How to access php variable using concatenation

Question

I want to access a variable that is either called $item1, $item2 or $item3.

I want to access this variable inside a for loop where $i is ++ every time. using $item.$i or something similar. However using that code means that I am trying to join the contents of two variables, and there is no variable called $item.

You can use the double $ symbol. This is expanded upon here: http://stackoverflow.com/questions/2715654/what-does-mean-in-php — Adam Thornton, May 16 '12 at 16:28
@FelixKling thanks, I'm looking into doing that now. You see $item1, 2 and 3 are already arrays, so it would need to be a multidimensional array — Claire, May 16 '12 at 16:31
Using multidimensional arrays is totally normal ;) Variable variables is really a bad and inflexible approach. — Felix Kling, May 16 '12 at 16:38
Just because you can, doesnt mean you should. try to avoid variable variables if possible for readability's sake. — Mike Graf, May 16 '12 at 16:38
@FelixKling thanks. Its good for me to learn, I don't know unless someone tells me! — Claire, May 16 '12 at 16:40

Sampson · Accepted Answer · 2012-05-16T16:47:56.113

Arrays: A Better Method

While PHP does permit you to build dynamic variable names from various other values, you probably shouldn't in this case. It seems to me that an array would be more appropriate for you:

$items = array( 0, 12, 34 );

You could then access each value individually:

echo $items[0]; // 0
echo $items[1]; // 12

Or loop over the entire set:

foreach ( $items as $number ) {
  echo $number; // 1st: 0, 2nd: 12, 3rd: 34
}

Merging Multiple Arrays

You indicated in the comments on the OP that $item1 through $item3 are already arrays. You could merge them all together into one array if you like with array_merge(), demonstrated below:

$item1 = array( 1, 2 );
$item2 = array( 3, 4 );
$item3 = array( 5, 6 );

$newAr = array_merge( $item1, $item2, $item3 );

print_r( $newAr );

Which outputs:

Array
(
  [0] => 1
  [1] => 2
  [2] => 3
  [3] => 4
  [4] => 5
  [5] => 6
)

If You Must: Dynamic Variable Names

For completeness, if you were to solve your problem by dynamically constructing variable names, you could do the following:

$item1 = 12;
$item2 = 23;
$item3 = 42;

for ( $i = 1; $i <= 3; $i++ ) {
  echo ${"item".$i} . PHP_EOL;
}

With array_merge be careful of key collisions (for string keys). see: php.net/array_merge — Mike Graf, May 16 '12 at 17:31

score 2 · Answer 2 · answered May 16 '12 at 16:31

2

build the variable name you want to access into another variable then use the variable variable syntax

<?php
   $item1 = 'a';
   $item2 = 'b';
   $item3 = 'c';

   for ($i = 1; $i<=3; $i++) {
       $varname = 'item' . $i;
       echo $$varname;
   }
?>

output:

abc

Note there are other ways to do this, see the manual.

answered May 16 '12 at 16:31

Mike Graf

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Finally something approaching a correct answer. – DaveRandom May 16 '12 at 16:33
Just noticed this answer after I posted mine. Eerie how we thought of the same variable values. – maiorano84 May 16 '12 at 16:48

Ibu · Answer 3 · 2012-05-16T16:36:06.063

0

for ($i =1;$i<4;$i++){
    $var = 'item'.$i;
    echo $$var;
}

Here you are using the the double $ to create a variable variable.

edited May 16 '12 at 16:36

answered May 16 '12 at 16:29

Ibu

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Too many `$`s (and I'm not talking about the double one ;-) ) – DaveRandom May 16 '12 at 16:30
@DaveRandom it's kinda hard not to use $ in php – Ibu May 16 '12 at 16:32
Run this code and you will find it does not work. Look at the line where you assign a value to `$var`. – DaveRandom May 16 '12 at 16:33
@DaveRandom ah thank you for noticing – Ibu May 16 '12 at 16:36

Samy Dindane · Answer 4 · 2012-05-16T16:35:18.433

0

Use ${'item'.$i}
If $i == 1, then you will access $item1.

But it's better to use arrays in your case.

edited May 16 '12 at 16:35

answered May 16 '12 at 16:30

Samy Dindane

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thanks, I'm looking into doing that now. You see $item1, 2 and 3 are already arrays, so it would need to be a multidimensional array – Claire May 16 '12 at 16:31

Brad · Answer 5 · 2012-05-16T16:40:31.983

0

Can you use an array instead of individual variables? then you can reference array elements by index value based in i.

$items = array();

$i = 1;
$items[$i] = "foo";

$i++;
$items[$i] = "bah";

echo $items[1], $items[2];     // gives "foobah"

edited May 16 '12 at 16:40

answered May 16 '12 at 16:31

Brad

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You seem to be in Javascript mode, `i` think... – DaveRandom May 16 '12 at 16:33
D'oh! thanks Dave. Who's says polyglot is the way forward? – Brad May 16 '12 at 16:39

score 0 · Answer 6 · answered May 16 '12 at 16:46

It's a little late, and the accepted answer is the proper way to do this, but PHP does allow you to access variable variable names in the way OP describes:

<?php
$item1 = 'a';
$item2 = 'b';
$item3 = 'c';
for($i=1;$i<=3;$i++)
    echo ${"item$i"}; //Outputs: abc
?>

How to access php variable using concatenation

6 Answers6

Arrays: A Better Method

Merging Multiple Arrays

If You Must: Dynamic Variable Names