How do I use SQL to order results by oldest first? I am using unix timestamps.
Thanks.
The oldest UNIX timestamp is the one that's smallest, so you want to ORDER BY my_timestamp_column ASC.
I have no idea why both the answers so far have said to order by the column DESC.
Unix time, or POSIX time, is a system for describing points in time, defined as the number of seconds elapsed since midnight proleptic Coordinated Universal Time (UTC) of January 1, 1970, not counting leap seconds
The ORDER BY clause can use ASC or DESC, if you sepcify none it will default to use ASC:
Most recent time stamps first:
SELECT * FROM tableName ORDER BY columnName DESC
Less recent time stamps first:
SELECT * FROM tableName ORDER BY columnName ASC
What's the problem you're having using ORDER BY 'unix-time-stamp-field' ASC;?
EDIT: jemfinch is right, it is ASC.
In order to get oldest first data when using unix_timetamp, run this query:
Select * FROM tablename order by FROM_UNIXTIME(ts) ASC
here:
ts respond to column of the table which has unix_timestamp.
Use ORDER BY clause with DESC modifier that reverses the results:
SELECT ... FROM ... ORDER BY timestampCol DESC;
Edit
Of course you should use ASC (or none, cause ASC is default)... ;)
Related
DO we have a way to get first record considering the time.
example
get first record today, get first record yesterday, get first record day before yesterday ...
Note: I want to get all records considering the time
sample expected output should be
first_record_today,
first_record_yesterday,..
As I understand the question, the "first" record per day is the earliest one.
For that, we can use RANK and do the PARTITION BY the day only, truncating the time.
In the ORDER BY clause, we will sort by the time:
SELECT sub.yourdate FROM (
SELECT yourdate,
RANK() OVER
(PARTITION BY DATE_TRUNC('DAY',yourdate)
ORDER BY DATE_TRUNC('SECOND',yourdate)) rk
FROM yourtable
) AS sub
WHERE sub.rk = 1
ORDER BY sub.yourdate DESC;
In the main query, we will sort the data beginning with the latest date, meaning today's one, if available.
We can try out here: db<>fiddle
If this understanding of the question is incorrect, please let us know what to change by editing your question.
A note: Using a window function is not necessary according to your description. A shorter GROUP BY like shown in the other answer can produce the correct result, too and might be absolutely fine. I like the window function approach because this makes it easy to add further conditions or change conditions which might not be usable in a simple GROUP BY, therefore I chose this way.
EDIT because the question's author provided further information:
Here the query fetching also the first message:
SELECT sub.yourdate, sub.message FROM (
SELECT yourdate, message,
RANK() OVER (PARTITION BY DATE_TRUNC('DAY',yourdate)
ORDER BY DATE_TRUNC('SECOND',yourdate)) rk
FROM yourtable
) AS sub
WHERE sub.rk = 1
ORDER BY sub.yourdate DESC;
Or if only the message without the date should be selected:
SELECT sub.message FROM (
SELECT yourdate, message,
RANK() OVER (PARTITION BY DATE_TRUNC('DAY',yourdate)
ORDER BY DATE_TRUNC('SECOND',yourdate)) rk
FROM yourtable
) AS sub
WHERE sub.rk = 1
ORDER BY sub.yourdate DESC;
Updated fiddle here: db<>fiddle
I have data with millisecond precision timestamp. I want to only filter for the most recent timestamp within a given second. Ie. records (2020-07-13 5:05.38.009, event1), (2020-07-13 5:05.38.012, event2) should only retrieve the latter.
I've tried the following:
SELECT
timestamp as time, event as value, event_type as metric
FROM
table
GROUP BY
date_trunc('second', time)
But then I'm asked to group by event as well and I see all the data (as if no group by was provided)
In Postgres, you can use distinct on:
select distinct on (date_trunc('second', time)) t.*
from t
order by time desc;
I have a table with some columns and a date column (that i made a partition with)
For example
[Amount, Date ]
[4 , 2020-4-1]
[3 , 2020-4-2]
[5 , 2020-4-4]
I want to get the latest Amount based on the Date.
