Question

我需要一些关于sql-joins的帮助。我有3个select语句，它给我一个这样的输出：

select-statement A:
location amount_A
7234     17
7456     2

select-statement B:
location number_x
7234     4455
7456     555

select-statement C:
location errors
7234     1
7456     44537

我希望将结果放在一个表中，如下所示：

location Amount_A  number_x  errors
7234     17        4455      1
7456     2         555       44537

实现这一目标的最佳和/或最简单的方法是什么？每个select语句都使用其他表。？！

这些是陈述：

A: select substring(column_a for 4) location, count(*) Amount_A from table_a where column_a like '7%' group by location  ;
B: select substring(e.column_xy for 4), count(*) number_x from table_b b, table_e e , table_c c where b.stationextension_id = e.id and b.id = c.id and ( c.column_h in ( 'value_a', 'value_b' ) ) group by substring(e.column_xy for 4)  ;
C: select substring(name from 1 for 4), count(*) from errors group by substring(name from 1 for 4) ;

Answer 1

使用Inner Join连接所有三个查询。

还使用Inner Join语法连接两个表，这些表比旧式逗号分隔连接更具可读性。

Substring函数有一些不同的参数希望那些是原始查询中的例子，你需要将适当的值传递给substring函数

select a.location,Amount_A ,number_x,errors from
(
SELECT substring(column_a for 4) location, count(*) Amount_A FROM table_a WHERE column_a LIKE '7%' GROUP BY location  
) A 
inner join
(
SELECT substring(e.column_xy for 4) location , count(*) number_x FROM table_b b inner join table_e e on  b.stationextension_id = e.id and b.stationextension_id = e.id 
inner join  table_c c on  b.id = c.id 
where c.column_h IN ( 'value_a', 'value_b' ) ) GROUP BY substring(e.column_xy FOR 4)  ;
) B on a.location = b.location 
inner join 
(
SELECT substring(name from 1 FOR 4) location, count(*) errors FROM errors GROUP BY substring(name FROM 1 FOR 4) ;
)
C on c.location = b.location

Answer 2

您可以将所有三个查询合并为一个

identical(coef(m1), coef(m2))
## [1] TRUE
identical(vcov(m1), vcov(m2))
## [1] TRUE

Answer 3

试试这个：

select 
  a.location, a.amount_A,
  b.number_x,
  c.errors
from (
  select_satement_A
) as a
join (
  select_satement_B
) as b on a.location = b.location
join (
  select_satement_C
) as c on a.location = c.location

如果select_satement_A的所有数据都需要retrieved使用left join。将使用select_satement_A和select_satement_B中的相应数据检索select_satement_C中的所有数据。如果找不到与join条件匹配的null将被替换。

   select 
      a.location, a.amount_A,
      b.number_x,
      c.errors
    from (
      select_satement_A
    ) as a
    left join (
      select_satement_B
    ) as b on a.location = b.location
    left join (
      select_satement_C
    ) as c on a.location = c.location

对于要检索的所有数据，请使用full join。

Answer 4

您总是可以将一些语句与一些左外连接组合在一起。我认为地点可能没有任何错误。

如果没有更好地理解基础数据，以正确的顺序获取连接有点棘手，但我希望这能指向正确的方向。

select  substring(e.column_xy for 4) location, 
        count(a.column_xy)           Amount_A, 
        count(e.ie)                  number_x, 
        count(e2.name)               errors
  from  table_b b
  inner join  table_e  e on b.stationextension_id        = e.id
  inner join  table_c  c on b.id                         = c.id
  left  join  table_a  a on substring(e.column_xy for 4) = substring(a.column_a from 1 for 4)
                        and a.column_a                like '7%'
  left  join  errors  e2 on substring(e.column_xy for 4) = substring(e2.name    from 1 for 4)
  where c.column_h in ( 'value_a', 'value_b' ) 
  group by substring(e.column_xy for 4);

SQL连接在不同的选择计数结果表上

4 个答案: