how run bash script with php shell_exec with passing 2 variable parameter

Question

i want running bash script with passing 2 parameter and show with bash script create file.txt

$email = "omdik";
$password = "123";

$page = shell_exec("/var/www/html/aktiva-development/test.sh $email, $password");

and this file test.sh

#!/bin/bash
echo "username : $1 dan password : $2" > /tmp/cekhasil.txt

why the file not create, like the bash command is not running, but if script bash like ls dir the dir list will be show, and the access permision for file and folder i was change to 777

thank you for help.

Did you try enclosing the arguments within double quotes? shell_exec(‘/var/www/html/aktiva-development/test.sh “$email”, “$password”’); — abhishek phukan
– abhishek phukan, Commented Jun 19, 2021 at 7:14
@Armali the algoritma, start from login form, and if the username and password is true the bach script will be run, else username and password is false nothing. — Dicki Rizki Amarullah
– Dicki Rizki Amarullah, Commented Jun 20, 2021 at 7:52
yeah i am already for trying that, but the script bash is the same not running, @abhishekphukan — Dicki Rizki Amarullah
– Dicki Rizki Amarullah, Commented Jun 20, 2021 at 7:53
hey i find that error if i runninh that @abhishekphukan the message error is PHP Parse error: syntax error, unexpected 'var' (T_VAR) in /var/www/html/aktiva-development/verifikasi_new.php on line 49, — Dicki Rizki Amarullah
– Dicki Rizki Amarullah, Commented Jun 20, 2021 at 7:55
in line 49 is bash script $page = shell_exec(‘/var/www/html/aktiva-development/test.sh “$email”, “$password”’); @Armali — Dicki Rizki Amarullah
– Dicki Rizki Amarullah, Commented Jun 20, 2021 at 8:44

Armali · Accepted Answer · 2021-06-20 08:54:04Z

1

in line 49 is bash script $page = shell_exec(‘/var/www/html/aktiva-development/test.sh “$email”, “$password”’);

You have wrong ‘, “, ” and ’ quotes there; you have to use ' and ".

answered Jun 20, 2021 at 8:54

Armali

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5 Comments

Dicki Rizki Amarullah Over a year ago

huahh my head hurts, the result is still the same can't and no error message what, idk again what i must doit

Armali Over a year ago

Proposal: add the line exec 2>&1 before the echo and inspect $page then after return in PHP.

Dicki Rizki Amarullah Over a year ago

i think the error is wording format shell exec passwing with 2 argument, because in file test.sh under line echo username and password i add echo string something it show in my website, do you now format wording with shell_exec from php and then get at the file test.sh the value

Dicki Rizki Amarullah Over a year ago

i already try this format but can't, do you have any reference? $page = shell_exec("/tmp/my_script.php $my_url $my_refer"); $page = shell_exec('/tmp/my_script.php "'.$my_url.'" "'.$my_refer.'"');

Armali Over a year ago

I don't understand what reference you mean.

Dicki Rizki Amarullah · Accepted Answer · 2021-06-21 09:12:11Z

1

Hei all i am already solve for this case, solve with change access permision folder and file to www-data:www-data for sh file and php file

-rwxrwxrwx 1 www-data www-data test.sh and 
-rw-r--r-- 1 www-data www-data file.php

thanks all for answer

answered Jun 21, 2021 at 9:12

Dicki Rizki Amarullah

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