在MySQL中選擇所有具有不同和條件的行

Question

嗨，大家好我是mySQL中的新成員，我在查詢時遇到問題。我試圖寫一些查詢，讓我從表中的所有記錄，如果我有兩個記錄具有相同的日期，我需要只有這兩個之間manual_selection = 1記錄。

所以結果應該是我的所有記錄除了ID表= 1401和id = 1549

my table

我想結合我怎麼能得到這個記錄是這樣的：

SELECT * FROM project.score WHERE project_id = 358 
AND crawled_at IN(SELECT crawled_at FROM project.score WHERE project_id = 358 
AND manual_selection = 1 GROUP BY crawled_at) 
ORDER BY crawled_at;  

SELECT * FROM project.score WHERE project_id = 358 
GROUP BY crawled_at HAVING manual_selection = 1; 

但我所有的方式總是隻與manual_selection = 1的行我有ent idea我怎麼能在manual_selection = 1的情況下使用重複的「crawled_at」來區分行。有人可以幫我嗎？

Answer 1

試試這個：

select main.id, main.project_id, main.crawled_at, main.score, main.manual_selection 
from dcdashboard.moz_optimization_keywords as main 
left join dcdashboard.moz_optimization_keywords as non_manual_selection on non_manual_selection.crawled_at = main.crawled_at and non_manual_selection.manual_selection != 1 
group by main.crawled_at; 

結果的數據與來自問題集：

+------+------------+---------------------+-------+------------------+ 
| id | project_id | crawled_at   | score | manual_selection | 
+------+------------+---------------------+-------+------------------+ 
| 807 |  360 | 2016-02-06 00:00:00 | 76 |    0 | 
| 1001 |  360 | 2016-02-20 00:00:00 | 76 |    0 | 
| 223 |  360 | 2016-11-28 00:00:00 | 76 |    0 | 
| 224 |  360 | 2016-12-05 00:00:00 | 76 |    0 | 
| 670 |  360 | 2016-12-19 00:00:00 | 76 |    0 | 
| 1164 |  360 | 2017-04-19 00:00:00 | 78 |    1 | 
| 1400 |  360 | 2017-09-13 00:00:00 | 96 |    1 | 
| 1548 |  360 | 2017-09-15 00:00:00 | 96 |    1 | 
+------+------------+---------------------+-------+------------------+ 
8 rows in set (0.00 sec)