PHP变量名包含索引

Question

$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.

Answer 1

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.