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how to pass commandline argument to C program from PHP code
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| balakrishnan.dinesh@gmail.com 2007-01-29, 3:59 am |
| hi frndz,
As we know that, we can pass command line agrument for C using
"scanf" commands, So as same as that, Is there any way to pass those
commandline arguments through php code to C and process them. If there
plz soon, or tel me any alternative for this
Rgrds
Dinesh..
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| lorento 2007-01-29, 3:59 am |
| Did you mean execute C program like in command line ?
<?php
exec ("yourprogram -n 1234");
?>
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On Jan 29, 11:28 am, balakrishnan.din...@gmail.com wrote:
> hi frndz,
>
> As we know that, we can pass command line agrument for C using
> "scanf" commands, So as same as that, Is there any way to pass those
> commandline arguments through php code to C and process them. If there
> plz soon, or tel me any alternative for this
>
> Rgrds
> Dinesh..
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| Ruben van Engelenburg 2007-01-29, 7:58 am |
| balakrishnan.dinesh@gmail.com wrote:
> hi frndz,
>
> As we know that, we can pass command line agrument for C using
> "scanf" commands, So as same as that, Is there any way to pass those
> commandline arguments through php code to C and process them. If there
> plz soon, or tel me any alternative for this
Hi Dinesh,
I'm not sure what you mean, because in C you don't use scanf to retrieve
commandline arguments. You use scanf to get user-input. You use the argc
and argv arguments in main() to access commandline arguments.
In PHP you have similar variables: $argc and $argv. $argc holds the
number of arguments and $argv is the array containing the values.
HTH.
Ruben.
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| bob.chatman@gmail.com 2007-01-31, 3:59 am |
| On Jan 29, 2:21 am, Ruben van Engelenburg <nos...@nospam.com> wrote:
> balakrishnan.din...@gmail.com wrote:
>
>
> Hi Dinesh,
>
> I'm not sure what you mean, because in C you don't use scanf to retrieve
> commandline arguments. You use scanf to get user-input. You use the argc
> and argv arguments in main() to access commandline arguments.
> In PHP you have similar variables: $argc and $argv. $argc holds the
> number of arguments and $argv is the array containing the values.
>
> HTH.
> Ruben.
What he passed you is a command forwarded to the command line of your
server. its exactly the same as if you were to go:
yourprogram -n 1234
Your program would have ARGC = 2 and ARGV would be {"-n", "1234"}
it has nothing to do with the scanf or anything like that.
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