Would running array_diff_assoc() twice on an array give me a开发者_StackOverflow社区ll non-unique entries?
$array3 = array_diff_assoc($array1, $array2);
$array4 = array_diff_assoc($array1, $array3);
var_dump($array4);
Given:
- A the set of entries in
$array1
, and - B the set of entries in
$array2
,
B would be composed of:
- B', all the entries in B that are in A, and
- B'' all the entries in B that are not in A.
$array3
, diff_assoc_array($array1, $array2)
, would be the operation A \ B, which would reduces as follows:
- (A \ B') ∩ (A \ B'')
- (A ∩ ¬B') ∩ A
- A ∩ ¬B'.
$array4
, diff_assoc_array($array1, $array3)
, would be the operation A \ (A ∩ ¬B'), which reduces as follows:
- A ∩ ¬(A ∩ ¬B')
- A ∩ (¬A ∪ B')
- A ∩ B
Therefore yes, the final result would be the items common to both arrays.
Solved...
<?php
$array1 = array(0, 1, 2);
$array2 = array("00", "01", 2);
$array3 = array_diff_assoc($array1, $array2);
$array4 = array_diff_assoc($array1, $array3);
var_dump($array3);
echo "<br><br>";
var_dump($array4);
?>
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