PHP Variables Defined As Other Variables

Question

I'm attempting to use PHP to make a configuration file. I have the variables set up to output to a final variable. However, when I'm trying to echo the output, it won't appear.

I've tried various solutions such as reformatting variables within the output variable.

<?php
   $var1 = "1";
   $var2 = "0";
   $var3 = "9";
   $varout = "{$var1}, {$var2}, {$var3};
   echo "Settings are {$varout}";
?>

Output:

Settings are {$varout}

it us working here : https://eval.in/1097799 – Niklesh Raut Apr 21 '19 at 03:18 — Niklesh Raut, Apr 21 '19 at 03:18
Closing double quote is missing for variable `$varout`. – Saji Apr 21 '19 at 03:21 — Saji, Apr 21 '19 at 03:21

score 0 · Answer 1 · answered Apr 21 '19 at 03:24

0

You are missing a closing double quote on your definition of $varout, add that and it will work.

answered Apr 21 '19 at 03:24

John Carpenter

47
3

score 0 · Answer 2 · answered Apr 21 '19 at 03:34

0

You can use concenat operator "."

answered Apr 21 '19 at 03:34

ankita

33
8

PHP Variables Defined As Other Variables

2 Answers2