数组合并问题

[session1983]

$a = Array
(
    '0' => Array
        (
            'user_id' => '40',
            'user_name' => '山山',
            'workno' => '1004'
       
        ),
    '1' => Array
        (
            'user_id' => '47',
            'user_name' => '好好',
            'workno' => '1014'
       
        ),
	 '2' => Array
	(
		   'user_id' => '20',
		   'user_name' => '战三',
		   'workno' => '1014'
   
	)


   
);


$b=Array
(
    '0' => Array
        (
            'user_id' => '40',
			'newbuy' => '280'
       
        ),
    '1' => Array
        (
            'user_id' => '47',
            'newbuy' => '56'
        )
 
   
);

我想实现的效果是：



$c = Array
(
'0' => Array
(
'user_id' => '40',
'user_name' => '山山',
'workno' => '1004'
'newbuy' => '280'

),
'1' => Array
(
'user_id' => '47',
'user_name' => '好好',
'workno' => '1014'
'newbuy' => '56'

),
'2' => Array
(
'user_id' => '20',
'user_name' => '战三',
'workno' => '1014'
'newbuy' => '0'

)



);

[ohmygirl]

$a和$b是一一对应的么，还是有什么对应关系？
如果是一一对应的：

<?php
  foreach($b as $k=>$v)
    foreach($v as $s=>$t)
    {
     if(isset($a[$k]))
       {
        if(!isset($a[$k][$s]))
          $a[$k][$s] = $t;

      }

    }

print_r($a);

[session1983]

不是一一对应的啊。你注意看。

[一起混吧]

 for($i=0;$i<count($a);$i++){
     if($b[$i]){
          	   $ar[$i]= array_merge($a[$i],$b[$i]);
	 }else {
	           $ar[$i]=$a[$i];
			   $ar[$i]['newbuy']='0';
	 }   

 }
 print_r($ar);

[liuqing124tangke]

试一下这样做吧，我测试了，通过了
$aa=array();
foreach($a as $key=>$value){
if (!isset($b[$key]))
$aa[$key]=array_merge($a[$key]);
else
$aa[$key]=array_merge($a[$key],$b[$key]);

}
print_r($aa);

[「已注销」]

感觉还是没有好的代码 再帮顶中

[foolbirdflyfirst]

$a = Array
(
    '0' => Array
        (   
            'user_id' => '40',
            'user_name' => '山山',
            'workno' => '1004'
           
        ),  
    '1' => Array
        (   
            'user_id' => '47',
            'user_name' => '好好',
            'workno' => '1014'
         
        ),  
     '2' => Array
    (   
           'user_id' => '20',
           'user_name' => '战三',
           'workno' => '1014'
   
    )   


   
);


$b=Array
(
    '9' => Array
        (   
            'user_id' => '40',
            'newbuy' => '280'
           
        ),  
    '10' => Array
        (   
            'user_id' => '47',
            'newbuy' => '56'
        )   
 
   
);
foreach($b as $v){$arr[$v['user_id']] = $v;}
foreach($a as $k=>$v)
{
        if($arr[$v['user_id']]) $a[$k] = array_merge($v,$arr[$v['user_id']]); 
        else $a[$k] = array_merge($v,array('newbuy'=>'0'));
}
echo "<pre/>";
print_r($a);

[「已注销」]

我这个代码怎么样 呵呵

<?php

$a = Array
    (
    '0' => Array
        (
        'user_id' => '40',
        'user_name' => '山山',
        'workno' => '1004'
    ),
    '1' => Array
        (
        'user_id' => '47',
        'user_name' => '好好',
        'workno' => '1014'
    ),
    '2' => Array
        (
        'user_id' => '20',
        'user_name' => '战三',
        'workno' => '1014'
    )
);


$b = Array
    (
    '0' => Array
        (
        'user_id' => '40',
        'newbuy' => '280'
    ),
    '1' => Array
        (
        'user_id' => '47',
        'newbuy' => '56'
    )
);

$k1 = array_keys($a['0']);
$k2 = array_keys($b['0']);
$k = array_merge($k1, $k2);

