php exec an echo with variables and string

Question

I want to to exchange "30" and "16" with two variables "hour" and "mins" in a php exec function

$hour = 30;
$mins = 16;
exec('echo "30 16 * * * sudo /home/pi/raspberry-remote/./send 11010 1 1" | crontab -');

and changed it like that:

exec('echo "$hour $mins * * * sudo /home/pi/raspberry-remote/./send 11010 1 1" | crontab -');

How can i get this right? Thank you

Just concatenate it like: 'echo "' . $hour . ' ' . $mins . ' — Rahil Wazir
– Rahil Wazir, Commented Jun 19, 2014 at 11:47

Jim · Accepted Answer · 2014-06-19 11:47:31Z

3

Variables inside single quoted strings are not populated. Try either

exec('echo "'.$hour.' '.$mins.' * * * sudo /home/pi/raspberry-remote/./send 11010 1 1" | crontab -');

or

exec("echo \"$hour $mins * * * sudo /home/pi/raspberry-remote/./send 11010 1 1\" | crontab -");

PHP String Documentation

answered Jun 19, 2014 at 11:47

Jim

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Barmar · Accepted Answer · 2014-06-19 11:47:16Z

2

Variables aren't expanded inside single-quoted strings, only inside double-quoted strings. So switch your quotes around.

exec("echo '$hour $mins * * * sudo /home/pi/raspberry-remote/./send 11010 1 1' | crontab -");

answered Jun 19, 2014 at 11:47

Barmar

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mpyw · Accepted Answer · 2014-06-19 11:54:48Z

1

I recommend you sprintf. That does not lose readability.

$format = 'echo "%d %d * * * sudo /home/pi/raspberry-remote/./send 11010 1 1" | crontab -';
exec(sprintf($format, $hour, $mins));

answered Jun 19, 2014 at 11:54

mpyw

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Arun Magar · Accepted Answer · 2014-06-19 11:58:38Z

1

You can do this using concat string

exec('echo "'.$hour.' ' .$mins.' * * * sudo /home/pi/raspberry-remote/./send 11010 1 1" | crontab -');exec('echo "$hour $mins * * * sudo /home/pi/raspberry-remote/./send 11010 1 1" | crontab -');

edited Jun 19, 2014 at 11:58

answered Jun 19, 2014 at 11:51

Arun Magar

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