Question

我在Postgres数据库中找到了下表：

表示例：

---------------------------
   name     |    number
---------------------------
DefaultName |       1
DefaultName |       2
DefaultName |       3
DefaultName |       4
Charlie     |       1
Charlie     |       3
Charlie     |       4
Charlie     |       5
Amanda      |       2
Amanda      |       3
Amanda      |       4
Amanda      |       5

我需要获取＆＃39;默认名称＆＃39;中存在的＆＃34;数字＆＃34;但是每个＆＃34;名称＆＃34; s中不存在＆＃39; defaultName中＆＃39 ;. 在这种情况下，它会返回：

---------------------------
   names    |    numbers
---------------------------
Charlie     |       2
Amanda      |       1

我正在尝试像下面这样的左连接，但是我无法找到一种方法来将DefaultName数字与其他名称相交并且＃39; ...

SELECT Test_Configs.name, Default_Configs.number

FROM Example AS Test_Configs  
    LEFT JOIN Example AS Default_Configs      
        ON Default_Configs.name = 'DefaultName'

Answer 1

我会为每个名称生成整个范围，并为基表生成LEFT JOIN以消除当前名称：

SELECT n.name, nr.number
FROM  (
   SELECT DISTINCT name
   FROM   example
   WHERE  name <> 'DefaultName'
   ) n                         -- all names except 'DefaultName'
CROSS  JOIN (
   SELECT number               -- assuming distinct numbers for 'DefaultName'
   FROM   example
   WHERE  name = 'DefaultName'
   ) nr                        -- combine with numbers from 'DefaultName'
LEFT   JOIN example x USING (name, number)
WHERE  x.number IS NULL;       -- minus existing ones

仅列出每个名称的间隙：

SELECT n.name, nr.number
FROM  (
   SELECT name, min(number) AS min_nr, max(number) AS max_nr
   FROM   example
   GROUP  BY 1
   ) n
 , generate_series(n.min_nr, n.max_nr) AS nr(number)
LEFT   JOIN example x USING (name, number)
WHERE  x.number IS NULL;

SQL Fiddle.

以下是排除另一个表中存在的行的基本技术（本例中的派生表）：

Select rows which are not present in other table

Answer 2

需要几次传递，选择默认值，组名不是默认值，然后是左连接并检查空值。查看SQLFiddle示例。

select Names.name, DefaultConfigs.number
from Example DefaultConfigs
  cross join (
    select name
    from Example
    where name != 'DefaultName'
    group by name
    ) Names
  left join Example Missing on Missing.name = Names.name
    and Missing.number = DefaultConfigs.number
where DefaultConfigs.name = 'DefaultName'
  and Missing.name is null
;

Answer 3

构建名称和默认号码的所有组合。然后删除那些在场的人。

select othernames.name, defaultnumbers.number
from (select number from example where name = 'DefaultName') defaultnumbers
cross join (select distinct name from example where name <> 'DefaultName') othernames
except
select name, number from example;

使用否定子句加入

3 个答案: