SQL Server。合并两个select语句，每个内部连接

Question

提前对不起，如果我的问题重复或类似于另一个，我已经检查了另一个类似这样的帖子，但我不明白。所以，这是我的问题。我有接下来的两个选择语句，并且我无法正确地将第二个查询放在第一个查询中，我尝试在连接之后放入select中，并且如果可以帮助我，给予和建议，它会很棒。

select distinct 
    movements.date, movements.productcode, 
    products.productname, 
    clasification.family, 
    clients.clientname, clients.clientcode, 
    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 

这是结果我从这个查询获得：

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

但我也想添加两列，一个用于客户端的clasification，在那里我可以得到，如果客户端是决赛用户,安装程序,生产者，等等，这是查询给了我。

select 
    invoices.classifications.classification_value, 
    client_name, client_code 
from 
    invoices.clients 
inner join 
    invoices.classifications on invoices.classifications.cid_classification = invoices.clients.cid_clasifclient1 
inner join 
    invoices.documents on invoices.documents.cid_client = invoices.clients.cid_client 

，第二列是发票系列，我需要合并到列进一个，但是当我把它添加到第一个查询不返回任何值。

select 
    convert(varchar(20), documents_series) + '' + 
     convert(varchar(6), document_number) as invoice_series 
from 
    invoices.documents 

所以我真的很喜欢，如果你能帮助我，抱歉，如果我不是很好的广告SQL。

Answer 1

因为它似乎是承上启下的是客户表主要idclient和cid_client领域暗示一个主/外键对（或一个一对多连接表），请考虑使用最后一次查询，之间第一个查询中的派生表加入这两个字段。另外，尝试使用table aliases如下图所示，以减少repetitiously长表名：

select distinct m.date, m.productcode, p.productname 
     , f.family, cl.clientname, cl.clientcode 
     , m.quantity, m.price, m.credit, m.total 
     , m.exchange,m.state, m.city 
     , sub.clasification_value, sub.invoice_series 

from invoices.movements m  
inner join invoices.products p 
    on m.idproduct = p.idproduct 
left outer join invoices.documents d 
    on m.iddocument = d.iddocument 
inner join invoices.coins c 
    on d.idcoin = c.idcoin 
inner join invoices.address a 
    on a.iddocument = d.iddocument 
inner join invoices.clasifications f 
    on f.family='aspen' 
inner join invoices.clients cl 
    on d.idclient = cl.idclient 

inner join 
    (select subcl.idclient, subf.clasification_value 
      , subcl.client_name, subcl.client_code 
      , convert(varchar(20), subd.documents_series)+''+ 
      convert(varchar(6), subd.document_number) as invoice_series 
    from invoices.clients subcl 
    inner join invoices.clasifications subf 
     on subf.cid_clasification = cl.cid_clasifclient1 
    inner join invoices.documents subd 
     on subd.cid_client = cl.cid_client) As sub 

    on sub.idclient = cl.idclient 

或者有CTE，而不是派生表：

with sub as 
(
    select subcl.idclient, subf.clasification_value 
     , subcl.client_name, subcl.client_code 
     , convert(varchar(20), subd.documents_series)+''+ 
      convert(varchar(6), subd.document_number) as invoice_series 
    from invoices.clients subcl 
    inner join invoices.clasifications subf 
    on subf.cid_clasification = cl.cid_clasifclient1 
    inner join invoices.documents subd 
    on subd.cid_client = cl.cid_client 
) 

select distinct m.date, m.productcode, p.productname 
     , f.family, cl.clientname, cl.clientcode 
     , m.quantity, m.price, m.credit, m.total 
     , m.exchange,m.state, m.city 
     , sub.clasification_value, sub.invoice_series 

from invoices.movements m  
inner join invoices.products p 
    on m.idproduct = p.idproduct 
left outer join invoices.documents d 
    on m.iddocument = d.iddocument 
inner join invoices.coins c 
    on d.idcoin = c.idcoin 
inner join invoices.address a 
    on a.iddocument = d.iddocument 
inner join invoices.clasifications f 
    on f.family='aspen' 
inner join invoices.clients cl 
    on d.idclient = cl.idclient 
inner join sub 
    on sub.idclient = cl.idclient