Calling php 4 from within php 5

[freak1321]

My hosting company offers both php 5 and 4.  I use php 5.  I want to know how to access a php 4 program (or class).  It's ffmpeg-php and I want to access this code in php 4

 

$ffmpegObj = new ffmpeg_movie($filetmp);

$duration = $ffmpegObj->getDuration();

 

ffmpeg-php is only available on php 4 so basically I need to call this function from within php 5.  I was guessing something like:

 

$test = exec("/usr/bin/php -r '$ffmpegObj = new ffmpeg_movie($filetmp);$duration = $ffmpegObj->getDuration();'");

echo $test;

 

but that doesn't work - any ideas anyone? Seems like a tough question...  :'(

[btherl]

You have a problem there in that the code in double quotes will have variable substitution done BEFORE being passed to php4.  Try enclosing it in single quotes instead.  You will need to escape single quotes within the string also.  Eg.

 

$test = exec('/usr/bin/php -r \'$ffmpegObj = new ffmpeg_movie($filetmp);$duration = $ffmpegObj->getDuration();\'');

[freak1321]

Your right about the problem with the quotation marks (or maybe its something else)... I can't even get any of these to work - could anyone come up with a simple working example?

 

$test = system('/usr/bin/php -r \'echo "hi";\'');

echo $test;

 

OR

$st = 'hello';

$test = system('/usr/bin/php -r \'echo $st;\'');

echo $test;

 

OR

 

$st = 'hello';

$test = system('/usr/bin/php -r \'echo '.$st.'\'');

echo $test;

 

[btherl]

Here you go:

 

$st = 'hello';
$command = '/usr/bin/php -r \'echo "'.$st.'";\'';
print "About to execute $command\n";
$test = system($command);
echo $test;

 

Much better to print out the command before executing, so you can visually inspect it.  Dealing with shell escaping is a hassle though, for any more complicated code.

 

[freak1321]

Your totally right - and that totally worked - thanks so much! Funny how things always seem so simple in hindsight!

[keeB]

How about porting that library to PHP5 ?

 

Shouldn't be too difficult