Question

我有两个表，一个表包含一些用户配置（#USERCONFIG），另一个表（#COMBINATIONS），我需要在第一个表中找到多个配置组合。

CREATE TABLE #COMBINATIONS (INDEX1 INT, MENU CHAR(10))
CREATE TABLE #USERCONFIG (USERID VARCHAR(10), MENU VARCHAR(10))


INSERT INTO #COMBINATIONS VALUES (1, 'ABC300')
INSERT INTO #COMBINATIONS VALUES (1, 'ABC400')
INSERT INTO #COMBINATIONS VALUES (2, 'ABC100')
INSERT INTO #COMBINATIONS VALUES (2, 'ABC500')
INSERT INTO #COMBINATIONS VALUES (2, 'ABC600')


INSERT INTO #USERCONFIG VALUES ('SMITHJ', 'ABC100')
INSERT INTO #USERCONFIG VALUES ('SMITHJ', 'ABC500')
INSERT INTO #USERCONFIG VALUES ('SMITHJ', 'ABC600')
INSERT INTO #USERCONFIG VALUES ('SMITHC', 'ABC100')
INSERT INTO #USERCONFIG VALUES ('SMITHC', 'ABC500')
INSERT INTO #USERCONFIG VALUES ('SMITHA', 'ABC100')
INSERT INTO #USERCONFIG VALUES ('SMITHA', 'ABC200')
INSERT INTO #USERCONFIG VALUES ('SMITHA', 'ABC300')
INSERT INTO #USERCONFIG VALUES ('SMITHA', 'ABC400')
INSERT INTO #USERCONFIG VALUES ('SMITHA', 'ABC600')

使用此示例数据，我希望结果集如下所示：

'SMITHJ', '2'
'SMITHA', '1'
'SMITHC', '2'

它将在组合表中返回所有配置匹配的用户。

任何帮助将不胜感激。

Answer 1

以下内容将列出用户及其所拥有的 complete 组合。如果有帮助，您可以将其视为配方成分和用户成分的教科书问题：

SELECT alluser.USERID, index_menu.INDEX1
FROM (SELECT DISTINCT USERID FROM #USERCONFIG) AS alluser
CROSS JOIN #COMBINATIONS AS index_menu
LEFT JOIN #USERCONFIG AS user_menu ON alluser.USERID = user_menu.USERID AND index_menu.MENU = user_menu.MENU
GROUP BY alluser.USERID, index_menu.INDEX1
HAVING COUNT(index_menu.MENU) = COUNT(user_menu.MENU)

Answer 2

此代码段将获得该结果：

IF OBJECT_ID('tempdb..#COMBINATIONS') IS NOT NULL DROP TABLE #COMBINATIONS;
IF OBJECT_ID('tempdb..#USERCONFIG') IS NOT NULL DROP TABLE #USERCONFIG;

CREATE TABLE #COMBINATIONS (INDEX1 INT, MENU VARCHAR(10));
CREATE TABLE #USERCONFIG (USERID VARCHAR(10), MENU VARCHAR(10));

INSERT INTO #COMBINATIONS (INDEX1, MENU) VALUES 
(1, 'ABC301'),
(1, 'ABC401'),
(2, 'ABC102'),
(2, 'ABC502'),
(2, 'ABC602');

INSERT INTO #USERCONFIG (USERID, MENU) VALUES 
('SMITHJ', 'ABC102'),
('SMITHJ', 'ABC502'),
('SMITHJ', 'ABC602'),
('SMITHC', 'ABC102'),
('SMITHC', 'ABC502'),
('SMITHA', 'ABC102'),
('SMITHA', 'ABC200'),
('SMITHA', 'ABC301'),
('SMITHA', 'ABC401'),
('SMITHA', 'ABC602');

SELECT USERID, INDEX1
FROM
(
    SELECT uconf.USERID, comb.INDEX1, 
     COUNT(*) AS Total,
     DENSE_RANK() OVER (PARTITION BY uconf.USERID ORDER BY COUNT(*) DESC, comb.INDEX1 ASC) AS Rnk
    FROM #USERCONFIG uconf
    INNER JOIN #COMBINATIONS comb
      ON comb.MENU = uconf.MENU
    GROUP BY uconf.USERID, INDEX1
) q
WHERE Rnk = 1
ORDER BY Total DESC, USERID;

返回：

USERID  INDEX1
SMITHJ  2
SMITHA  1
SMITHC  2

SQL在2个表中查找多个组合（包含所有记录）

2 个答案: