森栏
上面接受的答案对我不起作用,因为它按字母顺序而不是按时间顺序对星期进行排序:2012/12012/102012/11...2012/192012/2这是我按周进行计数和分组的解决方案:SELECT CONCAT(YEAR(date), '/', WEEK(date)) AS week_name, YEAR(date), WEEK(date), COUNT(*)FROM column_nameGROUP BY week_nameORDER BY YEAR(DATE) ASC, WEEK(date) ASC产生:YEAR/WEEK YEAR WEEK COUNT2011/51 2011 51 152011/52 2011 52 142012/1 2012 1 202012/2 2012 2 142012/3 2012 3 192012/4 2012 4 19