Question

ID  TELEFON     CULOARE    piesa1   piesa2  piesa3  piesa4
1   telefon1    culoare1    0N        0N      0N      0N
1   telefon1    culoare1    14Y       0N      0N      0N
2   telefon2    culoare2    0N        8Y      0N      0N
2   telefon2    culoare2    0N        0N      4Y      0N
3   telefon3    culoare3    0N        0N      0N      0Y
3   telefon3    culoare3    0N        0N      0N      0N
3   telefon3    culoare3    0N        0N      0N      0N
3   telefon3    culoare3    0N        0N      5Y      0N
4   telefon4    Neutru      8N        0N      0N      0N
4   telefon4    Neutru      0N        0N      1N      0N
4   telefon4    Neutru      0N        0N      0N      7Y

我有一个这样的结果集，但是我想以某种方式合并ID列重复的行，并且只留下值结束的单元格＆＃39; Y＆＃39;或者使用＆＃39; N＆＃39;但是前面的数字大于0。我设法只通过添加SUM agg函数和按行的唯一标识符分组来对数字执行此操作，在我的情况下将是TELEFON + CULOARE列。

查询此结果：

SELECT tt.id as `#`
,t.telefon as TELEFON
, IFNULL(c.culoare,'Neutru') as CULOARE
, (IF(p.piesa = 'piesa1', CONCAT(tt.total_piese,tt.is_stoc_min),'0N')) AS piesa1
,(IF(p.piesa = 'piesa2', CONCAT(tt.total_piese,tt.is_stoc_min),'0N')) AS piesa2
,(IF(p.piesa = 'piesa3', CONCAT(tt.total_piese,tt.is_stoc_min),'0N')) AS piesa3
,(IF(p.piesa = 'piesa4', CONCAT(tt.total_piese,tt.is_stoc_min),'0N')) AS piesa4 
,tt.total_piese
,tt.is_stoc_min
FROM telefon as t LEFT JOIN
total_piese as tt
ON t.id = tt.id_telefon
LEFT JOIN culoare as c
ON c.id  = tt.id_culoare
LEFT JOIN piesa as p
ON p.id = tt.id_piesa
ORDER BY t.id;

我的愿望结果是这样的：

TELEFON     CULOARE   piesa1    piesa2  piesa3  piesa4
telefon1    culoare1    14Y       0N      0N      0N
telefon2    culoare2    0N        8Y      4Y      0N
telefon3    culoare3    0N        0N      5Y      0Y
telefon4    Neutru      8N        0N      1N      7Y

我有一个返回合并行的查询，但没有＆＃39; Y＆＃39;或者＆＃39; N＆＃39;在价值的最后：

SELECT tt.id as `#`
,t.telefon as TELEFON
, IFNULL(c.culoare,'Neutru') as CULOARE
,SUM(IF(p.piesa = 'piesa1', tt.total_piese,0)) AS piesa1
,SUM(IF(p.piesa = 'piesa2', tt.total_piese,0)) AS piesa2
,SUM(IF(p.piesa = 'piesa3', tt.total_piese,0)) AS piesa3
,SUM(IF(p.piesa = 'piesa4',tt.total_piese,0)) AS piesa4 
,tt.stoc_min
 FROM telefon as t LEFT JOIN
 total_piese as tt
 ON t.id = tt.id_telefon
 LEFT JOIN culoare as c
 ON c.id  = tt.id_culoare
 LEFT JOIN piesa as p
 ON p.id = tt.id_piesa
 GROUP BY t.Telefon,c.Culoare
 ORDER BY t.id;

Answer 1

嗯，给你一个更好（或更优化）的答案有点困难，因为我们没有＆＃34;基本数据＆＃34;，但已经是结果。< / p>

无论如何，您可以在piesa字段上使用max将您的第一次尝试用作子查询。（因为任何数字都将大于0，并且Y大于N，因此MAX应该为您提供所需的输出）

select id as `#`, telefon, culoare,
max(piesa1) as piesa1,
max(piesa2) as piesa2,
max(piesa3) as piesa3,
max(piesa4) as piesa4
from
  (SELECT 
     tt.id 
    ,t.telefon as TELEFON
    ,IFNULL(c.culoare,'Neutru') as CULOARE
    ,(IF(p.piesa = 'piesa1', CONCAT(tt.total_piese,tt.is_stoc_min),'0N')) AS piesa1
    ,(IF(p.piesa = 'piesa2', CONCAT(tt.total_piese,tt.is_stoc_min),'0N')) AS piesa2
    ,(IF(p.piesa = 'piesa3', CONCAT(tt.total_piese,tt.is_stoc_min),'0N')) AS piesa3
    ,(IF(p.piesa = 'piesa4', CONCAT(tt.total_piese,tt.is_stoc_min),'0N')) AS piesa4 
    ,tt.total_piese
    ,tt.is_stoc_min
  FROM telefon as t LEFT JOIN
    total_piese as tt ON t.id = tt.id_telefon
  LEFT JOIN culoare as c ON c.id  = tt.id_culoare
  LEFT JOIN piesa as p ON p.id = tt.id_piesa) s
GROUP BY id, telefon, culoare
ORDER BY Id

或可能（未经测试）直接在查询中使用max

SELECT tt.id as `#`
,t.telefon as TELEFON
,IFNULL(c.culoare,'Neutru') as CULOARE
,max((IF(p.piesa = 'piesa1', CONCAT(tt.total_piese,tt.is_stoc_min),'0N'))) AS piesa1
,max((IF(p.piesa = 'piesa2', CONCAT(tt.total_piese,tt.is_stoc_min),'0N'))) AS piesa2
,max((IF(p.piesa = 'piesa3', CONCAT(tt.total_piese,tt.is_stoc_min),'0N'))) AS piesa3
,max((IF(p.piesa = 'piesa4', CONCAT(tt.total_piese,tt.is_stoc_min),'0N'))) AS piesa4 
FROM telefon as t LEFT JOIN
total_piese as tt
ON t.id = tt.id_telefon
LEFT JOIN culoare as c
ON c.id  = tt.id_culoare
LEFT JOIN piesa as p
ON p.id = tt.id_piesa
GROUP BY tt.Id, t.telefon, c.culoare
ORDER BY t.id;

在满足某些条件时合并唯一键上的字符串单元

1 个答案: