如果在php

Question

如果第一个字段liked_cnt , coupon_cnt相同，我想要将两个数组字段id相加 我怎么能这样做？

我这样做但是太慢了

if (end($like)['id'] > end($coupon)['id']) {
          $i = end($like)['id'];
          $j = end($coupon)['id'];
        }else {
          $i = end($coupon)['id'];
          $j = end($like)['id'];
        }

        for ($k=0; $k <= $i; $k++) {
          for ($l=0; $l  < $j; $l++) {
            if ($like['id'] == $coupon['id']) {
              $score[$like['id']] = ($coupon['coupon_cnt'] * 1000) + $like['liked_cnt'];
            }else {
              $score[$i] = 0;
            }
          }
        }



//first array $Like

Array ( [0] => Array ( [id] => 85 [liked_cnt] => 6 ) [1] => Array ( [id] => 86 [liked_cnt] => 14 ) [2] => Array ( [id] => 92 [liked_cnt] => 6 ) [3] => Array ( [id] => 93 [liked_cnt] => 6 ) 

//second array $coupon
Array ( [0] => Array ( [id] => 35 [coupon_cnt] => 2 ) [1] => Array ( [id] => 86 [coupon_cnt] => 1 ) [2] => Array ( [id] => 139 [coupon_cnt] => 1 ) )

Answer 1

没有内置于php的魔术算法能够以更有效的方式真正计算出您想要的结果。这可能稍微有点效率，但最重要的是它更具可读性：

content-length →0
content-type →application/json
status →200
via →1.1 swfbfbbaf3fb6c32bdccb152354539e473d.cloudfront.net (CloudFront)
x-amz-cf-id →K9V3XUxHOretrza0kCM5dk_G5eZgePrtrBziyVTxptrePD7wjsWqk-l0kCQQ==
x-amzn-requestid →5ac81024-5c27-11e7-af9a-9f3c8494c542
x-amzn-trace-id →Root=1-5953e77f-ed76d15b5bfre9374c9

输出显然是：

<?php
$input = [
    [
        ['id' => 85, 'liked_cnt' => 6],
        ['id' => 86, 'liked_cnt' => 14],
        ['id' => 92, 'liked_cnt' => 6],
        ['id' => 93, 'liked_cnt' => 6]
    ],
    [
        ['id' => 35, 'coupon_cnt' => 2],
        ['id' => 86, 'coupon_cnt' => 1],
        ['id' => 139, 'coupon_cnt' => 1]
    ]
];

$output = [];
foreach ($input as $set) {
    array_walk($set, function($entry) use (&$output) {
        $count = array_pop($entry);
        $id = array_pop($entry);
        if (array_key_exists($id, $output)) {
            $output[$id]['total_cnt'] += $count;
        } else {
            $output[$id] = ['id' => $id, 'total_cnt' => $count];
        }
    });
}
print_r(array_values($output));

更新：

考虑到下面的第三条评论，我添加了这个修改后的版本，将Array ( [0] => Array ( [id] => 85 [total_cnt] => 6 ) [1] => Array ( [id] => 86 [total_cnt] => 15 ) [2] => Array ( [id] => 92 [total_cnt] => 6 ) [3] => Array ( [id] => 93 [total_cnt] => 6 ) [4] => Array ( [id] => 35 [total_cnt] => 2 ) [5] => Array ( [id] => 139 [total_cnt] => 1 ) ) 属性的一般乘法引入因子1000：

coupon_cnt