Variable variables doesn't work as expected and is creating an array

Question

I'm using this line of code:

$var{++$counter} = $results['row'];

I've set this up with a goal of creating these variables:

$var1 = row 1
$var2 = row 2
$var3 = row 3

Why is it created an array for $var ? Instead of just defining three variables?

It states my question below the code and in the title.... Why is it created an array for $var instead of just defining the variables. — KDJ, May 28 '15 at 12:56
Because that's not the correct syntax for variable variables. (Which this would also be a poor use case for. What's wrong with using an array? What's the purpose here?) — mario, May 28 '15 at 12:57
Curly braces can be used for array position, just like square brackets. Darn you PHP! — Kevin Nagurski, May 28 '15 at 12:59
@Daan And you answered the last question: http://stackoverflow.com/a/30504089/3933332 :)?! — Rizier123, May 28 '15 at 13:00
@mario - No reason specifically..Just experimenting. This was showed me me and wanted to see how it responded and the use of it. — KDJ, May 28 '15 at 13:00

score 4 · Accepted Answer · answered May 28 '15 at 12:58

Simply because {} can also be used to access arrays as you can read from the manual:

Note: Both square brackets and curly braces can be used interchangeably for accessing array elements (e.g. $array[42] and $array{42} will both do the same thing in the example above).

Means the following 2 lines are the same:

$var{++$counter}
$var[++$counter]

What you want is variable variables, which would be this:

${"var" . ++$counter} = $results['row'];

Variable variables doesn't work as expected and is creating an array

1 Answers1