly what I expect it to.
>> But thanks :)
>
> Sorry, I jumped over that in your original mail! I amy as well try and be
> of help again then...
>
> Does the user on the system that your webserver runs as have permission to
> execute the program you are trying to execute?
>
> Do you expect the program to have any output, if so is it appearing in
> your $results array? is anything being put in the $results array?
>
> Also, you could try using the third optional paramater for the exec
> function to retrieve the return status of the command you are executing.
>
> David.
--
Cheers
Mike Morton
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