PHP 根据key 给二维数组分组

我们经常拿到一个二维数组出来,会发现结果和自己想要的有些偏差,可能需要根据二维数组里的某个字段对数组分组。

先来看以下数组,

Array
(
[0] => Array
(
[id] => 1
[wo_id] => 2
[evaluate_id] => 1
[type] => 分组1
[ctime] => 2016-12-02 11:39:34
)
[1] => Array
(
[id] => 2
[wo_id] => 31
[evaluate_id] => 2
[type] => 分组1
[ctime] => 2016-12-12 11:39:50
)
[2] => Array
(
[id] => 3
[wo_id] => 31
[evaluate_id] => 4
[type] => 分组2
[ctime] => 2016-12-02 11:40:01
)
[3] => Array
(
[id] => 4
[wo_id] => 2
[evaluate_id] => 5
[type] => 分组2
[ctime] => 2016-12-07 10:21:40
)
[4] => Array
(
[id] => 5
[wo_id] => 2
[evaluate_id] => 3
[type] => 分组2
[ctime] => 2016-12-07 10:21:51
)
[5] => Array
(
[id] => 6
[wo_id] => 2
[evaluate_id] => 2
[type] => 分组3
[ctime] => 2016-12-07 10:21:56
)
[6] => Array
(
[id] => 7
[wo_id] => 4
[evaluate_id] => 2
[type] => 分组3
[ctime] => 2016-12-07 10:22:10
)
)

这里是一个很典型的例子,如果我希望把 数组中的数据 根据 type(分组) 分开处理,如果不分组,可能循环上会非常冗余,所以有了下面的一个分组函数:

public static function array_group_by($arr, $key)
{
$grouped = [];
foreach ($arr as $value) {
$grouped[$value[$key]][] = $value;
}
// Recursively build a nested grouping if more parameters are supplied
// Each grouped array value is grouped according to the next sequential key
if (func_num_args() > 2) {
$args = func_get_args();
foreach ($grouped as $key => $value) {
$parms = array_merge([$value], array_slice($args, 2, func_num_args()));
$grouped[$key] = call_user_func_array('array_group_by', $parms);
}
}
return $grouped;
}

$arr : 二维数组

$key: 需要分组的key

得到的结果是:

Array
(
[分组2] => Array
(
[0] => Array
(
[id] => 1
[wo_id] => 2
[evaluate_id] => 1
[ctime] => 2016-12-02 11:39:34
[type] => 分组2
)
[1] => Array
(
[id] => 2
[wo_id] => 31
[evaluate_id] => 2
[ctime] => 2016-12-12 11:39:50
[type] => 分组2
)
[2] => Array
(
[id] => 3
[wo_id] => 31
[evaluate_id] => 4
[ctime] => 2016-12-02 11:40:01
[type] => 分组2
)
)
[分组1] => Array
(
[0] => Array
(
[id] => 4
[wo_id] => 2
[evaluate_id] => 5
[ctime] => 2016-12-07 10:21:40
[type] => 分组1
)
[1] => Array
(
[id] => 5
[wo_id] => 2
[evaluate_id] => 3
[ctime] => 2016-12-07 10:21:51
[type] => 分组1
)
)
)

郑重声明:本文版权包含图片归原作者所有,转载文章仅为传播更多信息之目的,如作者信息标记有误,请第一时间联系我们(delete@yzlfxy.com)修改或删除,多谢。

郑重声明:本文版权归原作者所有,转载文章仅为传播更多信息之目的,如作者信息标记有误,请第一时间联系我们修改或删除,多谢。

留言与评论(共有 0 条评论)
昵称:
匿名发表
   
验证码: