php数组问题

问题描述

问 题

$a = array(
  0 => array(
    'type' => 1,
    'business' => 106,
    'sum' => 9
    ),
  1 => array(
    'type' => 2,
    'business' => 104,
    'sum' => 881
    ),
  2 => array(
    'type' => 2,
    'business' => 105,
    'sum' => 280
    )
);
$b = array(
  0 => array(
    'type' => 2,
    'business' => 0,
    'sum' => 600
    ),
  1 => array(
    'type' => (int) 2,
    'business' => (int) 102,
    'sum' => (float) 20080
    ),
  2 => array(
    'type' => 2,
    'business' => 104,
    'sum' => 585
    ),
  3 => array(
    'type' => 2,
    'business' => 105,
    'sum' => 2300
    ),
  4 => array(
    'type' => 2,
    'business' => 107,
    'sum' => 1584
    ),
  5 => array(
    'type' => 2,
    'business' => 108,
    'sum' => 60
    )
);

---

两个数组，type和business的取值不固定，需要将type和business一样的sum相加，得到一个新的数组，不一样的加到新数组后面，新数组的格式类型是

array(
    (int) 0 => array(
        'business' => 106,
        'type1' => (float) 9,
        'type2' => (int) 0
    ),
    (int) 1 => array(
        'business' => 104,
        'type1' => (int) 0,
        'type2' => (float) 1466
    ),
    (int) 2 => array(
        'business' => 105,
        'type1' => (int) 0,
        'type2' => (float) 2580
    ),
    (int) 3 => array(
        'business' => 0,
        'type1' => (int) 0,
        'type2' => (float) 600
    ),
    (int) 4 => array(
        'business' => 102,
        'type1' => (int) 0,
        'type2' => (float) 20080
    ),
    (int) 5 => array(
        'business' => 107,
        'type1' => (int) 0,
        'type2' => (float) 1584
    ),
    (int) 6 => array(
        'business' => 108,
        'type1' => (int) 0,
        'type2' => (float) 60
    )
)

有几个type新数组的business下面就有几个值

解决方案

$newarr = array();
foreach($a as $akey=>$aval) {
    foreach($b as $bkey=>$bval) {
        if($aval['type'] == $bval['type'] && $aval['business'] == $bval['business']) {
            $addarr = array();
            $addarr['type'] = $aval['type'];
            $addarr['business'] = $aval['business'];
            $addarr['sum'] = $aval['sum'] + $bval['sum'];
            $newarr[] = $addarr;
            unset($a[$akey],$b[$bkey]);
            break;
         }
    }
}
$result = array_merge($newarr,$a,$b);var_dump($result);

字段可能你要自己改改

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