Using dynamic variable names in PHP

Question

I'm trying to use dynamic variable names. So inside this loop I want to create $file0, $file1, and $file2:

for($i=0; $i<=2; $i++) {
    $("file" . $i) = file($filelist[$i]);
}

var_dump($file0);

The return is null which tells me it's not working. I have no idea what the syntax or the technique I'm looking for is here, which makes it hard to research. $filelist is defined earlier on.

Don't! There is never a good reason to use them. They are, effectively, just an untidy array. Use a proper array instead. — Quentin
– Quentin, Commented Feb 13, 2012 at 8:37
Alright I'm sorry, I just went back and picked an answer for each question I've asked so far. Good thing it's only 7 :P — user1159454
– user1159454, Commented Feb 13, 2012 at 8:56
And Quentin, why are they bad practice?? There must be a reason they exist alongside arrays I'd think — user1159454
– user1159454, Commented Feb 13, 2012 at 8:57
@user1159454 — They are a disorganised mess without all the tools that can be applied to arrays available to them. They exist for ancient legacy reasons and crazy edge cases. — Quentin
– Quentin, Commented Feb 13, 2012 at 11:20
Crazy edge cases may be exactly why someone would want to ask this question. — Vincent
– Vincent, Commented Nov 6, 2014 at 19:50

Sarfraz · Accepted Answer · 2012-02-13 08:50:00Z

602

Wrap them in {}:

${"file" . $i} = file($filelist[$i]);

Working Example

Using ${} is a way to create dynamic variables, simple example:

${'a' . 'b'} = 'hello there';
echo $ab; // hello there

edited Feb 13, 2012 at 8:50

answered Feb 13, 2012 at 8:35

Sarfraz

384k81 gold badges561 silver badges613 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

dtakis Over a year ago

Is it also possible to create dynamic arrays with the same logic?

Marcio Mazzucato Over a year ago

I have the same doubt of @dtakis, can you help us? If possible, please take a look at this question.

BeNice Over a year ago

Jut a warning if you are using this to include a variable (the most useful way IMHO) don't forget to use double quotes. ${'fixedTime$i'} = $row['timeInstance']; gives you a not very useful $fixedTime$i instead of $fixedTime1, $fixedTime2 etc. (Fortunately spotted it almost straight away.)

John Slegers · Accepted Answer · 2022-05-27 00:00:22Z

116

Overview

In PHP, you can just put an extra $ in front of a variable to make it a dynamic variable :

$$variableName = $value;

While I wouldn't recommend it, you could even chain this behavior :

$$$$$$$$DoNotTryThisAtHomeKids = $value;

You can but are not forced to put $variableName between {} :

${$variableName} = $value;

Using {} is only mandatory when the name of your variable is itself a composition of multiple values, like this :

${$variableNamePart1 . $variableNamePart2} = $value;

It is nevertheless recommended to always use {}, because it's more readable.

Differences between PHP5 and PHP7

Another reason to always use {}, is that PHP5 and PHP7 have a slightly different way of dealing with dynamic variables, which results in a different outcome in some cases.

In PHP7, dynamic variables, properties, and methods will now be evaluated strictly in left-to-right order, as opposed to the mix of special cases in PHP5. The examples below show how the order of evaluation has changed.

Case 1 : `$$foo['bar']['baz']`

PHP5 interpetation : ${$foo['bar']['baz']}
PHP7 interpetation : ${$foo}['bar']['baz']

Case 2 : `$foo->$bar['baz']`

PHP5 interpetation : $foo->{$bar['baz']}
PHP7 interpetation : $foo->{$bar}['baz']

Case 3 : `$foo->$bar['baz']()`

PHP5 interpetation : $foo->{$bar['baz']}()
PHP7 interpetation : $foo->{$bar}['baz']()

Case 4 : `Foo::$bar['baz']()`

PHP5 interpetation : Foo::{$bar['baz']}()
PHP7 interpetation : Foo::{$bar}['baz']()

edited May 27, 2022 at 0:00

answered Mar 3, 2016 at 12:43

John Slegers

47.4k23 gold badges205 silver badges173 bronze badges

2 Comments

martinstoeckli Over a year ago

A good answer to a dubious topic. I'm really glad i don't have to debug such code.

Rid Iculous Over a year ago

best answer since it just answers the question without being judgemental

Joakim Johansson · Accepted Answer · 2012-02-13 08:36:17Z

22

Try using {} instead of ():

${"file".$i} = file($filelist[$i]);

answered Feb 13, 2012 at 8:36

Joakim Johansson

3,1801 gold badge29 silver badges43 bronze badges

1 Comment

Vladimir Ch Over a year ago

from phpNET manual php.net/manual/ru/language.variables.variable.php

$price_for_monday = 10; $price_for_tuesday = 20; $today = 'tuesday';  $price_for_today = ${ 'price_for_' . $today}; echo $price_for_today; // will return 20

Tom · Accepted Answer · 2013-01-16 06:03:13Z

8

I do this quite often on results returned from a query..

e.g.

// $MyQueryResult is an array of results from a query

foreach ($MyQueryResult as $key=>$value)
{
   ${$key}=$value;
}

Now I can just use $MyFieldname (which is easier in echo statements etc) rather than $MyQueryResult['MyFieldname']

Yep, it's probably lazy, but I've never had any problems.

answered Jan 16, 2013 at 6:03

Tom

811 silver badge1 bronze badge

2 Comments

DarkSide Over a year ago

Have you ever tried to use extract() function of PHP? php.net/manual/en/function.extract.php

Rid Iculous Over a year ago

handy one-liner for DB result sets: foreach($row as $k=>$v){$$k=$v}

corysus · Accepted Answer · 2013-05-30 19:14:11Z

3

Tom if you have existing array you can convert that array to object and use it like this:

$r = (object) $MyQueryResult;
echo $r->key;

answered May 30, 2013 at 19:14

corysus

6579 silver badges12 bronze badges

Comments

Vildan Bina · Accepted Answer · 2019-02-15 15:45:16Z

3

Try using {} instead of ():

${"file".$i} = file($filelist[$i]);

answered Feb 15, 2019 at 15:45

Vildan Bina

3303 silver badges13 bronze badges

Comments

Murad · Accepted Answer · 2015-05-12 16:58:42Z

1

i have a solution for dynamically created variable value and combined all value in a variable.

if($_SERVER['REQUEST_METHOD']=='POST'){
    $r=0;
    for($i=1; $i<=4; $i++){
        $a = $_POST['a'.$i];
        $r .= $a;
    }
    echo $r;
}

edited May 12, 2015 at 16:58

answered May 12, 2015 at 16:45

Murad

112 bronze badges

Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

I was in a position where I had 6 identical arrays and I needed to pick the right one depending on another variable and then assign values to it. In the case shown here $comp_cat was 'a' so I needed to pick my 'a' array ( I also of course had 'b' to 'f' arrays)

Note that the values for the position of the variable in the array go after the closing brace.

${'comp_cat_'.$comp_cat.'_arr'}[1][0] = "FRED BLOGGS";

${'comp_cat_'.$comp_cat.'_arr'}[1][1] = $file_tidy;

echo 'First array value is '.$comp_cat_a_arr[1][0].' and the second value is .$comp_cat_a_arr[1][1];

edited Jun 20, 2020 at 9:12

CommunityBot

11 silver badge

answered Nov 10, 2015 at 20:33

M61Vulcan

3373 silver badges7 bronze badges

Comments

Tom Friedrichs · Accepted Answer · 2022-10-14 09:53:02Z

0

After trying all the mentioned above, it worked the simple way: It's accessed as $_POST variable, as it comes directly out of a dynamic form.

$variableName= "field_".$id;
echo $_POST[$variableName];

answered Oct 14, 2022 at 9:53

Tom Friedrichs

436 bronze badges

Collectives™ on Stack Overflow

Using dynamic variable names in PHP

9 Answers 9

Working Example

3 Comments

Overview

Differences between PHP5 and PHP7

Case 1 : `$$foo['bar']['baz']`

Case 2 : `$foo->$bar['baz']`

Case 3 : `$foo->$bar['baz']()`

Case 4 : `Foo::$bar['baz']()`

2 Comments

1 Comment

2 Comments

Comments

Comments

Comments

Comments

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

9 Answers 9

3 Comments

Overview

Differences between PHP5 and PHP7

Case 1 : $$foo['bar']['baz']

Case 2 : $foo->$bar['baz']

Case 3 : $foo->$bar['baz']()

Case 4 : Foo::$bar['baz']()

2 Comments

1 Comment

2 Comments

Comments

Comments

Comments

Comments

Comments

Linked

Related

Case 1 : `$$foo['bar']['baz']`

Case 2 : `$foo->$bar['baz']`

Case 3 : `$foo->$bar['baz']()`

Case 4 : `Foo::$bar['baz']()`