[英]mysql distinct column on multiple row conditions
我有一個mysql數據庫(item_preset)與下表:
+-----+-----------+---------+-------+
| id | preset_id | item_id | value |
+-----+-----------+---------+-------+
| 1 | 1 | 1 | 2 |
| 2 | 1 | 2 | 1 |
| 3 | 1 | 4 | 60 |
| 4 | 1 | 3 | 16 |
| 5 | 1 | 3 | 17 |
| 6 | 1 | 3 | 18 |
| 7 | 1 | 3 | 25 |
| 8 | 1 | 3 | 26 |
| 9 | 1 | 3 | 27 |
| 10 | 2 | 1 | 3 |
| 11 | 2 | 2 | 0 |
| 12 | 2 | 4 | 0 |
| 13 | 2 | 3 | 16 |
| 14 | 2 | 3 | 17 |
| 15 | 2 | 3 | 19 |
| 16 | 2 | 3 | 20 |
| 17 | 2 | 3 | 21 |
| 18 | 3 | 1 | 2 |
| 19 | 3 | 2 | 0 |
| 20 | 3 | 4 | 0 |
| 21 | 3 | 3 | 25 |
| 22 | 3 | 3 | 28 |
| 23 | 4 | 1 | 1 |
| 24 | 4 | 2 | 1 |
| 25 | 4 | 4 | 120 |
| 26 | 4 | 3 | 16 |
| 27 | 4 | 3 | 17 |
| 28 | 4 | 3 | 18 |
| 29 | 4 | 3 | 22 |
| 30 | 4 | 3 | 23 |
| 31 | 4 | 3 | 24 |
| 32 | 6 | 1 | 3 |
| 33 | 6 | 2 | 1 |
| 34 | 6 | 4 | 90 |
| 35 | 6 | 3 | 18 |
| 36 | 6 | 3 | 22 |
| 37 | 6 | 3 | 23 |
| 38 | 6 | 3 | 24 |
| 39 | 6 | 3 | 29 |
| 40 | 6 | 3 | 30 |
+-----+-----------+---------+-------+
我想要做的是根據多行條件獲取不同的preset_id。 例如,要獲取preset_id 1我需要所有條件為真(item_id = 1和value_id = 2),(item_id = 2和value = 1)等...
我嘗試過以下方法:從item_preset中選擇distinct preset_id,其中(item_id = 1和value = 2)和(item_id = 2,value = 1)和(item_id = 4,value = 60);
但得到一個空集。 如果我嘗試使用Or而不是,我會獲得符合任何條件的所有preset_ids。
有任何想法嗎?
謝謝
你可能想試試這個,
SELECT preset_id
FROM tableName
WHERE (item_id = 1 and value = 2) OR
(item_id = 2 and value = 1) OR
(item_id = 4 and value = 60)
GROUP BY preset_id
HAVING COUNT(*) = 3
更多括號,
select distinct(preset_id) from item_preset where ((item_id = 1 and value = 2) or (item_id = 2 and value = 1) or (item_id = 4 and value = 60));
和value是keyword
- 不要將值用作column name
你可以嘗試這樣的事情:
select distinct preset_id from item_preset
where preset_id in (select preset_id from item_preset where item_id = 1 and value = 2)
and preset_id in (select preset_id from item_preset where item_id = 2 and value = 1)
and preset_id in (select preset_id from item_preset where item_id = 4 and value = 60);
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