php数组组合的问题-PHP中文网问答

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php数组组合的问题

PHP中文网 2017-04-11 10:30:27

[PHP讨论组]

664

下面的如何有效的使用php实现

数组1:


Array
(
    [0] => Array
        (
            [day] => 2017-03-28
            [reg_count] => 14
        )

    [1] => Array
        (
            [day] => 2017-03-27
            [reg_count] => 53
        )
        
)

数组2:

Array
(
    [0] => Array
        (
            [day] => 2017-03-29
            [pay_count] => 2
        )

    [1] => Array
        (
            [day] => 2017-03-28
            [pay_count] => 12
        )
}

根据day字段,组合成一个新数组

Array
(
    [0] => Array
        (
            [day] => 2017-03-29
            [pay_count] => 2
            [reg_count] => 0
        )

    [1] => Array
        (
            [day] => 2017-03-28
            [pay_count] => 12
            [reg_count] => 14
        )
        
    [2] => Array
        (
            [day] => 2017-03-27
            [pay_count] => 53
            [reg_count] => 0
        )
)

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全部回复(4)

迷茫2017-04-11 10:32:27 4楼

$a,$b分别是两个数组
$c =array_merge($a,$b);
$res = array();
foreach($c as $v){
    $res[$v['day']]['day'] =$v['day'];
    if(isset($v['reg_count'])){
        $res[$v['day']]['reg_count'] =$v['reg_count'];
    }
    if(isset($v['pay_count'])){
        $res[$v['day']]['pay_count'] =$v['pay_count'];
    }
}
$res = array_values($res);
var_dump($res);die;

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黄舟2017-04-11 10:32:27 3楼

$a = array(

    0=>array(
        "day"=>"2017-03-28",
        "reg_count"=>"14"
    ),
    1=>array(
        "day"=>"2017-03-27",
        "reg_count"=>"53"
    )
);

$b = array(
    0=>array(
        "day"=>"2017-03-29",
        "pay_count"=>"2"
    ),
    1=>array(
        "day"=>"2017-03-28",
        "pay_count"=>"12"
    )
);
$newarr = array();
foreach($a as $aval) {
    $newarr[$aval["day"]]["day"] = $aval["day"];
    $newarr[$aval["day"]]["reg_count"] = $aval["reg_count"];
    $newarr[$aval["day"]]["pay_count"] = 0;
}
foreach($b as $bval) {
    if(isset($newarr[$bval["day"]])) {
        $newarr[$bval["day"]]["pay_count"] = $bval["pay_count"];
    } else {
        $newarr[$bval["day"]]["day"] = $bval["day"];
        $newarr[$bval["day"]]["reg_count"] = 0;
        $newarr[$bval["day"]]["pay_count"] = $bval["pay_count"];
    }
}    
var_dump($newarr);

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迷茫2017-04-11 10:32:27 2楼

<?php
$arr1 = [ 
    ['day' => '2017-03-28', 'reg_count' => 14,], 
    ['day' => '2017-03-27', 'reg_count' => 53,]
];
$arr2 = [ 
    ['day' => '2017-03-29', 'pay_count' => 2,], 
    ['day' => '2017-03-28', 'pay_count' => 12,] 
];

$arr1 = array_column($arr1, 'reg_count', 'day');
$arr2 = array_column($arr2, 'pay_count', 'day');
$days = array_unique(array_merge(array_keys($arr1), array_keys($arr2)));

$result = [];
foreach($days as $day) {
    $result[] = [
        'day'       => $day,
        'reg_count' => isset($arr1[$day]) ? $arr1[$day] : 0,
        'pay_count' => isset($arr2[$day]) ? $arr2[$day] : 0,
    ];
}
rsort($result);

var_dump($result);

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ringa_lee2017-04-11 10:32:27 1楼

$arr1 = [['day'=>'2017-03-28','reg_count'=>14],['day'=>'2017-03-27','reg_count'=>53]];
$arr2 = [['day'=>'2017-03-29','pay_count'=>2],['day'=>'2017-03-28','pay_count'=>12]];

$arr1_1 = array_column($arr1,'reg_count','day');
$arr2_1 = array_column($arr2,'pay_count','day');
$return_arr = array();
foreach (array_unique(array_merge(array_keys($arr1_1),array_keys($arr2_1))) as $key =>$val){
    $return_arr[$key]['day'] = $val;
    $return_arr[$key]['reg_count'] = 0;
    $return_arr[$key]['pay_count'] = 0;
    if(key_exists($val, $arr1_1)){
        $return_arr[$key]['reg_count'] = $arr1_1[$val];
    }
    if(key_exists($val, $arr2_1)){
        $return_arr[$key]['pay_count'] = $arr2_1[$val];
    }
}
print_r($return_arr);

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