Question

我正在使用具有多个联接和group_concat的复杂查询，这是查询：

 select a.packagename as packagename, a.packageid as packageid, 
 sum(a.packageduration) as days, (a.packageprice + a.seater_4) as price,
 b.statename,
 substring_index(GROUP_CONCAT( DISTINCT (select f.placename )  SEPARATOR ', 
 '),',',4) placename,
 substring_index(GROUP_CONCAT( DISTINCT (select d.packagetypename )  
 SEPARATOR ', '),',',4) packagetypename
 from tblpackages a
 join tblstates b on b.stateid = a.stateid              
 join tblpackage_packagetypes c on c.packageid = a.packageid
 join tblpackagetype d on d.packagetypeid = c.packagetypeid
 join tblpackageplaces e on e.packageid = a.packageid
 join tblplaces f on f.placeid = e.placeid
                where d.placename = "placename"
                 group by a.packageid

我的packagetypename的group_concat每个查询显示4个值，而placename仅显示where子句中给出的1个值？

因为查询是按包裹ID分组的，对于任何包裹我都有N个地方。按逻辑查询工作正常，但我想显示所有与我对应的软件包ID的软件包

Answer 1

我建议将WHERE子句中的条件移动到ON子句中，将其移动到相应的联接，然后使该联接成为左联接：

SELECT ...
FROM tblpackages a
INNER JOIN tblstates b
    ON b.stateid = a.stateid              
INNER JOIN tblpackage_packagetypes c
    ON c.packageid = a.packageid
LEFT JOIN tblpackagetype d
    ON d.packagetypeid = c.packagetypeid AND d.placename = 'placename'
INNER JOIN tblpackageplaces e
    ON e.packageid = a.packageid
INNER JOIN tblplaces f
    ON f.placeid = e.placeid
GROUP BY
    a.packageid

Answer 2

将您的查询输出修改为

select a.packagename as packagename, a.packageid as packageid, 
a.packageduration as days, (a.packageprice + a.seater_4) as price,
b.statename,
substring_index(GROUP_CONCAT( DISTINCT (select f.placename )  SEPARATOR ', '),',',4) 
placename,
substring_index(GROUP_CONCAT( DISTINCT (select d.packagetypename )  SEPARATOR ', 
'),',',4) packagetypename
from tblpackages a
inner join tblstates b on b.stateid = a.stateid              
inner join tblpackage_packagetypes c on c.packageid = a.packageid
inner  join tblpackagetype d on d.packagetypeid = c.packagetypeid
inner join tblpackageplaces e on e.packageid = a.packageid
left join tblplaces f on f.placeid = e.placeid  and d.packagetypename = 'religious'
                 group by a.packageid

pname，pid，天，价格，州名，地名，包装类型名

test1、1、3、111，xyz（（katra，shiv khori），宗教
test2，2，3，444，xyz，（Katra），宗教

test3、3、3、445，xyz，NULL（（家庭，夫妇，冒险，朋友）

test4，4，555，xyz，（Katra，Patnitop，sanasar，nathatop），（宗教，家庭，情侣，冒险）

test6，666，abc，NULL（（家庭，冒险，朋友）

不需要包裹ID 3和6

使用where子句

2 个答案: