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我想采取以下数组,并生成一个新的数组,将所有具有相同ID的相同的键分组,同时获得视图的SUM和抓取仅限最后的DATE。生成一个新的数组按键分组,并获得另一个键的总和
原始数组
Array (
[0] => Array (
[id] => 10
[views] => 276
[date] => 2012-01-30 10:55:00
[total] => N
)
[1] => Array (
[id] => 40
[views] => 287
[date] => 2012-01-27 10:00:29
[total] => Y
)
[2] => Array (
[id] => 40
[views] => 824
[date] => 2012-01-29 14:40:45
[total] => Y
)
[3] => Array (
[id] => 42
[views] => 723
[date] => 2012-01-28 20:15:58
[total] => N
)
[4] => Array (
[id] => 43
[views] => 428
[date] => 2012-01-28 17:14:31
[total] => N
)
[5] => Array (
[id] => 45
[views] => 174
[date] => 2012-01-20 18:01:11
[total] => N
)
)
新阵列
Array (
[0] => Array (
[id] => 10
[views] => 276
[date] => 2012-01-30 10:55:00
[total] => N
)
[1] => Array (
[id] => 40
[views] => 1111
[date] => 2012-01-29 14:40:45
[total] => Y
)
[2] => Array (
[id] => 42
[views] => 723
[date] => 2012-01-28 20:15:58
[total] => N
)
[3] => Array (
[id] => 43
[views] => 428
[date] => 2012-01-28 17:14:31
[total] => N
)
[4] => Array (
[id] => 45
[views] => 174
[date] => 2012-01-20 18:01:11
[total] => N
)
)
什么 “总” 场? – cambraca 2012-02-01 17:10:05
那是我在尝试自己做这件事时添加的内容。总值与Y意味着它是一个多重ID和总元素。之后我有点不满意。会挑战你发布的答案。 – Tim 2012-02-01 17:13:10
我没有测试它,但它至少会给你一个出发点。 (然后,它可以正常工作:)) – cambraca 2012-02-01 17:24:15