小问题,
我有桌子：
我有桌子：
id | user_id | paper_update ------------------------------ 1 | 1 | 30-5-2011 2 | 2 | 30-5-2012 3 | 3 | 30-5-2012 4 | 1 | 30-5-2013 5 | 2 | 30-5-2013 6 | 3 | 30-5-2014 7 | 4 | 30-5-2014 8 | 5 | 30-5-2014 9 | 5 | 30-5-2015 10 | 5 | 30-5-2016 11 | 1 | 30-5-2010 -------------------------------
我想要做的是只选择具有相同user_id的记录之间paper_update最大的记录,实际上我想按user_id分组以达到最大paper_update.
所以对于这个表,它将返回：
id | user_id | paper_update ------------------------------ 4 | 1 | 30-5-2013 5 | 2 | 30-5-2013 6 | 3 | 30-5-2014 7 | 4 | 30-5-2014 10 | 5 | 30-5-2016 -------------------------------
解决方法
这是在laravel中获得groupBy Max Date的最简单,最短的方法.使用user_id获取具有最大日期的行.
$rows = DB::table('papers')
->select(DB::raw('id,max(paper_update) as year,user_id'))
->groupBy('user_id')
//->orderBy('paper_update','desc')
->get();
select id,max(paper_update) as year from `papers` group by `user_id`;
输出将是
Array
(
[0] => stdClass Object
(
[id] => 4
[year] => 30-5-2013
[user_id] => 1
)
[1] => stdClass Object
(
[id] => 5
[year] => 30-5-2013
[user_id] => 2
)
[2] => stdClass Object
(
[id] => 6
[year] => 30-5-2014
[user_id] => 3
)
[3] => stdClass Object
(
[id] => 7
[year] => 30-5-2014
[user_id] => 4
)
[4] => stdClass Object
(
[id] => 10
[year] => 30-5-2016
[user_id] => 5
)
)
