Using braces with dynamic variable names in PHP

Question

I'm trying to use dynamic variable names (I'm not sure what they're actually called) But pretty much like this:

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.

Answer 1

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

Answer 2

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']()

Answer 3

Try using {} instead of ():

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

Answer 4

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.

Answer 5

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

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

Answer 6

Try using {} instead of ():

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

Answer 7

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;
}

Answer 8

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];

Answer 9

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];