I have a table which looks like that:
As You see, there are some date duplicates, so how to select only one row for each date in that table?
the column 'id_from_other_table' is from INNER JOIN with the table above
2
-
2What database, and what version?LittleBobbyTables - Au Revoir– LittleBobbyTables - Au Revoir2010-09-12 15:40:37 +00:00Commented Sep 12, 2010 at 15:40
-
Hello Tony, I found this to be useful when I was trying to do a similar thing. It does not answer your question and remove the duplicates but it helps give a count which might be relevant: COUNT(*) - COUNT(DISTINCT name) AS 'duplicate name'JPK– JPK2016-05-04 10:27:09 +00:00Commented May 4, 2016 at 10:27
Add a comment
|
7 Answers
There are multiple rows with the same date, but the time is different. Therefore, DISTINCT start_date will not work. What you need is: cast the start_date to a DATE (so the TIME part is gone), and then do a DISTINCT:
SELECT DISTINCT CAST(start_date AS DATE) FROM table;
Depending on what database you use, the type name for DATE is different.
Comments
Do you need any other information except the date? If not:
SELECT DISTINCT start_date FROM table;
2 Comments
LittleBobbyTables - Au Revoir
The dates are the same, but the times are different; as such,
DISTINCT won't work on start_date without some formatting
Monty _s Flying Circus
What if we need the other information? I will try to remember to post a query when I get it
You mention that there are date duplicates, but it appears they're quite unique down to the precision of seconds.
Can you clarify what precision of date you start considering dates duplicate - day, hour, minute?
In any case, you'll probably want to floor your datetime field. You didn't indicate which field is preferred when removing duplicates, so this query will prefer the last name in alphabetical order.
SELECT MAX(owner_name),
--floored to the second
dateadd(second,datediff(second,'2000-01-01',start_date),'2000-01-01') AS StartDate
From MyTable
GROUP BY dateadd(second,datediff(second,'2000-01-01',start_date),'2000-01-01')
1 Comment
Tony
I consider year, month, date, and hh:mm:ss
Select Distinct CAST(FLOOR( CAST(start_date AS FLOAT ) )AS DATETIME) from Table
Comments
If you want to select any random single row for particular day, then
SELECT * FROM table_name GROUP BY DAY(start_date)
If you want to select single entry for each user per day, then
SELECT * FROM table_name GROUP BY DAY(start_date),owner_name
Comments
here is the solution for your query returning only one row for each date in that table here in the solution 'tony' will occur twice as two different start dates are there for it
SELECT * FROM
(
SELECT T1.*, ROW_NUMBER() OVER(PARTITION BY TRUNC(START_DATE),OWNER_NAME ORDER BY 1,2 DESC ) RNM
FROM TABLE T1
)
WHERE RNM=1
2 Comments
Dharmesh Porwal
hi @zessx how you edited it ,i put the formatted code as you edited but after posting,it did't appear as it is now ,how u do it ???
zessx
Pasting formatted code is not enough (in fact, it does nothing). You need to paste your code, select it and use
Ctrl+K (or click on the {} button in the StackOverflow editor). You also can do it manually, adding 4 spaces before each line of your code.You have to convert the "DateTime" to a "Date". Then you can easier select just one for the given date no matter the time for that date.
