我的查询:MySQL查询:更改表到另一个表现
select Fimp_date, Fadpos, sum(Fplay_num) as Fplay_num from tbl_cnt_display_num_by_adpos where Fimp_date between date_format(now(),'%Y%m%d')-interval 2 day and date_format(now(),'%Y%m%d') group by Fadpos, Fimp_date;
,这是结果: + ----------- + -------- + ----------- +
| Fimp_date | Fadpos | Fplay_num |
+ ----------- + -------- + ----------- +
| 20151020 | 0 | 1562156 |
| 20151020 | 1 | 19354 |
| 20151020 | 101 | 352796 |
| 20151020 | 102 | 232052 |
| 20151020 | 103 | 42324 |
| 20151020 | 104 | 214 |
| 20151020 | 105 | 167904 |
| 20151020 | 500 | 49 |
+ ----------- + -------- + ----------- +
这意味着每天作为一个行,不同的Fadpos作为不同的列,值是当天Fadpos的Fplay_num。
那么如何编写sql?
可能重复[mysql pivot/crosstab query](http://stackoverflow.com/questions/12382771/mysql-pivot-crosstab-query) –