不带分组依据的SQL max

Question

I would like to get one row with the maximum date. I cannot use group by as I need to retrieve all data in that row.

I have this:

ID     Date          Country
1      05/05/2019    US
2      05/06/2019    UK

I want to get this:

ID     Date          Country
2      05/06/2019    UK

I've tried the below but it didn't work for me

select TOP 1 ID, Date, country
from table
order by Date desc

Answer 1

I don't believe you. Here is a db<>fiddle that shows three different interpretations of the date in your sample data:

• as a string
• as mm/dd/yyyy
• as dd/mm/yyyy

All three of them produce the same result.

I suspect that your actual data is more complicated and you have oversimplified the example for the question. Further, my suspicion is that the date column is stored as a string rather than a date .

As a string, you might have some hidden characters that affect the sorting (such as leading spaces).

If this is the case, fix the data type and your code will work.

Answer 2

你可以试试这个吗？

select Top 1 ID,Date, country from table where date = max(date)

Answer 3

This depends on what DB system you are using.

In Microsoft SQL server, you can use row_number() function:

select top 1 *
from facts
order by ROW_NUMBER() over (order by dateKey)

Answer 4

首先在[Date]列中设置DATE or DATETIME数据类型，然后尝试以下代码：

SELECT TOP 1 ID, [Date] , country FROM TableName ORDER BY Date DESC