T-SQL，比较记录属性，然后比较后续日期

Question

I have a table in SQL, and it looks like this:

FOO   BAR   DateTime
---   ---   ------
FOO1  BAR1  4/25/2014
FOO2  BAR2  4/24/2014
........

to any n number of records in the above format.

I am trying to write a query that first returns all records that meet two property conditions, say all the records where FOO = FOO1 and BAR = BAR1, then with those results, return the record that has the most recent value out of the DateTime column. There can be multiple records where FOO = FOO1 and BAR = BAR1, and I am trying to get the most recent based on the datetime field value.

Answer 1

Well if I understood the request correctly (and it is SQL Server) it would be something like this:

SELECT TOP 1 FOO, BAR, DateTime
FROM Table 
WHERE FOO='FOO1' AND BAR='BAR1'
ORDER BY DateTime DESC

WHERE condition limits the set, ORDER BY sorts the set in descending order by date and TOP 1 selects the most recent record.

Answer 2

SELECT FOO, BAR, Datetime FROM table WHERE Foo = 'FOO1' and Bar = 'BAR1' 
and DateTime = (SELECT MAX(DATETIME) from table WHERE Foo = 'FOO1' and Bar = 'BAR1')

Probably this...or any other answer...haha

Answer 3

This should give you the answer in almost any dialect of SQL

select FOO, BAR, MAX(DateTime) from 
table_name
where 
FOO = 'FOO1'
and BAR = 'BAR1'
group by FOO, BAR

Answer 4

As you said you need just one data , the following query should do it. Since you have tagged your question with MySql so its mysql solution.

select FOO, BAR, MAX(DateTime) from 
table
where 
FOO = 'FOO1'
and BAR = 'BAR1'