CODEWITHSUNDEEP
phpshellexec
Ingrid Lambrechtse
Ingrid Lambrechtse

Reputation: 43

Passing args to sh via php. First argument is received but not the second

The first argument $htmlpageid can be referenced to by $1 in the sh script. The second argument should be referenced by $2. Right?

<?php
    $htmlpageid = uniqid();
    $days ="1";   
    $command = shell_exec("sudo ./createclientcert.sh $htmlpageid $days");    
?>

No matter what i try but i can't get the second argument $days over to the script. I tried several methods but none of them passes the second argument $days.

Upvotes: 1

Views: 55

Answers (1)

Ingrid Lambrechtse
Ingrid Lambrechtse

Reputation: 43

I made i mistake by referencing $days by $2 in the sh script because i referenced it inside a function definition.

Very sorry for this dumb thing. ;)

#!/bin/sh
newclient () {
    echo "# valid from "$(date) "till "$(date --date='+'$2' day') >> /root/$1.ovpn
}

Instead I should had coded the command outside a function or called the function with the arguments passed to the script:

#!/bin/sh
newclient () {
    echo "# valid from "$(date) "till "$(date --date='+'$2' day') >> /root/$1.ovpn
}

newclient $1 $2

Upvotes: 1

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