Dynamic variable names in php not working

Question

Suppose i have the following variable

$a="stack";

$b="overflow";

now i want to create another variable using the variable $a and $b .I want the new variable to be $stack_overflow ="The best place to learn programming";

I tried the following to create the variable $stack_overflow but it didn't work

${$a."_".$b}="The best place to learn programming";
echo $stack_overflow;

It worked fine for me: codepad.viper-7.com/XBlOHB

Jonathan Kuhn
– Jonathan Kuhn

2014-12-17 18:04:16 +00:00
Commented Dec 17, 2014 at 18:04 — Jonathan Kuhn
– Jonathan Kuhn, Commented Dec 17, 2014 at 18:04
Your updated code works!

Rizier123
– Rizier123

2014-12-17 18:06:09 +00:00
Commented Dec 17, 2014 at 18:06 — Rizier123
– Rizier123, Commented Dec 17, 2014 at 18:06

Rizier123 · Accepted Answer · 2014-12-17 18:03:30Z

2

This should work for you:

<?php

    $a = "stack";
    $b = "overflow";
    $name = $a . "_" . $b;

    $$name = "The best place to learn programming";
    echo $stack_overflow;

?>

Output:

The best place to learn programming

For more information see: http://php.net/manual/en/language.variables.variable.php

EDIT:

Your updated code:

${$a."_".$b}="The best place to learn programming";
echo $stack_overflow;

works too.

answered Dec 17, 2014 at 18:03

Rizier123

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Comments

user2650277 · Accepted Answer · 2014-12-17 18:15:39Z

0

Actually my code works and even @Rizier123 code does as well(in this simplifed example)

I did a mistake in naming my variables so it would have never worked in my application

answered Dec 17, 2014 at 18:15

user2650277

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1 Comment

Rizier123 Over a year ago

I would recommend you to edit your question(stackoverflow.com/posts/27531913/edit) and write this!(mark it as an edit if you do) Otherwise some people may down vote this.(from my experience)

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