SELECT DISTINCT值-多个具有相同ID的行-多个条件

Question

这是我的出发桌：

User_ID | FIELD_KEY    | VALUE
11      |  name        |  John
11      |  test1_score |  9
11      |  test2_score |  6
11      |  test3_score |  8
11      |  test5_score |  3
27      |  name        |  Peter
27      |  test1_score |  7
27      |  test3_score |  3
28      |  name        |  Nick
28      |  test1_score |  6
28      |  test2_score |  5
33      |  name        |  Felix
33      |  test1_score |  7
33      |  test2_score |  3

如何选择以下项的唯一User_ID

条件：

• 同时具有两个条目的所有用户：test1_score和test2_score
• test1_score> = 7的所有用户
• 按test1_score ASC排序

似乎是一个挑战...

一个查询甚至可能吗？

在此示例中，结果应该是两个具有User_ID的用户：＃11和＃33。 这是因为即使用户＃28同时具有test1_score和test2_score条目，也仅对于用户＃11和＃33 test1_score> = 7。

理想情况下，我会得到如下结果：

User_ID   |   NAME         | TEST1_SCORE  |  TEST2_SCORE    |
33        |     Felix      |    7         |  3              |
11        |     John       |    9         |  6              |

很感谢任何形式的帮助。

谢谢！

Answer 1

这就是我要做的

select distinct t.id, name, test1_score, test2_score from t
inner join (select id, value test1_score from t where field_key = 'test1_score' and value >= 7) t1 on (t1.id = t.id)
inner join (select id, value test2_score from t where field_key = 'test2_score' and value is not null) t2 on (t2.id = t.id)
inner join (select id, value name from t where field_key = 'name') t3 on (t3.id = t.id)
order by test1_score;

sqlfiddle

Answer 2

这对您有用吗？

SELECT A.USER_ID, 
       A.VALUE AS "NAME", 
       B.VALUE AS "TEST1_SCORE", 
       C.VALUE AS "TEST2_SCORE"
FROM MYTABLE A, MYTABLE B, MYTABLE C
WHERE A.USER_ID = B.USER_ID
  AND A.USER_ID = C.USER_ID
  AND A.FIELD_KEY = 'name'
  AND B.FIELD_KEY = 'test1_score'
  AND B.VALUE >= 7
  AND C.FIELD_KEY = 'test2_score'
ORDER BY 3 ASC
;