<?php
$arr1 = array(
'0' => 'zero',
'1' => 'one',
'2' => 'two',
'3' => 'three'
);
$arr2 = array(
'2' => 'two',
'three' => 'three',
'4' => 'four',
'5' => 'five'
);
//差集【對(duì)比返回在 arr1 中但是不在 arr2 及任何其它參數(shù)數(shù)組中的值?！?
$array_diff = array_diff($arr1,$arr2); 
echo "<pre>"; print_r($array_diff);
/* 輸出結(jié)果
*Array
*(
* [0] => zero
* [1] => one
*)
*/
//交集【對(duì)比返回既在 arr1 中也在 arr2 數(shù)組中的值?！?
# 注意事項(xiàng): 鍵值保留arr1數(shù)組中的鍵值不變
$array_intersect = array_intersect($arr1,$arr2); 
echo "<pre>"; print_r($array_intersect);
/* 輸出結(jié)果
*Array
*(
* [2] => two
* [3] => three
*)
*/
//并集【將arr1數(shù)組中的值附加在arr2數(shù)組的后面。返回作為結(jié)果的數(shù)組。 】
#注意事項(xiàng)：【如果輸入的數(shù)組中有相同的字符串鍵名，則該鍵名后面的值將覆蓋前一個(gè)值。
#然而，如果數(shù)組包含數(shù)字鍵名，后面的值將不會(huì)覆蓋原來的值，而是附加到后面?！?
$array_merge = array_merge($arr1,$arr2); 
echo "<pre>"; print_r($array_merge);
/* 輸出結(jié)果
* Array
* (
* [0] => zero
* [1] => one
* [2] => two
* [3] => three
* [4] => two
* [three] => three
* [5] => four
* [6] => five
* )
*/
?>