刚刚开发需求写了个SQL,记个笔记,学习下关于数据库的多表合并查询的用法


select t.* from A t 
UNION ALL/UNION/Intersect/MINUS
select s.* from B s;

##UNION ALL

772156-20170328165135826-1065021997.png

使用UNION ALL,表示取A、B的合集,不过滤重复的数据行

##UNION

772156-20170328170010639-1715535324.png

使用UNION,会将结果集A和结果集B进行UNION ALL运算,然后取两者交集的余集作为结果集

##Intersect
772156-20170328170456248-1010642438.png

使用Intersect,会将结果集A和结果集B进行UNION ALL运算,然后两者之间的集交集作为结果集和UNION刚好相反

##MINUS
772156-20170328171317045-1304191687.png

使用MINUS,取结果集A减去结果集B留下的差集,注:如果结果集A小于等于结果集B,返回空结果集.


好啦,下面进入实战阶段,我就直接将我写的SQL贴出来吧

select a.*
from (select t.c_fund_account_name as "fundAccountNo", --基金账号
tfp.project_code as "projectCode", --项目编号
tfp.project_name as "projectName", --项目名称
tfp.project_shortname as "projectShortName", --项目简称
c.c_fund_name as "fundName", --基金产品名称
c.c_fund_code as "fundCode", --基金产品代码
nvl(thold.subsistAssetsShare, 0) as "subsisAssetsShare", --份额(家族)
to_char(thold.updateDate, 'yyyy-MM-dd') as "updateDate", --日期
nvl(c.c_current_share, 0) as "currentShare", --份额(基金网站)
to_char(c.d_date, 'yyyy-MM-dd') as "dateDate", --日期
(nvl(thold.subsistAssetsShare, 0) - nvl(c.c_current_share, 0)) as "diffValue", --差值
CAST((CASE
WHEN (nvl(thold.subsistAssetsShare, 0) -
nvl(c.c_current_share, 0)) = 0 THEN
'1'
WHEN (nvl(thold.subsistAssetsShare, 0) -
nvl(c.c_current_share, 0)) <> 0 THEN
'0'
END) as nvarchar2(2)) as "identical" --是否一致
from t_fund_account t
inner join (select fhs.*,
row_number() over(partition by fhs.c_fund_account_no, fhs.c_project_code order by fhs.d_date desc) rn
from td_fund_holding_share fhs) c
on t.c_fund_account_name = c.c_fund_account_no
and t.c_project_code = c.c_project_code
LEFT JOIN (SELECT tha.project_code as projectCode,
sum(tha.current_share) as subsistAssetsShare, -- 持有份额
sum(tha.current_cost) as currentCost, --当前成本
sum(tha.accumulated_profit) as accumulatedProfit, --累积利润
max(tha.update_time) as updateDate
FROM t_hold_assets tha
left join t_polling_product p
on tha.c_product_code = p.c_product_code
WHERE 1 = 1
and tha.delete_flag = '0'
and p.c_stock_type_level1 = '0'
and p.c_stock_type_level2 = '01'
GROUP BY tha.project_code) thold
on thold.projectCode = t.c_project_code
left join t_family_project tfp
on tfp.project_code = t.c_project_code
and tfp.delete_flag = '0'
where rn = 1
AND t.c_fund_account_type = '1' --基金账户
AND t.delete_flag = '0'

UNION ALL

select t.c_fund_account_name as "fundAccountNo", --基金账号
tfp.project_code as "projectCode", --项目编号
tfp.project_name as "projectName", --项目名称
tfp.project_shortname as "projectShortName", --项目简称
CAST('' as nvarchar2(50)) as "fundName", --基金产品名称
CAST('' as nvarchar2(50)) as "fundCode", --基金产品代码
nvl(thold.subsistAssetsShare, 0) as "subsisAssetsShare", --份额(家族)
to_char(thold.updateDate, 'yyyy-MM-dd') as "updateDate", --日期
to_number(nvl('', 0)) as "currentShare", --份额(基金网站)
to_char(CAST('' as nvarchar2(50)), 'yyyy-MM-dd') as "dateDate", --日期
nvl(thold.subsistAssetsShare, 0) - nvl('', 0) as "diffValue", --差值
CAST((CASE
WHEN (nvl(thold.subsistAssetsShare, 0) - nvl('', 0)) = 0 THEN
'1'
WHEN (nvl(thold.subsistAssetsShare, 0) - nvl('', 0)) <> 0 THEN
'0'
END) as nvarchar2(2)) as "identical" --是否一致
from t_fund_account t
LEFT JOIN (SELECT tha.project_code as projectCode,
sum(tha.current_share) as subsistAssetsShare, -- 持有份额
max(tha.update_time) as updateDate
FROM t_hold_assets tha
left join t_polling_product p
on tha.c_product_code = p.c_product_code
WHERE 1 = 1
and tha.delete_flag = '0'
and p.c_stock_type_level1 = '0'
and p.c_stock_type_level2 = '01'
GROUP BY tha.project_code) thold
on thold.projectCode = t.c_project_code
left join t_family_project tfp
on t.c_project_code = tfp.project_code
and tfp.delete_flag = '0'
where t.c_fund_account_name not in
(select td.c_fund_account_no from td_fund_holding_share td)
and t.c_fund_account_type = '1'
and t.delete_flag = '0') a
order by a."diffValue" desc, a."projectCode" desc

上面sql具体意思是:查询出基金信息,A和基金对比先查询其中有的数据,再union all A有基金没有的数据,一起取个并集。OK,需求完成
哈哈哈哈,其实只要你SQL写得牛逼,然后了解下业务流程,什么都好说哈哈哈

哦对了,最后啰嗦一句。 对于这些并集计算之后,需要排序 则语法为:
select t.* from (语句1 union all 语句2)
t order by t.id;

44aba224bc315c60b71173b480b1cb13485477bf.jpg