Question

提前抱歉，如果我的问题重复或类似于另一个问题，我已经检查过类似的其他帖子，但我无法做到正确。所以，这是我的问题。我有接下来的两个select语句，我在第一个中正确放置第二个查询时遇到了麻烦，我已经尝试在连接之后放入select中，如果你可以帮助我或给予建议它会很棒。

select distinct movements.date, movements.productcode, products.productname
,clasification.family, clients.clientname, clients.clientcode, documents.invoicereference
,movements.quantity, movements.price, movements.credit, movements.total, movements.exchange
,movements.state, movements.city
from invoices.movements

  inner join invoices.products
  on invoices.movements.idproduct=invoices.products.idproduct
  left outer join invoices.documents
  on invoices.movements.iddocument=invoices.documents.iddocument
  inner join invoices.coins
  on invoices.documents.idcoin=invoices.coins.idcoin
  inner join invoices.address
  on invoices.address.iddocument=invoices.documents.iddocument
  inner join invoices.clasifications
  on invoices.clasifications.family='aspen'
  inner join invoices.clients
  on invoices.documents.idclient=invoices.clients.idclient

这是我从上一个查询得到的。

+----+------------+------------+-----------+-----------+-----------------+--------+----------+------+-----+--------+-----+----+
|date|product_code|product_name|client_name|client_code|invoice_reference|quantity|unit_price|credit|total|exchange|state|city|
+----+------------+------------+-----------+-----------+-----------------+--------+----------+------+-----+--------+-----+----+

但我还想添加另一个引用客户端类型的列，它们有两个分类，如果它们的client_code就像这个'N00000'（或者更具体，如果它们以零结尾），它们是< strong>“休闲”客户如果不是（比如'N00013'，他们是“固定”客户。这是查询。

select 
  case
   when invoices.clients.idclient='N00000' then 'Casual'
   else 'Fixed'
    end as client_type,
     clientcode, invoices.clients.clientname
from invoices.documents.
inner join invoices.clients
on invoices.documents.idclient=invoices.clients.idclient

我的问题是我似乎无法将查询正确地放在第一个内容中，如果有人可以帮助我，我真的很感激。

Answer 1

在我看来，您已经在第一个查询中完成了必要的连接，因此您只需要在第一个查询的列列表中添加CASE表达式，如下所示：

select distinct movements.date, movements.productcode, products.productname
,clasification.family, clients.clientname, clients.clientcode, documents.invoicereference
,movements.quantity, movements.price, movements.credit, movements.total, movements.exchange
,movements.state, movements.city, case
   when invoices.clients.idclient='N00000' then 'Casual'
   else 'Fixed'
    end as client_type
from invoices.movements

  inner join invoices.products
  on invoices.movements.idproduct=invoices.products.idproduct
  left outer join invoices.documents
  on invoices.movements.iddocument=invoices.documents.iddocument
  inner join invoices.coins
  on invoices.documents.idcoin=invoices.coins.idcoin
  inner join invoices.address
  on invoices.address.iddocument=invoices.documents.iddocument
  inner join invoices.clasifications
  on invoices.clasifications.family='aspen'
  inner join invoices.clients
  on invoices.documents.idclient=invoices.clients.idclient

Answer 2

别名是你的朋友，代码格式也是如此。

（或者更具体地说，如果它们以零结尾），他们会＆＃34;休闲＆＃34;

select distinct 
    im.date
  , im.productcode
  , ip.productname
  , icla.family
  , icli.clientname
  , icli.clientcode
  , id.invoicereference
  , im.quantity
  , im.price
  , im.credit
  , im.total
  , im.exchange
  , im.state
  , im.city
  , case when id.idclient like '%0' then 'Casual'
         when id.idclient is not null then 'Fixed'
         end as client_type
from invoices.movements im
  inner join invoices.products ip
    on im.idproduct=ip.idproduct
  left outer join invoices.documents id
    on im.iddocument=id.iddocument
  inner join invoices.coins ico
    on id.idcoin=ico.idcoin
  inner join invoices.address
    on invoices.address.iddocument=id.iddocument
  inner join invoices.clasifications icla
    on icla.family='aspen'
  inner join invoices.clients icli
    on id.idclient=icli.idclient

SQL Server。如何将选择案例放在具有多个联接的另一个选择中

2 个答案: