Question

我正在进行此查询

SELECT zip, COUNT(*) AS amount 
FROM table
WHERE group_id = 19
GROUP BY zip ORDER BY amount DESC

这给了我一个列表，其中包含每个邮政编码的group_id 19次出现次数。但是我也想要每个zip的所有groups_id的总数（没有WHERE子句）。像这样（虚拟代码）

SELECT zip, COUNT(*) AS amount, COUNT(*) FOR THIS ZIP AS total, (amount/total * 100) AS percent
FROM table
WHERE group_id = 19
GROUP BY zip ORDER BY amount DESC

这可能吗？

Answer 1

您可以使用条件聚合：

SELECT zip, COUNT(*) as amount, SUM(group_id = 19) AS grp19_amount,
       AVG(group_id = 19) as percent
FROM table
GROUP BY zip
ORDER BY grp19_amount DESC

Answer 2

在SELECT查询中，您只能包含GROUP BY子句中包含的列或聚合函数的结果。所以COUNT（*）应该没问题，但是，（金额/总数* 100）将不起作用，除非你包括一个添加聚合函数，如SUM

在SQL查询中获取几个不同的计数

2 个答案: