仅对具有多行的组的值进行求和

Question

我有这个问题：

Font Book.app

结果如下：

只有当有多行SELECT extract(year from date1), extract(month from date1), spending FROM ( SELECT *, COUNT(*) OVER(PARTITION BY CONCAT(extract(year FROM date1), extract(month FROM date1))) N FROM table) as A WHERE N > 1 GROUP BY date1 ORDER BY date1 ASC; 和spending时，我才需要对字段year求和。期望的结果：

month

Answer 1

好的，我找到了解决方案：HAVING：

SELECT extract(year from date1), extract(month from date1), spending    
FROM table  
GROUP BY extract(month from date1)), extract(year from date1), extract(month from date1)
HAVING count (CONCAT(extract(year from date1), extract(month from date1))) > 1
ORDER BY extract(year from date1), extract(month from date1) ASC; 

以防万一。

Answer 2

使用date_trunc()和一些简化可以更简单，更快捷：

SELECT date_trunc('month', date1)::date AS month
     , sum(spending)                    AS sum_spending
     , count(*)                         AS count_rows  -- optional addition
FROM   table
GROUP  BY 1
HAVING count(*) > 1
ORDER  BY 1;

仅返回多行的支出总和。

如果您需要显示单独的年份和月份数字，您可以在子查询中使用上述查询，但速度更快：

SELECT extract(year  FROM month)::int AS year
     , extract(month FROM month)::int AS month
     , sum_spending, count_rows
FROM (
   SELECT date_trunc('month', date1)::date AS month
        , sum(spending)                    AS sum_spending
        , count(*)                         AS count_rows  -- optional
   FROM   table
   GROUP  BY 1
   HAVING count(*) > 1
   ORDER  BY 1
   ) sub;

或者直接在解决方案中提取数字，但只需在count(*)子句中使用更快HAVING：

SELECT extract(year  FROM date1)::int AS year
     , extract(month FROM date1)::int AS month
     , sum(spending)                  AS sum_spending
     , count(*)                       AS count_rows  -- optional
FROM   table
GROUP  BY 1, 2
HAVING count(*) > 1
ORDER  BY 1, 2;

1, 2是（完全可选的）位置引用以缩短语法，因此我们不必重复SELECT列表中的表达式。例如：

• Select first row in each GROUP BY group?

强制转换为整数（::int）也是可选的。提取的泛型返回类型是双精度，但年和日期可以安全地转换为整数。更小，更快，更充足。

Answer 3

试试这个

SELECT extract(year from date1), extract(month from date1), sum(spending)    
FROM ( SELECT *, COUNT(*) OVER(PARTITION BY CONCAT(extract(year FROM date1), extract(month FROM date1))) N
       FROM table) as A    
WHERE N > 1    
GROUP BY extract(year from date1),extract(month from date1)
ORDER BY extract(year from date1),extract(month from date1) ASC;