Mysql 选择内部联接只返回一个结果

从数据库中的两个表中选择某些数据时遇到问题

这些表是

psthostess

和

psttoduit

在my

PSHostess

中，我想选择字段

code

和

Name

在my

PSTODO uit

中，我要选择：

• PB之和=1
• PG之和=1
• PA之和=1
• h.GoedkuringdoorNew的总和=GF
• h.GoedkuringdoorNew的总和=SB
• h.GoedkuringdoorNew的总和=通过
• h.Blanco之和

这是我的疑问：

SELECT p.Code, p.Name, sum(h.PB = 1), sum(h.PG = 1), sum(h.PA = 1), 
       sum(h.GoedkeuringDoorNew = 'GF'), sum(h.GoedkeuringDoorNew = 'SB'),  
       sum(h.GoedkeuringDoorNew = 'VIA'), sum(h.Blanco)
   FROM psthostess p
   INNER JOIN `psttodo-uit` h ON h.`Hostess Code` = p.Code
   WHERE p.Indienst = 1

---

问题是我总是得到一行的结果。但是在

pshostess

中有多行

Indienst=1

（其中）。如何修复此问题？

您需要按非聚合列对其进行分组

SELECT p.Code, p.Name, sum(h.PB = 1), 
sum(h.PG = 1), sum(h.PA = 1), sum(h.GoedkeuringDoorNew = 'GF'), 
sum(h.GoedkeuringDoorNew = 'SB'), sum(h.GoedkeuringDoorNew = 'VIA'), sum(h.Blanco)
FROM psthostess p
INNER JOIN `psttodo-uit` h ON h.`Hostess Code` = p.Code
WHERE p.Indienst = 1
group by p.Code, p.Name

---

您需要按非聚合列对其进行分组

SELECT p.Code, p.Name, sum(h.PB = 1), 
sum(h.PG = 1), sum(h.PA = 1), sum(h.GoedkeuringDoorNew = 'GF'), 
sum(h.GoedkeuringDoorNew = 'SB'), sum(h.GoedkeuringDoorNew = 'VIA'), sum(h.Blanco)
FROM psthostess p
INNER JOIN `psttodo-uit` h ON h.`Hostess Code` = p.Code
WHERE p.Indienst = 1
group by p.Code, p.Name

---

您需要按非聚合列对其进行分组

SELECT p.Code, p.Name, sum(h.PB = 1), 
sum(h.PG = 1), sum(h.PA = 1), sum(h.GoedkeuringDoorNew = 'GF'), 
sum(h.GoedkeuringDoorNew = 'SB'), sum(h.GoedkeuringDoorNew = 'VIA'), sum(h.Blanco)
FROM psthostess p
INNER JOIN `psttodo-uit` h ON h.`Hostess Code` = p.Code
WHERE p.Indienst = 1
group by p.Code, p.Name

---

您需要按非聚合列对其进行分组

SELECT p.Code, p.Name, sum(h.PB = 1), 
sum(h.PG = 1), sum(h.PA = 1), sum(h.GoedkeuringDoorNew = 'GF'), 
sum(h.GoedkeuringDoorNew = 'SB'), sum(h.GoedkeuringDoorNew = 'VIA'), sum(h.Blanco)
FROM psthostess p
INNER JOIN `psttodo-uit` h ON h.`Hostess Code` = p.Code
WHERE p.Indienst = 1
group by p.Code, p.Name