我们所有使用关系数据库的人都知道(或正在学习)SQL是不同的。获得期望的结果,并有效地这样做,涉及到一个乏味的过程,其部分特征是学习不熟悉的范例,并发现一些我们最熟悉的编程模式在这里不起作用。常见的反模式是什么?


当前回答

没有使用With子句或适当的连接并依赖子查询。

反模式:

select 
 ...
from data
where RECORD.STATE IN (
          SELECT STATEID
            FROM STATE
           WHERE NAME IN
                    ('Published to test',
                     'Approved for public',
                     'Published to public',
                     'Archived'
                    ))

好: 我喜欢使用with子句使我的意图更易于阅读。

with valid_states as (
          SELECT STATEID
            FROM STATE
           WHERE NAME IN
                    ('Published to test',
                     'Approved for public',
                     'Published to public',
                     'Archived'
                    )
select  ... from data, valid_states
where data.state = valid_states.state

最好的:

select 
  ... 
from data join states using (state)
where 
states.state in  ('Published to test',
                     'Approved for public',
                     'Published to public',
                     'Archived'
                    )

其他回答

像这样将冗余表连接到查询中:

select emp.empno, dept.deptno
from emp
join dept on dept.deptno = emp.deptno;

我见过太多人死死抓着IN(…),却完全忘记了EXISTS(存在)。有关一个好例子,请参见Symfony Propel ORM。

SELECT FirstName + ' ' + LastName as "Full Name", case UserRole when 2 then "Admin" when 1 then "Moderator" else "User" end as "User's Role", case SignedIn when 0 then "Logged in" else "Logged out" end as "User signed in?", Convert(varchar(100), LastSignOn, 101) as "Last Sign On", DateDiff('d', LastSignOn, getDate()) as "Days since last sign on", AddrLine1 + ' ' + AddrLine2 + ' ' + AddrLine3 + ' ' + City + ', ' + State + ' ' + Zip as "Address", 'XXX-XX-' + Substring(Convert(varchar(9), SSN), 6, 4) as "Social Security #" FROM Users

或者,把所有内容都塞进一行。

重新使用一个“死”字段来做一些它不打算做的事情(例如在“传真”字段中存储用户数据)-尽管作为一个快速修复非常诱人!

同一查询中的相同子查询。