我有三个不同的查询，它们有一个共同的行（user_id），应该在一个查询中组合以应用排序。如果一个或多个查询结果为空集，它也应该有效。
查询1：
SELECT
user_id,
COUNT(answer) AS total_predictions
FROM
prognose_predictions pc
INNER JOIN prognose_prognose pp
ON pp.prognose_id = '9'
AND pc.prognose_id = '9'
AND pc.prognose_id = pp.prognose_id
GROUP BY
user_id
ORDER BY
user_id ASC
查询2：
SELECT
user_id,
COUNT(*) AS comments,
ccomments
FROM
(
SELECT
COUNT(*) AS ccomments
FROM
prognose_ccomments cc
LEFT JOIN prognose_comments p
ON cc.post_id = p.p_id
) AS tmp_table,
prognose_comments c
INNER JOIN prognose_prognose x
ON x.prognose_id = c.prognose_id AND c.prognose_id = 9
GROUP BY
c.user_id
ORDER BY
c.user_id ASC
查询3：
SELECT
user_id,
COUNT(*) as logins
FROM
prognose_activitylog a
WHERE
login_time BETWEEN '33333333333' AND '4444444444'
GROUP BY
user_id
我想有这个：
user_id | total_predictions | comments | ccomments | logins
1 | 3 | 0 | 0 | 7
6 | 6 | 1 | 3 | 4
7 | 0 | 0 | 0 | 1
其中“0”表示表中没有数据，但user_id仍然可用。
通过这种逻辑，我可以在PHP中更好地处理这些数据。
如何实现这一目标？
===================================
@Tad：错误的最新查询“字段列表中的列'user_id'不明确”
SELECT
user_id,
total_predictions,
IFNULL(comments, '0') AS comments ,
IFNULL(ccomments, '0') AS ccomments,
IFNULL(logins,'0') AS logins
FROM
(
SELECT
user_id,
total_predictions,
(
-- Get comments from Query 2
SELECT
comments
FROM
(
SELECT
COUNT(*) AS comments,
user_id
FROM
(
SELECT
COUNT(*) AS ccomments
FROM
prognose_ccomments cc
LEFT JOIN
prognose_comments p
ON
cc.post_id = p.p_id ) AS tmp_table,
prognose_comments c
INNER JOIN
prognose_prognose x
ON
x.prognose_id = c.prognose_id
AND c.prognose_id = '9'
GROUP BY
c.user_id ) AS Query2
WHERE
Query1.user_id = Query2.user_id ) AS comments,
(
-- Get ccomments from Query2
SELECT
ccomments
FROM
(
SELECT
ccomments,
user_id
FROM
(
SELECT
COUNT(*) AS ccomments
FROM
prognose_ccomments cc
LEFT JOIN
prognose_comments p
ON
cc.post_id = p.p_id ) AS tmp_table,
prognose_comments c
INNER JOIN
prognose_prognose x
ON
x.prognose_id = c.prognose_id
AND c.prognose_id = '9'
GROUP BY
c.user_id )AS Query2
WHERE
Query1.user_id = Query2.user_id ) AS ccomments,
(
-- Get logins from Query3
SELECT
logins
FROM
(
SELECT
user_id,
COUNT(*) AS logins
FROM
prognose_activitylog a
WHERE
login_time BETWEEN '1332284401' AND '1333058399'
GROUP BY
user_id )AS Query3
WHERE
Query1.user_id = Query3.user_id ) AS logins
FROM
(
-- Get product_Num logins and total_predictions from Query 1
SELECT
user_id,
COUNT(answer) AS total_predictions
FROM
prognose_predictions pc
INNER JOIN
prognose_prognose pp
ON
pp.prognose_id = '9'
AND pc.prognose_id = '9'
AND pc.prognose_id = pp.prognose_id
GROUP BY
user_id )AS Query1 )AS FinalQuery
INNER JOIN prognose_users ON prognose_users.User_ID = FinalQuery.user_id
ORDER BY
total_predictions DESC,
comments DESC,
ccomments DESC,
logins DESC LIMIT 0,10 ;
答案 0 :(得分：2)
SELECT A.user_id,COALESCE(A.total_predictions,0),B.comments,COALESCE(B.ccomments,0),COALESCE(C.logins,0) FROM
(
SELECT
user_id,
COUNT(answer) as total_predictions
FROM
prognose_predictions pc
INNER JOIN
prognose_prognose pp
ON
pp.prognose_id = '9'
AND pc.prognose_id = '9'
AND pc.prognose_id = pp.prognose_id
GROUP BY user_id
ORDER BY user_id ASC
) A
INNER JOIN （ 选择 用户身份， COUNT（）AS评论， ccomments 从 （ 选择 COUNT（）AS ccomments 从 prognose_ccomments cc LEFT JOIN prognose_comments p 上 cc.post_id = p.p_id）AS tmp_table， prognose_comments c 内部联接 prognose_prognose x 上 x.prognose_id = c.prognose_id AND c.prognose_id = 9 通过...分组 c.user_id
ORDER BY c.user_id ASC
) B
ON A.user_id = B.user_id
RIGHT OUTER JOIN
(
SELECT
user_id,
COUNT(*) as logins
FROM
prognose_activitylog a
WHERE login_time BETWEEN '33333333333' AND '4444444444'
GROUP BY user_id
) C
ON B.user_id = C.user_id
ORDER BY A.user_id;
答案 1 :(得分：1)
Hope the following works fine except it's lengthy:
SELECT [user_id], total_predictions, ISNULL(comments, ''), ISNULL(ccomments, ''), ISNULL(logins, '')
FROM
(
SELECT [user_id], total_predictions,
(
--Get comments from Query 2
SELECT comments
FROM
(
SELECT COUNT(*) AS comments
FROM
(
SELECT COUNT(*) AS ccomments
FROM prognose_ccomments cc
LEFT JOIN prognose_comments p ON cc.post_id = p.p_id
) AS tmp_table, prognose_comments c
INNER JOIN prognose_prognose x ON x.prognose_id = c.prognose_id AND c.prognose_id = 9
GROUP BY c.user_id
--ORDER BY c.user_id ASC
) AS Query2
WHERE Query1.[user_Id] = Query2.[user_id]
) AS comments,
(
--Get ccomments from Query2
SELECT ccomments
FROM
(
SELECT ccomments
FROM
(
SELECT COUNT(*) AS ccomments
FROM prognose_ccomments cc
LEFT JOIN prognose_comments p ON cc.post_id = p.p_id
) AS tmp_table, prognose_comments c
INNER JOIN prognose_prognose x ON x.prognose_id = c.prognose_id AND c.prognose_id = 9
GROUP BY c.user_id
--ORDER BY c.user_id ASC
)AS Query2
WHERE Query1.[user_Id] = Query2.[user_id]
) AS ccomments,
(
--Get logins from Query3
SELECT logins
FROM
(
SELECT user_id, COUNT(*) as logins
FROM prognose_activitylog a
WHERE login_time BETWEEN '33333333333' AND '4444444444'
GROUP BY user_id
)AS Query3
WHERE Query1.[user_Id] = Query3.[user_id]
) AS logins
FROM
(
--Get [user_id] logins and total_predictions from Query 1
SELECT [user_id], COUNT(answer) AS total_predictions
FROM prognose_predictions pc
INNER JOIN prognose_prognose pp ON pp.prognose_id = '9' AND pc.prognose_id = '9' AND pc.prognose_id = pp.prognose_id
GROUP BY [user_id]
--ORDER BY [user_id] ASC
)AS Query1
)AS FinalQuery
ORDER BY [user_id]
抱歉，我在MS SQL Server 2005中测试了该查询。但是，应该可以将此相同的查询转换为其MySQL等价物。为此，可以从[从中导出上述查询]下面的简化测试查询开始：
SELECT user_id, total_predictions, ISNULL(comments, '') AS Comments, ISNULL(ccomments, '') AS ccomments, ISNULL(logins,'') AS logins
FROM
(
SELECT user_id, total_predictions,
(
--Get comments from Query 2
SELECT comments
FROM
(
SELECT '1' AS user_id, 'comments' AS comments --Later on replace this with your Query 2 after removing the ORDER BY
