Executing multiple scripts in the background...

[Jabop]

I have a script written that will convert video upon upload. I made it do it in the background to save the user the wait of the conversion process as well. I have two execs(); being called, both of them are crucial parts in the video converting process. However, since the initial conversion process is sent to the background, the second one gets called immediately.

 

Example:

 

exec(do video converting >& /dev/null & disown);
exec(update metadata >& /dev/null & disown);

 

This presents a problem: the metadata update execution gets done immediately after the previous process is put into the background, so it tries getting the data from the video that hasn't converted completely just yet.

 

My question is, is there a way I can delay the second execution by a certain amount of time? I tried sleep(x), but that hangs the http request til the sleep is finished as well.

 

 

 

 

Thanks everyone

[jonsjava]

have the second script called by the first.

[Jabop]

have the second script called by the first.

 

I don't execute an actual php script, but a bash script...

[jonsjava]

I understand. do something like this:

make a 3rd script:

#!/bin/bash
do video converting >& /dev/null;
update metadata >& /dev/null

then just call the 3rd script, assigning variables as needed in that one.

[mushroom]

I tried sleep(x), but that hangs the http request til the sleep is finished as well.

I don't think you can avoid that, but you can finish the page.

after "</html>" and before sleep(x); add

ob_flush();
flush();

[Jabop]

I understand. do something like this:

make a 3rd script:

#!/bin/bash
do video converting >& /dev/null;
update metadata >& /dev/null

then just call the 3rd script, assigning variables as needed in that one.

 

How can I call phpvars into the bash script?

 

I tried sleep(x), but that hangs the http request til the sleep is finished as well.

I don't think you can avoid that, but you can finish the page.

after "</html>" and before sleep(x); add

ob_flush();
flush();

 

I may give this a shot

[Jabop]

Double post

[jonsjava]

you pass it like any other bash script

<?php
$var1 = "var1";
shell_exec("yourscript.sh ".$var1);

[Jabop]

jonsjava - I'm not familiar with passing any variables to bash actually. There are many that need to be given to the query.

 

So would the 3rd script look something like

 

#!/bin/bash
do video converting -i $InputFile -o $OutputFile >& /dev/null;
update -i $FinalFile metadata >& /dev/null

 

Etc... ? Forgive me for the monotony.

[jonsjava]

looks right.  That's how I'd write it. (I'm a linux admin first, php programmer distant 2nd)

[Jabop]

Well that's good - I just didn't know that I could pass the variables like that. I'll give this a shot when I get the chance and will keep you updated. Thanks for the help

[jonsjava]

just got to thinking. Might be better to merge both of your scripts into one single script, that way, you won't have all that headache.

[Jabop]

So there is one script now:

do video converting -i $InputFile -o $OutputFile >& /dev/null;
update -i $FinalFile metadata >& /dev/null

 

I call the script by exec('whatever.sh'.$InputFile,$OutputFile,$Foo);?

 

I know that is failcode, but I don't know how it'd be passed.

[jonsjava]

in your script, just have it set so that it reads the variables passed:

$var1 = $1;
$var2 = $2;
$var3 = $3;

etc, etc.  Seeing how you know bash, you know that $1, $2, $3, etc refers to variables you passed to it in the command line.

[Jabop]

So when calling the script it is proper to separate variables by commas?

 

exec('script.sh'.$var1,$var2,$var3);

[jonsjava]

So when calling the script it is proper to separate variables by commas?

 

exec('script.sh'.$var1,$var2,$var3);

exec("script.sh $var1 $var2 $var3");

That should work