PHP: can't exec some programs when run through apache. why?

BrianK · 04-19-2007, 09:22 PM

I can do this, no problem:

PHP Code:


			
if (!exec('ls',$output)){

  echo "bad exec<br>\n";

}else{

  echo "good exec<br>\n";

  foreach ($output as $i){

    echo "$i<br>\n";

}

but I can't do this:

PHP Code:


			
if (!exec('/usr/bin/tcprobe',$output)){

  echo "bad exec<br>\n";

}else{

  echo "good exec<br>\n";

  foreach ($output as $i){

    echo "$i<br>\n";

}

... unless I do it on the command line, i.e. "php foo.php" - in which case it does what I expect, which is spit out what tcprobe spits out.

If I try to do it through a browser, I get "bad exec" for the latter, however, the former (exec ls) works fine.

any ideas why this would be happening?

j-ray · 04-20-2007, 03:06 AM

maybe a permission problem unless /usr/bin/tcprobe is executable for everybody including apache user

BrianK · 04-20-2007, 03:26 AM

Quote:

Originally Posted by j-ray

maybe a permission problem unless /usr/bin/tcprobe is executable for everybody including apache user

hmm.. I thought I put that in my original post, but I guess I edited it out. Yes, tcprobe is executable by all.

I've made a bit of progress, however, but it has uncovered more mysteries -

if I make stderr go to stdout, I get the expected results even through the browser:

PHP Code:


			
if (!exec('/usr/bin/tcprobe 2>&1',$output,$error)){

  echo "bad exec<br>\n";

  echo "error: $error<br>\n";

}else{

  echo "good exec<br>\n";

  $count = 0;

  echo "<br>";

  foreach ($output as $i){

    echo "$count - $i<br>\n";

    $count++;

  }

}

which returns this:

Code:

good exec

0 - tcprobe (transcode v1.0.2) (C) 2001-2003 Thomas Oestreich
1 -
2 - Usage: tcprobe [options] [-]
3 - -i name input file/directory/device/host name [stdin]
4 - -B binary output to stdout (used by transcode) [off]
5 - -H n probe n MB of stream [1]
6 - -s n skip first n bytes of stream [0]
7 - -T title probe for DVD title [off]
8 - -b bitrate audio encoder bitrate kBits/s [128]
9 - -f seekfile seek/index file [off]
10 - -d verbosity verbosity mode [1]
11 - -v print version

However, if I pass a real arg to tcprobe, I only get the first line of output. The following code outputs the next bit:

PHP Code:


			
if (!exec("/usr/bin/tcprobe -i $vid 2>&1",$output,$error)){

  echo "bad exec<br>\n";

  $count = 0;

  echo "error: $error<br>\n";

}else{

  echo "good exec<br>\n";

  $count = 0;

  echo "<br>";

  foreach ($output as $i){

    echo "$count - $i<br>\n";

    $count++;

  }

Code:

good exec

0 - [tcprobe] Apple QuickTime movie file

... but it *should* output this:

Code:

[tcprobe] Apple QuickTime movie file
[probe_mov.c] video codec=jpeg
[tcprobe] summary for Hummer.mov, (*) = not default, 0 = not detected
import frame size: -g 720x486 [720x576] (*)
       frame rate: -f 30.000 [25.000] frc=5 (*)
   no audio track: use "null" import module for audio
           length: 901 frames, frame_time=33 msec, duration=0:00:30.033

... which is what it does output if I run the php from the command line.

Any ideas?

BrianK · 09-20-2007, 04:10 PM

I know this is an old thread, but in the interest of answering all my questions that I find an answer to, here's the deal:

tcprobe appears to use multiple programs to accomplish what it's trying to do, i.e., it's a wrapper (i didn't know this until I looked through the source code, as tcprobe is a binary). That being the case, the tcprobe output (which is just the first line) outputs fine, however, the output from the programs that tcprobe calls goes elsewhere - I don't know where. To solve this problem, I used ffmpeg which is not a wrapper, but a full program in and of itself. The last code that I posted works fine with ffmpeg & does what I would expect it to.