$tmp = array('0', '0', '0', '0', '0');
$c = array_combine($k, $tmp);

for ($i = 0; $i < count($a); $i++) {
    if (isset($b[$i])) {
        $ar[$i] = array_merge($c, $a[$i]);
        $ar[$i] = array_merge($a[$i], $b[$i]);
    } else {
        $ar[$i] = array_merge($c, $a[$i]);
    }
}
print_r($ar['2']);

?>

[session1983]

不好意思，你的测试没有达到我的效果啊。每个都要有newbuy的。你的最后一个就没有啊。楼上的对。[Quote=引用 4 楼 liuqing124tangke 的回复:]

试一下这样做吧，我测试了，通过了
$aa=array();
foreach($a as $key=>$value){
if (!isset($b[$key]))
$aa[$key]=array_merge($a[$key]);
else
$aa[$key]=array_merge($a[$key],$b[$key]);

}
print_r($aa);
……
[/Quote]

[session1983]

感谢各位。假如是任意多个数组又如何处理呢？

[session1983]

假如我现在又加入一个数组呢？

$c=Array
(
    '0' => Array
        (
            'user_id' => '40',
			'oldbuy' => '280'
       
        ),
    '1' => Array
        (
            'user_id' => '47',
            'oldbuy' => '56'
        ),
	'2' => Array
        (
            'user_id' => '46',
            'oldbuy' => '536'
        )
 
   
);

[net_yuhai]

for($i=0;$i<count($a);$i++){
	if(@$b[$i]){
		$a[$i] =  array_merge($a[$i],$b[$i]);
	}else{
		$a[$i]['newbuy'] = '0';
	}
}
print_r($a);

[session1983]

最后一个应该： '2' => Array
(
'user_id' => '20',
'oldbuy' => '536'
)

[net_yuhai]

for($i=0;$i<count($a);$i++){
	if(@$b[$i]){
		$a[$i] =  array_merge($a[$i],$b[$i]);
	}else{
		$a[$i]['newbuy'] = '0';
	}
	if(@$c[$i]){
		$a[$i] =  array_merge($a[$i],$c[$i]);
	}else{
		$a[$i]['oldbuy'] = '0';
	}
}
print_r($a);

[net_yuhai]

我搞错了

[foolbirdflyfirst]

问问题咋不一次说完呢。。

function multiMerge(){
        $args = func_get_args();
        foreach($args as $k=>$v)
        {   
                foreach($v as $value)
                {   
                        if($t[$value['user_id']]) 
                        {   
                                $t[$value['user_id']] = array_merge($t[$value['user_id']],$value);
                        }   
                        else $t[$value['user_id']] = $value;
                }   
        }   
        //至于key补全，你最好自己手动弄个key列表，根据这个列表，没有的自动补上，默认值为0.
        return array_values($t);             
}
echo "<pre/>";
print_r(multiMerge($a,$b,$c));

[An4Greenhand]

顶一下 数组合并 出啦array_merge()以外 还有别的吗 我就知道这一个函数

[liuqing124tangke]

昨天没有看清你的效果图，不好意思！再修改

[q652820916]

header('Content-type: text/html; charset=utf-8');

$a = Array (
'0' => Array ('user_id' => '40', 'user_name' => '山山', 'workno' => '1004' ),
'1' => Array ('user_id' => '47', 'user_name' => '好好', 'workno' => '1014' ),
'2' => Array ('user_id' => '20', 'user_name' => '战三', 'workno' => '1014' ) );

$b = Array (
'0' => Array ('user_id' => '40', 'newbuy' => '280' ) ,
'1' => Array ('user_id' => '47', 'newbuy' => '56' ) );

foreach($a as $k=>$v){
$a[$k]["newbuy"]=isin($b,$v["user_id"]);
}

function isin($arr,$num){
$text=0;
foreach($arr as $k=>$v){
if($v["user_id"]!=$num){
continue;
}
$text=$v["newbuy"];
}
return $text;
}

var_dump($a);