Re: system, exec, shell_exec, passthru

Hi, I can run any external program like ls, cp, uptime... But if a try to run my shell script a get error.

         $passwdexe = "sudo /usr/bin/webpasswd";
         $user=$_POST[name];
         $passwd="$passwdexe $user 123456";
         echo $passwd;
         $result = system($passwd);

Resultat from the web, "sudo /usr/sbin/webpasswd billy 12345678" and from the error_log, "couldn't read file "./usr/sbin/webpasswd.": no such file or directory". .Mike Adam Zey wrote:

p.willis@telus.net wrote:

Hello, I am trying the run an external application with command line arguments using PHP under linux. ie: $command="myprog $arg1 $arg2 > textfile.txt"; system("echo \"$command\" > test.txt"); system($command); $handle=fopen("textfile.txt","r"); if($handle!=NULL) {

    while(!feof($handle))
    {
        ...
    }
    fclose($handle);

} I test my input arguments for the 'system' call by dumping the command into a text file. I can then test the command in the console. The commands work fine when run from the console. The commands don't work when run through the system command. I have tried system, exec, passthru, and shell_exec to no avail. Am I missing some permissions thing in my php.ini file? Thanks for any insight, Peter

Looking at your $command, there is no path in front of it. It's likely that PHP's shell doesn't have the program in its PATH. Try manually specifying the full path to the command: $command="/home/peter/myprog $arg1 $arg2 > textfile.txt"; (or whatever the path is) Keep in mind that under linux, (unlike DOS), you CANNOT execute applications in the same directory without a path. Even if the program is in the same directory as your present working directory (which is usually where the PHP script is located), you'd have to do "./myprog" as a relative path. Regards, Adam Zey.