I have to fetch data from a table in such a way that my start date to fetch data is based on the date where the last data was inserted.
For example I have data from 24/01/2011 to now. But for some specific id we have inserted the last data on 24/01/2012. In this case for that id I have to fetch the data from 14/01/2012 to 24/01/2012.
Because I don't know the last date when data was inserted , so first I have to fetch the max of date and based on that I can find the start date. Is there any fast way to do that. So everything is handled in single and fast query.
If I understand your question, answer may look like this:
SELECT *
FROM table_1 t1
WHERE table_1.start_date >= to_date('14/01/2012', 'DD-MM-YYYY')
AND table_1.start_date <=
(SELECT MAX(t11.insert_date) FROM table_1 t11 WHERE t1.id = t11.id)
The below query selects all rows where the start date is greater than or equal to the last start date - 10 days. I believe it is correct for firebird
SELECT *
FROM table t1
WHERE t1.startdate >= dateadd(-10 day to MAX(t1.startdate)
Related
I have a DateTime column (timestamp 2022-05-22 10:10:12) with a batch of stamps per each day.
I need to filter the rows where stamp is before 9am (here is no problem) and I'm using this code:
SELECT * FROM tickets
WHERE date_part('hour'::text, tickets.date_in) < 9::double precision;
The output is the list of the rows where the time in timestamp is less than 9 am (50 rows from 2000).
date_in
2022-05-22 08:10:12
2022-04-23 07:11:13
2022-06-15 08:45:26
Then I need to find all the days where at least one row has a stamp before 9 am - and here I'm stuck. Any idea how to select all the days where at least one stamp was before 9 am?
The code I'm trying:
SELECT * into temp1 FROM tickets
WHERE date_part('hour'::text, tickets.date_in) < 9::double precision
ORDER BY date_part('day'::text, date_in);
Select * into temp2
from tickets, temp1
where date_part('day'::text, tickets.date_in) = date_part('day'::text, temp1.date_in);
Update temp2 set distorted_route = 1;
But this is giving me nothing.
Expected output is to get all the days where at least one route was done before 9am:
date_in
2022-05-22 08:10:12
2022-05-22 10:11:45
2022-05-22 12:14:59
2022-04-23 07:11:13
2022-04-23 11:42:25
2022-06-15 08:45:26
2022-06-15 15:10:57
Should I make an additional table (temp1) to feed it with the first query result (just the rows before 9am) and then make a cross table query to find in the source table public.tickets all the days which are equal to the public.temp1?
Select * from tickets, temp1
where TO_Char(tickets.date_in, 'YYYY-MM-DD')
= TO_Char(temp1.date_in, 'YYYY-MM-DD');
or like this:
SELECT *
FROM tickets
WHERE EXISTS (
SELECT date_in FROM TO_Char(tickets.date_in, 'YYYY-MM-DD') = TO_Char(temp1.date_in, 'YYYY-MM-DD')
);
Ideally, I'd want to avoid using a temporary table and make a request just for one table.
After that, I need to create a view or update and add some remarks to the source table.
Assuming you mean:
How to select all rows where at least one row exists with a timestamp before 9 am of the same day?
SELECT *
FROM tickets t
WHERE EXISTS (
SELECT FROM tickets t1
WHERE t1.date_in::date = t.date_in::date -- same day
AND t1.date_in::time < time '9:00' -- time before 9:00
AND t1.id <> t.id -- exclude self
)
ORDER BY date_id; -- optional, but typically helpful
id being the PK column of your undisclosed table.
But be aware that ...
... typically you'll want to work with timestamptz instead of timestamp. See:
Ignoring time zones altogether in Rails and PostgreSQL
https://wiki.postgresql.org/wiki/Don%27t_Do_This#Don.27t_use_timestamp_.28without_time_zone.29
... this query is slow for big tables, because it cannot use a plain index on (date_id) (not "sargable"). Related:
How do you do date math that ignores the year?
There are various ways to optimize performance. The best way depends on undisclosed information for performance questions.
i have this table:
COD (Integer) (PK)
ID (Varchar)
DATE (Date)
I just want to get the new ID's from today, compared with yesterday (the ID's from today that are not present yesterday)
This needs to be done with just one query, maximum efficiency because the table will have 4-5 millions records
As a java developer i am able to do this with 2 queries, but with just one is beyond my knowledge so any help would be so much appreciated
EDIT: date format is dd/mm/yyyy and every day each ID may come 0 or 1 times
Here is a solution that will go over the base data one time only. It selects the id and the date where the date is either yesterday or today (or both). Then it GROUPS BY id - each group will have either one or two rows. Then it filters by the condition that the MIN date in the group is "today". Those are the id's that exist today but did not exist yesterday.
DATE is an Oracle keyword, best not used as a column name. I changed that to DT. I also assume that your "dt" field is a pure date (as pure as it can be in Oracle, meaning: time of day, which is always present, is 00:00:00).
select id
from your_table
where dt in (trunc(sysdate), trunc(sysdate) - 1)
group by id
having min(dt) = trunc(sysdate)
;
Edit: Gordon makes a good point: perhaps you may have more than one such row per ID, in the same day? In that case the time-of-day may also be different from 00:00:00.
If so, the solution can be adapted:
select id
from your_table
where dt >= trunc(sysdate) - 1 and dt < trunc(sysdate) + 1
group by id
having min(dt) >= trunc(sysdate)
;
Either way: (1) the base table is read just once; (2) the column DT is not wrapped within any function, so if there is an index on that column, it can be used to access just the needed rows.
The typical method would use not exists:
select t.*
from t
where t.date >= trunc(sysdate) and t.date < trunc(sysdate + 1) and
not exists (select 1
from t t2
where t2.id = t.id and
t2.date >= trunc(sysdate - 1) and t2.date < trunc(sysdate)
);
This is a general solution. If you know that there is at most one record per day, there are better solutions, such as using lag().
Use MINUS. I suppose your date column has a time part, so you need to truncate it.
select id from mytable where trunc(date) = trunc(sysdate)
minus
select id from mytable where trunc(date) = trunc(sysdate) - 1;
I suggest the following function index. Without it, the query would have to full scan the table, which would probably be quite slow.
create idx on mytable( trunc(sysdate) , id );
I want to select rows for a field MRD which is declared as date where it is prior for that date only.
So
(case when sum (transPoints) > 4 and MRD is that same date then 4
So if a row has a date of today, I want the case when to be triggered when the transaction points are bigger than 4 against all columns with the same date.
As you can imagine the date field will be different against many rows.
Based on what I can understand from your question, it seems that the GROUP BY clause may be what you're looking for. If your date column is in the correct format then you may have to use something like:
SELECT CAST(DateColumn as DATE)
FROM YourTable
GROUP BY CAST(DateColumn as DATE)
I have a table that has (among others) a timestamp column (named timestamp; it's a standard Oracle DATE datatype). The records are about 4-11 minutes apart, about 7 or 8 records every hour, and I'm trying to determine if there is any pattern to them.
Is there an easy way to see each record, and the number of minutes that record occurred after the previous record?
Thanks,
AndyDan
This is Oracle 9i+, using the LAG function to get the previous timestamp value without needing to self join:
SELECT t.timestamp - LAG(t.timestamp) OVER (ORDER BY t.timestamp) AS diff
FROM YOUR_TABLE t
...but because whole numbers represent the number of days in the result, a difference of less than 24 hours will be a fraction. Also, the LAG will return NULL if there's no earlier value -- same as if having used an OUTER JOIN.
To see minutes, use the ROUND function:
SELECT ROUND((t.timestamp - LAG(t.timestamp) OVER (ORDER BY t.timestamp)) *1440) AS diff_in_minutes
FROM YOUR_TABLE t
If the records have sequential id's you could do a self join like this:
SELECT t2.*, t2.timestamp - t1.timestamp AS timediff
FROM foo t1 inner join foo.t2 on t1.id = t2.id-1
You'd probably need to tweak this to handle the first and last records, but that's the basics.
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)?
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.
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);
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?