Dynamically Combine Arrays With Array_Intersect()

[cyberRobot]

I’m using array_intersect() to combine three arrays.

 

$matchingIDs = array_intersect($array1, $array2, $array3);

 

Everything works until one of the arrays is empty. Is there a way to dynamically feed arrays to the function? I could use several if statements and array_intersect() calls, but it seems like there should be a better way.

[kicken]

look into call_user_func_array

[Barand]

What do the arrays look like and what do you want to end up with?

[ManiacDan]

Read the manual for array_intersect:

 

array array_intersect ( array $array1 ' date=' array $array2 [, array $ ... '] )

array_intersect() returns an array containing all the values of array1 that are present in all the arguments. Note that keys are preserved.

It doesn't combine arrays, it's used to filter arrays. If you're combining arrays, you want array_merge.

 

What you say you're doing doesn't match the function you're using, yet you say it's working. Answering Barand's question will help us figure out what you really want.

[cyberRobot]

What do the arrays look like and what do you want to end up with?

 

Here's a quick code sample:

 

<?php
$array1 = array(1, 2, 3, 4, 5, 6, 7, ;
$array2 = array(2, 3, 5, 7);
$array3 = array(1, 2, 3, 7, 9);

$matchingIDs = array_intersect($array1, $array2, $array3);

print_r($matchingIDs);
?>

 

This gives me a list of IDs which appear in all three arrays:

 

Array
(
[1] => 2
[2] => 3
[6] => 7
)

 

If one of the arrays is empty, however, the result set is also empty.

 

<?php
$array1 = array(1, 2, 3, 4, 5, 6, 7, ;
$array2 = array();
$array3 = array(1, 2, 3, 7, 9);
?>

[Jessica]

Yes. That makes perfect sense. What is the problem?

[ManiacDan]

Right, you've just described a functioning array_intersect function. Step 2: What do you want instead of what you're doing? It was the part of Barand's question after "and" and before the question mark.

[cyberRobot]

Yes. That makes perfect sense. What is the problem?

 

If $array1, $array2, $array3 are empty, I want to ignore that array.

 

Note that the reason an array would be empty is because a user chose an option which results in that array containing every ID from the database. Instead of running a query to get everything, I want to ignore that array.

[Jessica]

So, write some logic that does that. count.

[ManiacDan]

better yet: empty()

Edited October 8, 2012 by ManiacDan

[cyberRobot]

Just to clarify, there isn't a cleaner way of doing this without a bunch of if statements? If so, I may be better off with running the query...

 

 

For reference, this is basically what I would be looking at:

 

<?php
//IF FIRST AND SECOND ARRAY CONTAIN IDS, FIND THE INTERSECT
if(!empty($array1) && !empty($array2)) {
    $matchingIDs = array_intersect($array1, $array2);

//ELSE...STORE MATCHES SO FAR
} else {
    if(!empty($array1)) { $matchingIDs = $array1; }
    elseif(!empty($array2)) { $matchingIDs = $array2; }
    else { $matchingIDs = array(); }
}

//IF THE MATCHING IDS SO FAR AND THE THRID ARRAY CONTAIN IDS, FIND THE INTERSECT
if(!empty($matchingIDs) && !empty($array3)) {
    $matchingIDs = array_intersect($matchingIDs, $array3);

//ELSE...UPDATE MATCHES
} else {
    if(!empty($array3)) { $matchingIDs = $array3; }
}
?>

[ManiacDan]

so you want the intersection of all 3 arrays, and any one or two of the three may be empty, and therefore you shouldn't use it?

 

$args = array();
if ( !empty($arr1) ) $args[] = &$arr1;
if ( !empty($arr2) ) $args[] = &$arr2;
if ( !empty($arr3) ) $args[] = &$arr3;

if ( count($args) == 1 ) {
 $new = $args[0];
} else {
 $new = call_user_func_array('array_intersect', $args);
}

[Barand]

or if the number of array is variable

 

<?php
$a = array (
 array(1,2,3,4,5,6,7,,
 array(2, 3, 5, 7),
 array()
 );

$res = array();

foreach ($a as $B) {
if (empty($res))
 $res = $B;
else if (!empty($B)) {
 $res = array_intersect($res, $B);
}
}

echo '<pre>'.print_r($res, 1).'</pre>';
?>

Edited October 8, 2012 by Barand

[cyberRobot]

Thanks for all the responses.

 

I forgot all about looking into kicken suggestion. Thanks for showing me an implementation ManiacDan. I haven't totally wrapped my mind around what's going on, but the code looks much cleaner than mine.

Edited October 8, 2012 by cyberRobot