mysql sum() 和双重结果使用多个连接

Posted 2023-03-31

【中文标题】mysql sum() 和双重结果使用多个连接

【英文标题】：mysql sum() and double result Using Multiple Joins

【发布时间】：2021-01-08 15:43:52

【问题描述】：

我有表 tb_orders

id_order	lot	rekanan	no_order	pec
5555	1	GOZNAK	390	y
6666	1	KOMSCO	391	y

我还有一张桌子 tb_pengambilans_detail

id	id_order	tanggal	jumlah
137	5555	2021-01-05 09:36:16	100000
136	6666	2021-01-05 09:35:57	100000
94	6666	2020-12-15 15:00:47	100000
138	5555	2021-01-05 09:37:51	100000

表格 tb_lini_pengambilans_detail

id	id_order	tanggal	deret5
38	6666	2020-12-08 09:26:36	30000
39	5555	2020-12-08 09:31:49	10000
40	6666	2020-12-14 07:55:36	80000
41	6666	2020-12-14 07:57:34	20000

表 tb_produksi_cutpacks_detail

id	tanggal	id_order	hcts	unfit
30	2021-01-04 13:02:10	5555	5000	0
32	2021-01-04 14:10:05	5555	0	10000
34	2021-01-04 15:11:57	5555	0	2000
35	2021-01-04 15:14:10	6666	9000	0

表 tb_lini_produksi_cutpacks_detail

id	tanggal	id_order	xbaik	xderet5
16	2021-01-04 14:52:38	5555	4000	1000
18	2021-01-04 14:53:21	6666	8000	0
19	2021-01-04 14:53:36	6666	0	2000
21	2021-01-04 14:57:41	5555	0	15000

我分别做了，所有结果都是正确的，不重复

SELECT a.rekanan,
    SUM(COALESCE(b.jumlah,0)) AS `pengambilan`
 FROM tb_orders a
LEFT JOIN tb_pengambilans_detail b USING(id_order)
WHERE a.pec='y'
GROUP BY a.rekanan

结果

rekanan	pengambilan
GOZNAX	200000
KOMSCO	200000

SELECT a.rekanan,
  SUM(COALESCE(c.unfit + c.hcts,0)) AS prod_hcts
FROM tb_orders a
LEFT JOIN tb_produksi_cutpacks_detail c USING(id_order)
WHERE a.pec='y'
GROUP BY a.rekanan

结果

rekanan	prod_hcts
GOZNAX	17000
KOMSCO	9000

SELECT a.rekanan,
  SUM(COALESCE(d.deret5 ,0)) AS `pengambilanlini`
FROM tb_orders a
LEFT JOIN tb_lini_pengambilans_detail d USING(id_order)
LEFT JOIN tb_lini_produksi_cutpacks_detail e USING(id_order)
WHERE a.pec='y'
GROUP BY a.rekanan

结果

rekanan	pengambilanlini
GOZNAX	20000
KOMSCO	260000

最后

SELECT a.rekanan,
  SUM(COALESCE(e.xbaik + e.xderet5 ,0)) AS prod_hctslini
FROM tb_orders a
LEFT JOIN tb_lini_produksi_cutpacks_detail e USING(id_order)
WHERE a.pec='y'
GROUP BY a.rekanan

结果

rekanan	prod_hctslini
GOZNAX	20000
KOMSCO	10000

在我确定一切之后，我加入了所有表格，结果都是双倍的......

SELECT a.rekanan,
SUM(COALESCE(b.jumlah,0)) AS `pengambilan`,
SUM(COALESCE(c.unfit + c.hcts,0)) AS prod_hcts,
SUM(COALESCE(d.deret5 ,0)) AS `pengambilanlini`,
SUM(COALESCE(e.xbaik + e.xderet5 ,0)) AS prod_hctslini
FROM tb_orders a
LEFT JOIN tb_pengambilans_detail b USING(id_order)
LEFT JOIN tb_produksi_cutpacks_detail c USING(id_order)
LEFT JOIN tb_lini_pengambilans_detail d USING(id_order)
LEFT JOIN tb_lini_produksi_cutpacks_detail e USING(id_order)
WHERE a.pec='y'
GROUP BY a.rekanan

结果

rekanan	pengambilan	prod_hcts	pengambilanlini	prod_hctslini
GOZNAX	1200000	68000	120000	120000
KOMSCO	1200000	108000	520000	60000

有什么办法让SQL结果不是double.. 请告诉我，以便解决问题.. 非常感谢

【参考方案1】：

----实际的单SQL解决方案是---------------

 SELECT a.rekanan,
SUM(COALESCE(b.jumlah,0)) AS `pengambilan`,
SUM(COALESCE(c.unfit + c.hcts,0)) AS prod_hcts,
SUM(COALESCE(d.deret5 ,0)) AS `pengambilanlini`,
SUM(COALESCE(e.xbaik + e.xderet5 ,0)) AS prod_hctslini
FROM tb_orders a
LEFT JOIN (select id_order ,SUM(jumlah) jumlah from  tb_pengambilans_detail group by id_order)  b USING(id_order)
LEFT JOIN (select id_order ,SUM(unfit) unfit ,SUM(hcts) hcts  from  tb_produksi_cutpacks_detail group by id_order)  c  USING(id_order)
LEFT JOIN  (select id_order ,SUM(deret5) deret5  from  tb_lini_pengambilans_detail group by id_order) d USING(id_order)
LEFT JOIN  (select id_order ,SUM(xbaik) xbaik , sum(xderet5) xderet5  from  tb_lini_produksi_cutpacks_detail group by id_order) e USING(id_order)
WHERE a.pec='y'
GROUP BY a.rekanan

【讨论】：

非常感谢您的回答..我的问题终于解决了..它的工作..问候

对不起，我从来没有投票过……我该如何投票？

没问题...只需点击答案的左侧向上箭头

哦好吧..对不起..我投给你了..再次感谢你