求生成新数组好的办法

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RT。如何让2个结构完全一样的数组合并成一个新的数组。用array_merge后面的数组会覆盖前面的。
数组结构为

[疯狂的狙击手]

数组结构为.....so 你是想让我们猜是吧

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[xuzuning]

我猜你是这个意思

$a = array('a' => 1);
$b = array('a' => 1);

print_r(array_merge_recursive($a, $b));

Array ( [a] => Array ( [0] => 1 [1] => 1 ) )

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[「已注销」]

唠叨哥的方法我用过。我数组有7列，结果被分为了7列。 A数组

array(
[0] =>
array(

['id'] =>
365199916911903
['request_type'] =>
7
['receive_uid'] =>
100004520460114
['type'] =>
2
['obj_id'] =>
50013
['send_uid'] =>
100004155300493
['send_time'] =>
'2013/01/31'
)
[1] =>
array(

['id'] =>
599130963436697
['request_type'] =>
7
['receive_uid'] =>
100004268022279
['type'] =>
2
['obj_id'] =>
50013
['send_uid'] =>
100004155300493
['send_time'] =>
'2013/01/31'
)
)

B数组

array(
['id'] =>
409399785810773
['request_type'] =>
7
['receive_uid'] =>
100004251090199
['type'] =>
2
['obj_id'] =>
50013
['send_uid'] =>
100004155300493
['send_time'] =>
'2013/01/31'
)

合并后的C数组

array(
[0] =>
array(

['id'] =>
365199916911903
['request_type'] =>
7
['receive_uid'] =>
100004520460114
['type'] =>
2
['obj_id'] =>
50013
['send_uid'] =>
100004155300493
['send_time'] =>
'2013/01/31'
)
[1] =>
array(

['id'] =>
409399785810773
['request_type'] =>
7
['receive_uid'] =>
100004251090199
['type'] =>
2
['obj_id'] =>
50013
['send_uid'] =>
100004155300493
['send_time'] =>
'2013/01/31'
)
)

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[「已注销」]

是还试过 $list[] = array($a,$b); 结果本来平级的数组分为了 array( 0=>array(0=>array,1=>array) 1=>array() )

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[疯狂的狙击手]

$a[] = $b;
$c = $a;
print_r($c);

即是你要的C数组

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[疯狂的狙击手]

$a = array(
	0=>array( 
		'id' =>365199916911903,
		'request_type'=>7,
		'receive_uid' =>100004520460114,
		'type' =>2,
		'obj_id' =>50013,
		'send_uid' =>100004155300493,
		'send_time' =>'2013/01/31'
	),
	1=>array(
		'id' =>599130963436697,
		'request_type' =>7,
		'receive_uid' =>100004268022279,
		'type' =>2,
		'obj_id' =>50013,
		'send_uid' =>100004155300493,
		'send_time' =>'2013/01/31'
	)
);
$b=array(
	'id' =>409399785810773,
	'request_type' =>7,
	'receive_uid' =>100004251090199,
	'type' =>2,
	'obj_id' =>50013,
	'send_uid' =>100004155300493,
	'send_time' =>'2013/01/31'
);
$a[] = $b;
echo "<pre>";
print_r($a);

Array ( [0] => Array ( [id] => 3.651999169119E+14 [request_type] => 7 [receive_uid] => 1.0000452046011E+14 [type] => 2 [obj_id] => 50013 [send_uid] => 1.0000415530049E+14 [send_time] => 2013/01/31 ) [1] => Array ( [id] => 5.991309634367E+14 [request_type] => 7 [receive_uid] => 1.0000426802228E+14 [type] => 2 [obj_id] => 50013 [send_uid] => 1.0000415530049E+14 [send_time] => 2013/01/31 ) [2] => Array ( [id] => 4.0939978581077E+14 [request_type] => 7 [receive_uid] => 1.000042510902E+14 [type] => 2 [obj_id] => 50013 [send_uid] => 1.0000415530049E+14 [send_time] => 2013/01/31 ) )

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[「已注销」]

说明下，a数组是累加的值。一开始没数据的时候，只有a数组，然后传入数组，我当它是b数组，然后新数组等于a数组加上b数组，然后在传入值的时候。依次往a数组后面加，数组的键和结构完全一致

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[ImN1]

搞不清楚规则……为何只替换[1]而不是替换[0]？这个要说清楚才行

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[xuzuning]

你这个人真有趣 你总得把你原始的数据和期望的结果告诉我们吧？

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额。简单来说，我先$memcache->get(当天的key);if(不存在这个key)则把传入的任何数组都合并， 数组格式

array(
['id'] =>
256676594464467
['request_type'] =>
7
['receive_uid'] =>
100004236761398
['type'] =>
2
['obj_id'] =>
50014
['send_uid'] =>
100004155300493
['send_time'] =>
'2013/01/31'
) 

所以说一开始是没有数组的，然后数组按顺序合并起来，因为数组的键都相同，所以无法使用array_merge

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[疯狂的狙击手]

引用 7 楼 yao970953039 的回复:

说明下，a数组是累加的值。一开始没数据的时候，只有a数组，然后传入数组，我当它是b数组，然后新数组等于a数组加上b数组，然后在传入值的时候。依次往a数组后面加，数组的键和结构完全一致

引用 10 楼 yao970953039 的回复:

额。简单来说，我先$memcache->get(当天的key);if(不存在这个key)则把传入的任何数组都合并， 数组格式 PHP code?12345678910111213141516array(['id'] =>256676594464467['request_type'] =>7['receive_uid'] =>100004236761398['type'……

搞了半天那你直接把单个的数组赋给一个空的二维数组不就得了 去

$b = array('a'=>'1'); 
$a[] = $b;

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[xuzuning]

合并成什么样子？ 新数组和老数组有什么关系？

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[「已注销」]

用array_merge_recursive。。然后
$arr = array();

        $count =  count($data['id']);
        for($i=0;$i<$count;$i++){
            $arr[] = array(
                'id'=>$data['id'][$i],
                'request_type'=>$data['request_type'][$i],
                'receive_uid'=>$data['receive_uid'][$i],
                'type'=>$data['type'][$i],
                'obj_id'=>$data['obj_id'][$i],
                'send_uid'=>$data['send_uid'][$i],
                'send_time'=>$data['send_time'][$i],
            );
        }

哎。解决~

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[疯狂的狙击手]

引用 13 楼 yao970953039 的回复:

PHP code?123456789101112131415用array_merge_recursive。。然后$arr = array(); $count = count($data['id']); for($i=0;$i<$count;$i++){ $arr[] = array( 'i……

妹纸 你就像一朵奇葩在风中颤抖着......

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引用 14 楼 insanity_sniper 的回复:

引用 13 楼 yao970953039 的回复:PHP code?123456789101112131415用array_merge_recursive。。然后$arr = array(); $count = count($data['id']); for($i=0;$i<$count;$i++){ $arr[] =……

这样做不好嘛

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[疯狂的狙击手]

引用 15 楼 yao970953039 的回复:

引用 14 楼 insanity_sniper 的回复:引用 13 楼 yao970953039 的回复:PHP code?123456789101112131415用array_merge_recursive。。然后$arr = array(); $count = count($data['id']); for($i=0;$i<$coun……

可以 结贴吧

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