Question

我在SQL Server中运行以下查询，返回计数和城市：

JBHL/40/16-17

我得到以下结果：

SELECT 
    COUNT(*), city 
FROM 
    address 
WHERE
    address1 LIKE '%[0-9]%' 
    AND phone1 LIKE '%[0-9]%'
GROUP BY
    city
HAVING 
    COUNT(*) > 250
ORDER BY
    COUNT(*) DESC

如何编写查询以获取计数结果的总和（上例中的1232 + 345？）

我已经尝试了子查询，但由于该组似乎导致了一些错误。我不熟悉SQL，并为一个糟糕的问题道歉。

Answer 1

试试这个：

select sum(c)
from (
    select COUNT(*) c
    from address
    where address1 like '%[0-9]%'
        and phone1 like '%[0-9]%'
    group by city
    having count(*) > 250
    );

Answer 2

您可以定义CTE然后使用它：

WITH cte AS (
    SELECT city, COUNT(*) AS total
    FROM address
    WHERE address1 LIKE '%[0-9]%' AND
          phone1 LIKE '%[0-9]%'
    GROUP BY city
    HAVING COUNT(*) > 250
)

SELECT t.city,
       t.total,
       (SELECT SUM(total) FROM cte) AS grand_total
FROM cte t
ORDER BY t.total DESC

如果您只想要总计，请使用以下查询：

SELECT SUM(total) AS grand_total
FROM cte

Answer 3

这是使用SUM Over()窗口聚合函数

的另一种方法

SELECT city,
       Count(*) AS total,
       Sum(Count(*))OVER() grand_total
FROM   address
WHERE  address1 LIKE '%[0-9]%'
       AND phone1 LIKE '%[0-9]%'
GROUP  BY city
HAVING Count(*) > 250

Answer 4

您可以使用GROUP的ROLLUP，例如：

    ;WITH address(address1,phone1,city) AS
    (
        SELECT '1','2','Atlanta' UNION ALL 
        SELECT '2','2','Atlanta' UNION ALL 
        SELECT '3','2','Atlanta' UNION ALL 
        SELECT '4','2','Atlanta' UNION ALL 
        SELECT '5','2','Chicago' UNION ALL 
        SELECT '6','0','Chicago' UNION ALL 
        SELECT '7','0','Chicago'
    )
    SELECT COUNT(*), ISNULL(city,'Summary') FROM address where
    address1 LIKE '%[0-9]%'
    and phone1 LIKE '%[0-9]%'
    group by city WITH ROLLUP
    HAVING count(*) > 2
    order BY CASE WHEN city IS NULL THEN 1 ELSE 0 END , COUNT(*) DESC


----------- -------
4           Atlanta
3           Chicago
7           Summary

通过查询求和组的结果

4 个答案: