我有两个查询需要合并成一个查询。问题是他们在 where 子句中使用相同的列,具体取决于我需要获取的内容。
查询 1
SELECT c.fullname, COUNT( DISTINCT sst.id ) AS 'Liczba rozpoczetych szkolen'
FROM mdl_course c
INNER JOIN mdl_scorm s ON s.course = c.id
INNER JOIN mdl_scorm_scoes_track sst ON s.id = sst.scormid
INNER JOIN mdl_user u ON u.id = sst.userid
WHERE sst.element = 'x.start.time' AND u.deleted =0
GROUP BY c.fullname ORDER BY `Liczba rozpoczetych szkolen` ASC
查询 2
SELECT c.fullname, COUNT(DISTINCT sst.userid ) AS 'Liczba_ukonczonych_szkolen'
FROM mdl_course c
INNER JOIN mdl_scorm s ON s.course=c.id
INNER JOIN mdl_scorm_scoes_track sst ON s.id = sst.scormid
INNER JOIN mdl_user u ON sst.userid=u.id
where `element`='cmi.core.score.raw' and `value` = 100 and u.deleted = 0
GROUP BY c.fullname ORDER BY `Liczba_ukonczonych_szkolen` DESC
它们依赖于名为"元素"的同一列。
我如何将结果显示为
fullname Liczba rozpoczetych szkolen Liczba_ukonczonych_szkolen
A1 34 4
A2 5 3
A3 34 33
我想出了这个,白巫作品。感谢您的建议@HoneyBadger
SELECT t1.fullname, t1.Liczba_rozpoczetych_szkolen, t2.Liczba_ukonczonych_szkolen,
round(((t2.Liczba_ukonczonych_szkolen /t1.Liczba_rozpoczetych_szkolen)*100),2) as procentowo
FROM
(SELECT c.fullname, COUNT( DISTINCT sst.id ) AS Liczba_rozpoczetych_szkolen
FROM mdl_course c
INNER JOIN mdl_scorm s ON s.course = c.id
INNER JOIN mdl_scorm_scoes_track sst ON s.id = sst.scormid
INNER JOIN mdl_user u ON u.id = sst.userid
WHERE sst.element = 'x.start.time' AND u.deleted =0
GROUP BY c.fullname ) AS t1,
(SELECT c.fullname, COUNT(DISTINCT sst.userid ) AS Liczba_ukonczonych_szkolen
FROM mdl_course c
INNER JOIN mdl_scorm s ON s.course=c.id
INNER JOIN mdl_scorm_scoes_track sst ON s.id = sst.scormid
left join mdl_user u ON sst.userid=u.id
where `element`='cmi.core.score.raw' and `value` = 100 and u.deleted = 0
GROUP BY c.fullname) as t2
WHERE t1.fullname = t2.fullname
您可以像这样将它们连接在一起:
SELECT COALESCE(R.fullname, U.fullname) AS fullname
, COALESCE(R.Liczba_rozpoczetych_szkolen, 0) AS Liczba_rozpoczetych_szkolen
, COALESCE(R.Liczba_ukonczonych_szkolen, 0) AS Liczba_ukonczonych_szkolen
FROM (
SELECT c.fullname
, COUNT( DISTINCT sst.id ) AS Liczba_rozpoczetych_szkolen
FROM mdl_course c
INNER JOIN mdl_scorm s
ON s.course = c.id
INNER JOIN mdl_scorm_scoes_track sst
ON s.id = sst.scormid
INNER JOIN mdl_user u
ON u.id = sst.userid
WHERE sst.element = 'x.start.time'
AND u.deleted =0
GROUP BY c.fullname
) AS R
FULL JOIN (
SELECT c.fullname
, COUNT(DISTINCT sst.userid ) AS Liczba_ukonczonych_szkolen
FROM mdl_course c
INNER JOIN mdl_scorm s
ON s.course=c.id
INNER JOIN mdl_scorm_scoes_track sst
ON s.id = sst.scormid
INNER JOIN mdl_user u
ON sst.userid=u.id
where `element`='cmi.core.score.raw'
AND `value` = 100
AND u.deleted = 0
GROUP BY c.fullname
) AS U
ON R.fullname = U.fullname
尝试使用 OR 谓词进行查询,并使用SELECT
列表中的CASE
进行额外筛选。有点
SELECT c.fullname, COUNT(DISTINCT CASE WHEN sst.element = 'x.start.time' THEN sst.id END) AS 'Liczba rozpoczetych szkolen'
, COUNT(DISTINCT CASE WHEN sst.element = 'cmi.core.score.raw' and `value` = 100 THEN sst.userid END) AS 'Liczba_ukonczonych_szkolen'
FROM mdl_course c
INNER JOIN mdl_scorm s ON s.course = c.id
INNER JOIN mdl_scorm_scoes_track sst ON s.id = sst.scormid
INNER JOIN mdl_user u ON u.id = sst.userid
WHERE sst.element = IN( 'x.start.time', 'cmi.core.score.raw') AND u.deleted =0
GROUP BY c.fullname ORDER BY `Liczba rozpoczetych szkolen` ASC