Share a PHP function to find the intersection of 5 Arrays_PHP Tutorial

	/**
		數組碰撞,找出多個數組的重疊值
		返回:
			重整之後的Array,如下:
			$arr=Array(
				0=>'',	//5個數組重疊的部份
				1=>'',	//第1個數組去掉[0]的值后
				2=>'',	//第2個數組去掉[0]的值后
				3=>'',	//第3個數組去掉[0]的值后
				4=>'',	//第4個數組去掉[0]的值后
				5=>'',	//第5個數組去掉[0]的值后
			);
		參數:
			要重整的5個數組
	/**/
	Public Static Function JiaoJi($arr1,$arr2,$arr3,$arr4,$arr5) {
		IF(!Is_array($arr1) Or !Is_array($arr2) Or !Is_array($arr3) Or !Is_array($arr4) Or !Is_array($arr5)) {
			throw New Exception('Error:'.__LINE__.',參數錯誤!');Die();
		}

		$arr=Array();
		//找出這5個數組中重複的部份,賦值給$arr[0]
		{
			//合併數組
			$narr=Array_Merge($arr1,$arr2,$arr3,$arr4,$arr5);

			//降維
			Foreach($narr as $v){
				$v = Implode('`',$v);	//降维
				$temp[] = $v;
			}unSet($narr);

			//去重复
			$unique_arr = Array_Unique($temp);

			//取差集
			$repeat_arr = Array_Diff_Assoc($temp,$unique_arr);unSet($temp,$unique_arr);

			//去重複
			$repeat_arr=Array_Unique($repeat_arr);

			//升維
			Foreach($repeat_arr as $k=>$v){
				$v = Explode('`',$v);
				$v['gid']=$v[0];unSet($v[0]);
				$v['cod']=$v[1];unSet($v[1]);
				$v['tid1']=$v[2];unSet($v[2]);
				$v['tid2']=$v[3];unSet($v[3]);
				$v['stime']=$v[4];unSet($v[4]);

				$temp[] = $v;
			}
			$arr[0]=$temp;
			unSet($temp,$repeat_arr);
		}

		/**
		去掉5個數組中包含$arr[0]的部份
		array(5) {
		  ["gid"]=>string(1) "2"
		  ["cod"]=>string(4) "4002"
		  ["tid1"]=>string(3) "184"
		  ["tid2"]=>string(3) "199"
		  ["stime"]=>string(19) "2013-07-25 02:19:00"
		}
		/**/
		$max=Max(Count($arr1),Count($arr2),Count($arr3),Count($arr4),Count($arr5));
		Foreach($arr[0] as $k=>$v) {
			For($i=0;$i<$max;$i++) {
				//去掉arr1和$arr[0]的重複部份
				IF(isSet($arr1[$i])) {
					IF($arr1[$i]['gid']==$v['gid'] And $arr1[$i]['cod']==$v['cod'] And $arr1[$i]['tid1']==$v['tid1'] And $arr1[$i]['tid2']==$v['tid2'] And $arr1[$i]['stime']==$v['stime']) {
						unSet($arr1[$i]);
					}
				}

				//去掉arr2和$arr[0]的重複部份
				IF(isSet($arr2[$i])) {
					IF($arr2[$i]['gid']==$v['gid'] And $arr2[$i]['cod']==$v['cod'] And $arr2[$i]['tid1']==$v['tid1'] And $arr2[$i]['tid2']==$v['tid2'] And $arr2[$i]['stime']==$v['stime']) {
						unSet($arr2[$i]);
					}
				}

				//去掉arr3和$arr[0]的重複部份
				IF(isSet($arr3[$i])) {
					IF($arr3[$i]['gid']==$v['gid'] And $arr3[$i]['cod']==$v['cod'] And $arr3[$i]['tid1']==$v['tid1'] And $arr3[$i]['tid2']==$v['tid2'] And $arr3[$i]['stime']==$v['stime']) {
						unSet($arr3[$i]);
					}
				}

				//去掉arr4和$arr[0]的重複部份
				IF(isSet($arr4[$i])) {
					IF($arr4[$i]['gid']==$v['gid'] And $arr4[$i]['cod']==$v['cod'] And $arr4[$i]['tid1']==$v['tid1'] And $arr4[$i]['tid2']==$v['tid2'] And $arr4[$i]['stime']==$v['stime']) {
						unSet($arr4[$i]);
					}
				}

				//去掉arr5和$arr[0]的重複部份
				IF(isSet($arr5[$i])) {
					IF($arr5[$i]['gid']==$v['gid'] And $arr5[$i]['cod']==$v['cod'] And $arr5[$i]['tid1']==$v['tid1'] And $arr5[$i]['tid2']==$v['tid2'] And $arr5[$i]['stime']==$v['stime']) {
						unSet($arr5[$i]);
					}
				}
			}
		}unSet($max);

		$arr[1]=$arr1;unSet($arr1);
		$arr[2]=$arr2;unSet($arr2);
		$arr[3]=$arr3;unSet($arr3);
		$arr[4]=$arr4;unSet($arr4);
		$arr[5]=$arr5;unSet($arr5);

		Return $arr;
	}

http://www.bkjia.com/PHPjc/440321.htmlwww.bkjia.comtruehttp: //www.bkjia.com/PHPjc/440321.htmlTechArticle/**Array collision, find the overlapping values ​​of multiple arrays and return: Array after reorganization, as follows: $arr=Array(0='',//The overlapping part of the 5 arrays 1='',//The first array removes the value of [0] 2='',//The second array.. .