Question

[JsonProperty]

详细信息：

动物表将拥有属于一个群体的动物动物中的每只动物可以在位置表中具有多个位置对于位置表中的每只动物，可以有person_id或organization_id。

我需要获取该ID并查看人员或组织表中的详细信息。

对于位置表中的每只动物，他们可能需要培训成本和供应成本，我需要获得总数

我现在所做的是，从动物表中获取所有动物：

Table Animal:
id, group_id and name

Table Location:
id, group_id,animal, supplies, training, person_id,organization_id,date

Table Person
id, firstname, lastname

Table orgnization
id,  org_name

这将为我提供一组动物名单

select * from animal where group_id=70

这将给出一组5或6个条目，并在代码中循环通过每只动物获得总供应量，以及每只动物的总培训成本

select * from location where group_id=70 and animal_id = 60905442.

这是为了获取动物的最新位置，因为动物可能位于多个地方。

目标是找到动物的最新位置。这将为该动物提供select * from location where animal_id = ? and group_id = ? order by date或person_id

然后如果上述查询有organizagoin_id，请查看person_id表格，查看Person和first_name 否则会在last_name表中查看organization。

我知道这可以在一个查询中完成，但需要一些帮助。

样本表数据：

org_name

组织表 id organization_name 80 Microsoft LLC 85 Apple LLC

期望的输出： animal_id group_id name totalsupplies totaltraining personfirstname personlastname orgname

Answer 1

我找到了您的问题的查询，但这是t-sql语法。我希望在查看查询是如何构建时，这可以帮助您进一步。我猜这个row_number函数在MySQL中不可用。

declare @animal table (id int, group_id varchar(50), name varchar(50))
insert into @animal values
(6923,'A','harry'),
(6924,'A','larry'),
(6925,'A','Marry'),
(6888,'B','Eddy')

declare @location table (id int, group_id varchar(50), animal_id int, supplies int, training int, person_id int, org_id int, loc_date date)
insert into @location values
(1,'A',6923,10,null,90,0,'20170510'),
(2,'A',6923,10,10,90,0,'20170517'),
(3,'A',6924,0,10,0,80,'20170512'),
(4,'A',6924,10,10,0,85,'20170515'),
(5,'A',6925,0,0,95,0,'20170513'),
(6,'B',6888,20,20,95,0,'20170514')

declare @person table (id int, firstname varchar(50), lastname varchar(50))
insert into @person values
(90,'Test','Tester'),
(95,'Sam','Tester')

declare @organization table (id int, name varchar(50))
insert into @organization values
(80, 'Microsoft'),
(85, 'Apple')

select a.id, a.group_id, a.name, ttl.totalsupplies, ttl.totaltraining, 
    case when lastloc.person_id <> 0 then p.firstname + ' ' + p.lastname else '' end as person, 
    case when lastloc.org_id <> 0 then o.name else '' end as organization
from 
   (select m.id, m.group_id, m.animal_id, m.supplies, m.training, m.person_id, m.org_id, m.loc_date
    from 
       (select id, group_id, animal_id, supplies, training, person_id, org_id, loc_date,
            ROW_NUMBER() over(partition by animal_id order by loc_date desc) AS r
        from @location) AS m
    where m.r = 1) AS lastloc
 inner join
   (select animal_id, sum(supplies) as totalsupplies, sum(training) as totaltraining
    from @location
    group by animal_id) AS ttl
      on ttl.animal_id = lastloc.animal_id
 inner join
   @animal as a on a.id = lastloc.animal_id
 left outer join
   @person as p on p.id = lastloc.person_id
 left outer join
   @organization as o on o.id = lastloc.org_id

这将导致

id    group_id  name    totalsupplies   totaltraining   person        organization
----------------------------------------------------------------------------------
6923  A         harry   20              10              Test Tester 
6924  A         larry   10              20                            Apple
6925  A         Marry   0               0               Sam Tester  
6888  B         Eddy    20              20              Sam Tester

MySQL多表连接和总数

1 个答案: