每天分组

Question

我希望得到每天获得结果的选择结果。林计数错误，并具有以下非工作声明：

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 

当我运行这个它说“天时”：无效indentifier。

Answer 1

应该GROUP BY之后列名。

TRY：

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

您的GROUP BY中的任何物品也必须在您的SELECT中。

如果您按当天分组，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，那么结果集将如何显示这两个不同的时间？它将无法。

下面的查询使用在SELECT和GROUP BY意味着的dd/MM/yyyy的格式将被显示在结果集中，你的记录将被这个分组CONVERT(VARCHAR(10), time, 103)。

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 SERVER？

纠正这一点，

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