SQL: need only 1 row per particular timestamp

Question

i have some SQL code that is inserting values from another (non sql-based) system. one of the values i get is a timestamp.

i can get multiple inserts that have the same timestamp (albeit different values for other fields).

my problem is that i am trying to get the first insert happening every day (based upon timestamp) since a particular day (i.e. give me the first insert of each day since January 28, 2007...)

my code to get the first timestamp of every day is as follows:

SELECT MIN(my_timestamp) AS first_timestamp
FROM my_schema.my_table
WHERE my_col1 = 'WHATEVER'
AND my_timestamp > timestamp '2010-Jul-27 07:45:24' - INTERVAL '365 DAY'
GROUP BY DATE (my_timestamp);

This delivers me the list of times available. But when I join against these times, I can get several rows, as there are lots of rows that mach these times. So for 365 days, I may get 5,000 rows (I could be inserting 100 rows at 00:00:00 every day).

Assuming, in the example above, my_table has columns my_col1 and my_col2, how can I get exactly 365 rows that contain my_col1 & my_col2? it doesn't matter which row i get back if there are multiple rows for a date; any row will suffice.

it's an odd question. the overall problem is: given a timestamp, how can one get 1-row-per-timestamp even if there are multiple rows that have said timestamp (assuming there is no other priority)?

thanks for the help in advance.

EDIT: So, let's say for example, this table has the following columns: my_col1, my_col2, and my_timestamp.
Here are example values (in order of my_col1 - my_col2 - my_timestamp):

• 'my_val1' - 10 - '2010-07-01 01:01:01'
• 'my_val2' - 11 - '2010-07-01 01:01:01'
• 'my_val3' - 12 - '2010-07-01 01:01:01'
• 'my_val4' - 13 - '2010-07-01 01:01:02'
• 'my_val5' - 14 - '2010-07-02 01:01:01'
• 'my_val6' - 15 - '2010-07-02 01:01:01'
• 'my_val7' - 16 - '2010-07-03 01:01:01'

in the end, i would want only 3 rows, 1 with a timestamp with '2010-07-01 01:01:01', one with '2010-07-02 01:01:01', and one with '2010-07-03 01:01:01'. the third one is easy, since there is only 1 row with that last timestamp. but the first two are the tricky ones. the sql i posted above will ignore the row with 'my_val4'.

i need a query that will return me all of the columns, not just the dates.

how would i get sql to give me either the first or last of the values that would match that timestamp (it doesn't matter either way. i just need to get 1-per first-day's timestamp matching)?

Answer 1

The solution is to use the SQL's DISTINCT statement (http://www.sql-tutorial.com/sql-distinct-sql-tutorial/):

SELECT DISTINCT MIN(my_timestamp) AS first_timestamp FROM my_schema.my_table WHERE my_col1 = 'WHATEVER' AND my_timestamp > timestamp '2010-Jul-27 07:45:24' - INTERVAL '365 DAY' GROUP BY DATE (my_timestamp);

Answer 2

select distinct on (date(my_timestamp)) *
from my_table
order by date(my_timestamp), my_timestamp

This selects all columns, exactly one row per date(my_timestamp). The single row per day is the first row for the group, as determined by order by (so that's the row with minimal my_timestamp).

Of course you can add whatever joins, wheres etc. you need. But this is the stub you're looking for.

Answer 3

I know you already have an answer, but I still don't understand why you have mentioned a join in your question. Why not just include the rest of the columns in your query, like this:

SELECT MIN(my_timestamp) AS first_timestamp, my_col1, my_col2
FROM my_table
GROUP BY DATE(my_timestamp);

This works in MySQL. Does it not return the expected result in PostgreSQL?