I have the following TSQL query:
SELECT DISTINCT MyTable1.Date
FROM MyTable1
INNER JOIN MyTable2
ON MyTable1.Id = MyTable2.Id
WHERE Name = 'John' ORDER BY MyTable1.Date DESC
It retrieves a long list of Dates, but I only need the first one, the one in the first row.
How can I get it?
Thanks a ton!
While the table name is selected type CTRL + 3 and you will notice that the query will run and will return a single row as a resultset. Now developer just has to select the table name and click on CTRL + 3 or your preferred shortcut key and you will be able to see a single row from your table.
To do that, you can use the ROW_NUMBER() function. In OVER() , you specify the groups into which the rows should be divided ( PARTITION BY ) and the order in which the numbers should be assigned to the rows ( ORDER BY ).
I want to select the last 3 rows of an sql table. I know I should use SELECT * FROM table ORDER BY DESC LIMIT 3 , but the problem with this code is that it selects the rows from the end. For example, it selects 30, then 29, then 28.
In SQL Server you can use TOP
:
SELECT TOP 1 MyTable1.Date
FROM MyTable1
INNER JOIN MyTable2
ON MyTable1.Id = MyTable2.Id
WHERE Name = 'John'
ORDER BY MyTable1.Date DESC
If you need to use DISTINCT
, then you can use:
SELECT TOP 1 x.Date
FROM
(
SELECT DISTINCT MyTable1.Date
FROM MyTable1
INNER JOIN MyTable2
ON MyTable1.Id = MyTable2.Id
WHERE Name = 'John'
) x
ORDER BY x.Date DESC
Or even:
SELECT MAX(MyTable1.Date)
FROM MyTable1
INNER JOIN MyTable2
ON MyTable1.Id = MyTable2.Id
WHERE Name = 'John'
--ORDER BY MyTable1.Date DESC
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