Php exec() return code is -1 when in a forked process, but 0 in a normal script

Question

I am using exec() inside a a script that runs as a daemon and forks child processes using the pear class Net_Server.

I am getting a strange issue whereby the return code (the third param of of exec) comes back as -1. When I run the command on the command line, or with exec in a normal php script the return code is 0 as it should be. Anyone have any idea why this is happening, and how to fix it?

Does that code use pcntl_signal() by any chance?

Wrikken
– Wrikken

2010-06-17 16:40:49 +00:00
Commented Jun 17, 2010 at 16:40 — Wrikken
– Wrikken, Commented Jun 17, 2010 at 16:40

Wrikken · Accepted Answer · 2010-06-17 16:53:55Z

1

 pcntl_signal(SIGCHLD, SIG_DFL);
 exec('...');
 pcntl_signal(SIGCHLD, 'whatever it was');

answered Jun 17, 2010 at 16:53

Wrikken

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3 Comments

WayneC Over a year ago

Thanks, that worked. The pear class was setting pcntl_signal(SIGCHLD, SIG_IGN);

gabo Over a year ago

Could you please give a bit more info about the 'whatever it was' part? How do I know what should I put there? Thanks

Wrikken Over a year ago

@gabo: whatever it was it just that, the problem of the original asker was that something was overriding the default signal handler for SIGCHLD with 'something'. The solution for him was going back to the default SIG_DFL for what he needed to do, and after that set the signal handler for SIGCHLD back to whatever it was overriden with in the first place (likely for a good reason). No way to tell what that was in the code. If you use a specific signal handler instead of the deafult, you most will know what that is in your code unless your using undocumented 3rd party code.

Sjoerd · Accepted Answer · 2010-06-17 16:30:47Z

0

I guess it is an issue with environment variables. Maybe the thing you are trying to execute is normally in your PATH, but not when you spawn a daemon. Supply the full path to the program.

answered Jun 17, 2010 at 16:30

Sjoerd

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