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up vote 0 down vote favorite 
$arr[0]=123;
$a="arr[0]";
echo $$a;

gives me error

Notice: Undefined variable: arr[0]

on the last line.

What should I do to make it work?

EDIT:

Above is the simplification of what I want to do. If someone wants to know why I want to do this, then here's the explanation:

This is something like what I want to do:

if(condition){
    $a=$arr1[0][0];
    $b=$arr1[0][1];
    $c=$arr1[0][2];
}
else{
    $a=$arr2[0];
    $b=$arr2[1];
    $c=$arr2[2];
}

I can compact it like this:

if(condition)
    $arr=$arr1[0];
else
    $arr=$arr2;
$a=$arr[0];
$a=$arr[1];
$a=$arr[2];

But I wanted to try doing this using variable variable:

if(condition)
    $arr="$arr1[0]";
else
    $arr="$arr2";
$a={$$arr}[0];
$b={$$arr}[1];
$c={$$arr}[2];

Sure, we don't need variable variables as we can still code without them. I want to know, for learning PHP, why the code won't work.

share|improve this question
1  
You don't need variable variables – Your Common Sense 2 days ago
I need them to compact the code. – user1444680 2 days ago
1  
compact the code ?? you must be joking right ? – Baba 2 days ago
5  
If you want to compact it, use echo 123;. – Gumbo 2 days ago
1  
I guess you mean that it should be $a = $arr[0] ? and then echo $a ? – bestprogrammerintheworld 2 days ago
show 4 more comments

5 Answers

up vote 1 down vote accepted

Now that you said what you’re actually trying to accomplish: Your code doesn’t work because if you look at $arr1[0][0], only arr is the variable name; the [0] are special accessors for certain types like strings or arrays.

With variable variables you can only specify the name but not any accessor or other operation:

A variable variable takes the value of a variable and treats that as the name of a variable.

Your solution with the additional variable holding the array to access later on would be the best solution to your problem.

share|improve this answer
Now that you said what you’re actually trying to accomplish my initial code was enough to convey the problem. – user1444680 2 days ago
up vote 1 down vote

What you are trying to do just won't work - the code $arr[0] is referencing a variable called $arr, and then applying the array-access operator ([$key]) to get the element with key 0. There is no variable called $arr[0], so you cannot reference it with variable-variables any more than you could the expression $foo + 1 .

The real question is why you want to do this; variable variables are generally a sign of very messy code, and probably some poor choices of data structure. For instance, if you need to select one of a set of variables based on some input, you probably want a hash, and to look up an item using $hash[$item] or similar. If you need something more complex, a switch statement can often cover the cases you actually need.

If for some reason you really need to allow an arbitrary expression like $arr[0] as input and evaluate it at runtime, you could use eval(), but be very very careful of where the input is coming from, as this can be a very easy way of introducing security holes into your code.

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up vote 0 down vote

As pointed out you don't need variable variables. To get a PHP variable variable name containing index (a key) use array_keys() or array_search() or other array parsers. From php's site:

$array = array(0 => 'blue', 1 => 'red', 2 => 'green', 3 => 'red');
$key = array_search('green', $array); // $key = 2;
$key = array_search('red', $array);   // $key = 1;

You could also use the following (using $var= instead of echo):

$arr[0]=123;
$arr[1]=456;
foreach ($arr as $key => $value) {
    echo "arr[{$key}] = {$value} \r\n";
}

Which outputs:

arr[0] = 123
arr[1] = 456

But I don't see why you'd do that, since the whole point of the array is not doing that kind of stuff.

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up vote 0 down vote

FROM PHP DOC

In order to use variable variables with arrays, you have to resolve an ambiguity problem. That is, if you write $$a[1] then the parser needs to know if you meant to use $a[1] as a variable, or if you wanted $$a as the variable and then the [1] index from that variable. The syntax for resolving this ambiguity is: ${$a[1]} for the first case and ${$a}[1] for the second.

Use

echo ${$a}[0]; // 123

Edit : Based on your edit you can simply have 

list($a, $b, $c) = (condition) ? $arr1[0] : $arr2;

Or 

$array =  (condition) ? $arr1[0] : $arr2;
$a = $array[0];
$b = $array[1];
$c = $array[2];
share|improve this answer
up vote -1 down vote 
$arr[0]=123;
$a="arr";
$i="0";
echo ${$a}{$i};
share|improve this answer

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