Question

我想从表中获取两个不同的值。我想获得，重复的重复数量，并希望获得总数。

WITH contrato_group AS 
(
    SELECT CONTRATO as u_contract, COUNT(*) AS count
    FROM `table` 
    GROUP BY CONTRATO
)
SELECT 
    COUNT(DISTINCT u_contract) AS at_least_one_count 
FROM
    contrato_group
WHERE 
    count > 1

这没关系，它至少返回一个合同计数。但我也想从原始表中选择总数。

SELECT COUNT(CONTRATO) as total_contract 
FROM `table`

我试过，把他们联合起来。

WITH contrato_group AS
(
    SELECT CONTRATO AS u_contract, COUNT(*) AS count
    FROM `table` 
    GROUP BY CONTRATO
) 
SELECT 
    COUNT(DISTINCT u_contract) AS at_least_one_count 
FROM
    contrato_group
WHERE 
    count > 1

UNION ALL

SELECT 
    COUNT(CONTRATO) AS total_contracts 
FROM
    `table`

它返回了我想要的内容，但这些值只在一列中。像这样：

Row     at_least_one_count   
--------------------------
  1     83084    
  2     22894

我想得到这样的结果：

Row   at_least_one_count    total_count
----------------------------------------
 1         22894                83084

Answer 1

尝试子选择

with contrato_group as (SELECT CONTRATO as u_contract, count(*) as count
FROM `table` 
GROUP BY CONTRATO) 

SELECT count(distinct u_contract) as at_least_one_count,
(SELECT count(CONTRATO) from `table`) as total_count
from contrato_group
WHERE count > 1

Answer 2

如果您的 DBMS 支持，您也可以通过 window 函数在单个语句中执行：

select t.*
from (select distinct contrato as u_contract, 
             count(*) over (partition by contrato) as at_least_one_count,  
             count(*) over () as total_count
      from table 
     ) t
where at_least_one_count  > 1;

Answer 3

这样的事情怎么样。

with   contrato_group as (SELECT CONTRATO as u_contract, count(*) as count
       FROM test_table
       GROUP BY CONTRATO)
SELECT count(case when count > 1 then 1 end) as duplicate_count, 
       sum(count) as total_count
from   contrato_group;

Here's a SQLFiddle for you to play around with

SQL，在一个SQL查询中从不同的表中获取行

3 个答案: