In contrast to what others have posted, you cannot use the min()
/max()
functions for this problem as these functions do not understand the datastructure (array) which are passed in. These functions only work for scalar array elements.
BEGIN EDIT
The reason why the use of min()
and max()
seem to yield the correct answer is related to type-casting arrays to integers which is an undefined behaviour:
The behaviour of converting to integer
is undefined for other types. Do not
rely on any observed behaviour, as it
can change without notice.
My statement above about the type-casting was wrong. Actually min()
and max()
do work with arrays but not in the way the OP needs them to work. When using min()
and max()
with multiple arrays or an array of arrays elements are compared element by element from left to right:
$val = min(array(2, 4, 8), array(2, 5, 1)); // array(2, 4, 8)
/*
* first element compared to first element: 2 == 2
* second element compared to second element: 4 < 5
* first array is considered the min and is returned
*/
Translated into the OP's problem this shows the reason why the direct use of min()
and max()
seems to yield the correct result. The arrays' first elements are the id
-values, therefore min()
and max()
will compare them first, incidentally resulting in the correct result because the lowest id
is the one with the lowest count
and the highest id
is the one with the highest count
.
END EDIT
The correct way would be to use a loop.
$a = array(
array('id' => 117, 'name' => 'Networking', 'count' => 16),
array('id' => 188, 'name' => 'FTP', 'count' => 23),
array('id' => 189, 'name' => 'Internet', 'count' => 48)
);
$min = PHP_INT_MAX;
$max = 0;
foreach ($a as $i) {
$min = min($min, $i['count']);
$max = max($max, $i['count']);
}