Question

Sorry for the confusing title…

I need to perform an array_intersect() against a variable number of arrays. To do this it seems I need to use the call_user_func_array() function, however, this doesn’t seem to be working and gives me the error:

Warning: array_intersect() [function.array-intersect]: Argument #1 is not an array in...

But, if I “print_r” the array to make sure then I see that it is an array:

Array ( [0] => arr_0 [1] => arr_1 ) 

My code (trimmed to just show the broken part):

$i = 0;
$arr_results = array();
foreach($arr_words as $word) {
    $arrayname = "arr_".$i;
    $$arrayname = array();
    while ($row = mysql_fetch_assoc($search)) {
        array_push($$arrayname, $row['id']);
    }
    array_push($arr_results, "$arrayname");
    $i++
}
$matches = call_user_func_array('array_intersect',$arr_results);

In the full code I’m populating the arrays in the foreach loop with data obtained from sql queries.

Answer 1

From my comments:

"$arrayname" is a string, not an array. call_user_func_array will pass each element in $arr_results as argument to array_intersect. array_intersect expects arrays as arguments, but each item in $arr_results is a string, not an array.

All you have to do is create an array of arrays instead of array names:

$arr_results = array();
foreach($arr_words as $word) {
    $ids = array();
    while ($row = mysql_fetch_assoc($search)) {
        $ids[] = $row['id'];
    }
    $arr_results[] = $ids;
}
$matches = call_user_func_array('array_intersect',$arr_results);

I turn $arrayname into an array with $$arrayname = array();

Right, you create a variable, lets say arr_0 which will point to array. But there is still a difference between the variable name arr_0 and the string containing the variable name "arr_0". You create an array of strings, and that just won’t work. PHP does not know that the string contains a name of a variable. For example, consider this:

$arr = "arr_0";
echo $arr[0];

Based on your logic, it should output the first element of the array, but it does not, because $arr is a string, not an array, although it contains the name of a variable.

You’d have to use eval, but you really should not. You could also use variable variables again:

array_push($arr_results, $$arrayname);

that would work as well, but as I said, variable variables are confusing and in 99% of the cases, you are better of with an array.