sql - 每天分组

Question

我想要一个选择结果，我可以在其中获得每天的结果。我计算错误，并有以下非工作声明：

SELECT message, time, personid, count(*)
FROM errorlog
WHERE time BETWEEN TO_DATE(todays date - 10 days) AND TO_DATE(todays date) AND SUBSTR(message,0,3) = 'ERR'
GROUP BY DAY(time), personid, message
ORDER BY 4

当我运行它时，它显示“DAY”：无效标识符。

Answer 1

它应该是 GROUP BY 之后的列名。

尝试：

SELECT message, DAY(time) dayTime, personid, count(*)
FROM errorlog
WHERE time BETWEEN TO_DATE(todays date - 10 days) AND TO_DATE(todays date) AND SUBSTR(message,0,3) = 'ERR'
GROUP BY dayTime, personid, message
ORDER BY 4

Answer 2

您中的任何项目GROUP BY也必须在您的SELECT.

如果您按天分组，并且显示time（我怀疑它是日期时间数据类型），并且您在哪一天有 2 条记录，那么您的结果集将如何显示这两个不同的时间？本来就不行。

以下查询使用和 的CONVERT(VARCHAR(10), time, 103)含义，即在您的结果集中显示一个格式，并且您的记录将按此分组。SELECTGROUP BYdd/MM/yyyy

SELECT message, CONVERT(VARCHAR(10), time, 103), personid, count(*)
FROM errorlog
WHERE time BETWEEN TO_DATE(todays date - 10 days) AND TO_DATE(todays date) AND SUBSTR(message,0,3) = 'ERR'
GROUP BY CONVERT(VARCHAR(10), time, 103), personid, message
ORDER BY 4

Answer 3

SQL 服务器？

纠正这个：

SELECT message, DAY(time) AS [DAY], personid, count(*)