) AS Query2
WHERE Query1.user_id = Query2.user_id
) AS comments,
(
--Get ccomments from Query2
SELECT ccomments
FROM
(
SELECT '1' AS user_id, 'ccomments' AS ccomments --Later on replace this with your Query 2 after removing the ORDER BY
)AS Query2
WHERE Query1.user_id = Query2.user_id
) AS ccomments,
(
--Get logins from Query3
SELECT logins
FROM
(
SELECT '1' AS user_id, 'logins' AS logins --Later on replace this with your Query 3 after removing the ORDER BY
)AS Query3
WHERE Query1.user_id = Query3.user_id
) AS logins
FROM
(
--Get product_Num logins and total_predictions from Query 1
SELECT '1' AS user_id, 'total_predictions' AS total_predictions -- Later on replace this with your Query 1 after removing the ORDER BY
)AS Query1
)AS FinalQuery
ORDER BY user_id
从MySQL命令行代码执行的MySQL查询版本：
mysql> SELECT user_id, total_predictions, IFNULL(comments, '') AS Comments
, IFNULL(ccomments, '') AS ccomments, IFNULL(logins,'') AS logins
-> FROM
-> (
-> SELECT user_id, total_predictions,
-> (
-> -- Get comments from Query 2
-> SELECT comments
-> FROM
-> (
-> SELECT '1' AS user_id, 'comments' AS comments -- Later on replace this with your Query 2 after removing the ORDER BY
-> ) AS Query2
-> WHERE Query1.user_id = Query2.user_id
-> ) AS comments,
-> (
-> -- Get ccomments from Query2
-> SELECT ccomments
-> FROM
-> (
-> SELECT '1' AS user_id, 'ccomments' AS ccomments -- Later on replace this with your Query 2 after removing the ORDER BY
-> )AS Query2
-> WHERE Query1.user_id = Query2.user_id
-> ) AS ccomments,
-> (
-> -- Get logins from Query3
-> SELECT logins
-> FROM
-> (
-> SELECT '1' AS user_id, 'logins' AS logins -- Later on replace this with your Query 3 after removing the ORDER BY
-> )AS Query3
-> WHERE Query1.user_id = Query3.user_id
-> ) AS logins
-> FROM
-> (
-> -- Get product_Num logins and total_predictions from Query 1
-> SELECT '1' AS user_id, 'total_predictions' AS total_predictions -- Later on replace this with your Query 1 after removing the ORDER BY
-> )AS Query1
-> )AS FinalQuery
-> ORDER BY user_id
->
-> ;
+---------+-------------------+----------+-----------+--------+
| user_id | total_predictions | Comments | ccomments | logins |
+---------+-------------------+----------+-----------+--------+
| 1 | total_predictions | comments | ccomments | logins |
+---------+-------------------+----------+-----------+--------+
1 row in set (0.06 sec)
仅供参考，为了将简化的测试查询转换为MySQL查询，我做了两个小的改动： 1）ISNULL到IFNULL（如前所述）;和 2）MySQL注释中的第二个破折号后面应至少跟一个空格。示例： - 从查询2获取注释。在转换之前，“ - ”和“Get”之间没有空格，因为空格在MS SQL Server中是可选的。
答案 2 :(得分：1)
很高兴知道它适合你。
对于您的最新要求，我将使用INNER JOIN，如下所示：
SELECT user_id, user_name, email, phone, total_predictions, IFNULL(comments, '0') AS Comments , IFNULL(ccomments, '0') AS ccomments, IFNULL(logins,'0') AS logins
FROM
(
-- All the sub queries go here
)AS FinalQuery
INNER JOIN prognose_users ON prognose_users.user_id = FinalQuery.user_id