SQL如何计算每天相同事件的数量

我有一张有日期和问题的桌子。它看起来像：

Date                    | Issue
2018-01-01 09:32:33.000 | Problem 1
2018-01-01 19:47:11.000 | Problem 1
2018-01-01 14:25:16.000 | Problem 2
2018-01-02 11:07:20.000 | Problem 1
2018-01-02 19:18:51.000 | Problem 2
2018-01-03 20:45:41.000 | Problem 1
2018-01-03 04:27:56.000 | Problem 1
2018-01-03 15:27:56.000 | Problem 2
2018-01-03 09:27:56.000 | Problem 2
2018-01-03 22:27:56.000 | Problem 3

我想计算每天每个问题的总数量。我希望它看起来像：

Date        | Issue       | Amount
2018-01-01  | Problem 1   | 2
2018-01-01  | Problem 2   | 1
2018-01-02  | Problem 1   | 1
2018-01-02  | Problem 2   | 1
2018-01-03  | Problem 1   | 2
2018-01-03  | Problem 2   | 2
2018-01-03  | Problem 3   | 1

现在我有了这个，但显然它不起作用：

SELECT CONVERT(VARCHAR, CONVERT(DATE TIME, Date), 23) AS Date,
       Issue,
       COUNT(Date) AS Amount
FROM Issues
GROUP BY Date, Issue
ORDER BY Date, Issue

提前感谢。

你很接近了：

SELECT CONVERT(date,[Date]) AS [Date], --I suggest a better name for your column, as that's confusing
       Issue,
       COUNT(Date) AS Amount
FROM Issues
GROUP BY CONVERT(date,[Date]), Issue
ORDER BY [Date], Issue; --[Date] here will reference the alias, rather than the column

---

您还需要

组中的
表达式。
非常感谢！现在效果很好。