PHP: Sort set based on another table

 I have visited similar question here but not getting what i want in php. Suppose i have 2 arrays. All checking should be case insensitive. say Field0 is same as field0 or fiEld1 is same as Field1.

array1 is array(
"0"=>"field0",
"1"=>"field1",
"2"=>"field2",
"3"=>"field3",
);

array2 is array(
"0"=>"field3",
"1"=>"field2",
"2"=>"field0",
"3"=>"field1",
"4"=>"field6",
"5"=>"field5",
);

Now I want array2 to be sorted based on array1 like the following:

array2 is array(
"0"=>"field0",
"1"=>"field1",
"2"=>"field2",
"3"=>"field3",
"4"=>"field6",
"5"=>"field5",
);

look here array1 has 4 elements so array2 should be sorted based exactly 4 elements of array1 and rest 2 element (index 4,5 of array2) should as it is in array2 before.

---

Try array_intersect combined with array_diff:

<?php
$array1 = array(
    "0"=>"field0",
    "1"=>"field1",
    "2"=>"field2",
    "3"=>"field3",
);

$array2 = array(
    "0"=>"field3",
    "1"=>"field2",
    "2"=>"field0",
    "3"=>"field1",
    "4"=>"field6",
    "5"=>"field5",
);

$array3 = array_merge(
    array_intersect($array1, $array2),
    array_diff($array2, $array1)
);

var_dump($array3);

Update

For a case insensitive approach, use array_map to guarantee all entries in both arrays are lower-cased:

$array3 = array_map('strtolower', $array1);
$array4 = array_map('strtolower', $array2);

$array5 = array_merge(
    array_intersect($array3, $array4),
    array_diff($array4, $array3)
);

var_dump($array5);