I thought about doing a LIMIT 1 with ORDER BY, but, is that optimized by BigQuery or it will scan my entire table?
I want to avoid costs at all possible, I thought about doing a query based on the date today, and if nothing found search for yesterday, but I don't know how to do it in only one query.
Below is for BigQuery Standard SQL
#standardSQL
SELECT ARRAY_AGG(amount ORDER BY `date` DESC LIMIT 1)[SAFE_OFFSET(0)]
FROM `project.dataset.table`
WHERE `date` >= DATE_SUB(CURRENT_DATE(), INTERVAL 1 DAY)
Note: above assumes your date field is of DATE data type.
If your date field is a partition, you can use it in WHERE clause to filter which partitions should be read in your query.
In your case, you could do something like:
SELECT value
FROM <your-table>
WHERE Date >= DATE_SUB(CURRENT_DATE(), INTERVAL 1 DAY)
ORDER BY Data DESC
LIMIT 1
This query basically will:
Filter only today's and yesterday's partitions
Order the rows by your Date field, from the most recent to the older
Select the first element of the ordered list
If the table has a row with today's date, the query will return the data for today. If it dont't, the query will return the data for yesterday.
Finally, I would like to attach here this reference regarding querying partitioned tables.
I hope it helps
The LIMIT order stops the query whet it gets the amount of results indicated.
I think the query should be something like this, I'm not sure if "today()-1" returns
SELECT Amount
FROM <table> as t
WHERE date(t.Date) = current_date()
OR date(t.Date) = makedate(year(current_date()), dayofyear(current_date())-1);
Edited: Sorry, my answer is for MariaDB I now see you ask for Google-BigQuery which I didn't even know, but it looks like SQL, I hope it has some functions like the ones I posted.
I have "tasks" table with 3 fields:
date
priority (0,1,2)
done (0,1)
What I am trying to achieve is with the whole table sorted by done flag, tasks that are not done should be sorted by priority, while tasks that are done should be sorted by date:
Select * from tasks order by done asc
If done=0 additionally order by priority desc
If done=1 additionally order by date desc
Is it possible to do this in MySQL without unions?
Thanks.
You could try ORDER BY (done asc, aux desc) where aux is computed with a CASE to yield either priority or date based on the value of done (you may have to cast them to the same type to fit in the same expression, e.g. cast the date to a suitable integer day number).
For example:
SELECT * FROM tab
ORDER BY done desc,
case done
when 0 then prio
else to_days(thedate)
end desc;
Taken from Alex Martelli just shortened a bit with IF() and fixed the ASC/DESC ordering
SELECT * FROM tab
ORDER BY done ASC, IF(done, to_days(thedate), prio) DESC;
I have a bunch of tasks in a MySQL database, and one of the fields is "deadline date". Not every task has to have to a deadline date.
I'd like to use SQL to sort the tasks by deadline date, but put the ones without a deadline date in the back of the result set. As it is now, the null dates show up first, then the rest are sorted by deadline date earliest to latest.
Any ideas on how to do this with SQL alone? (I can do it with PHP if needed, but an SQL-only solution would be great.)
Thanks!
Here's a solution using only standard SQL, not ISNULL(). That function is not standard SQL, and may not work on other brands of RDBMS.
SELECT * FROM myTable
WHERE ...
ORDER BY CASE WHEN myDate IS NULL THEN 1 ELSE 0 END, myDate;
SELECT * FROM myTable
WHERE ...
ORDER BY ISNULL(myDate), myDate
SELECT foo, bar, due_date FROM tablename
ORDER BY CASE ISNULL(due_date, 0)
WHEN 0 THEN 1 ELSE 0 END, due_date
So you have 2 order by clauses. The first puts all non-nulls in front, then sorts by due date after that
The easiest way is using the minus operator with DESC.
SELECT * FROM request ORDER BY -date DESC
In MySQL, NULL values are considered lower in order than any non-NULL value, so sorting in ascending (ASC) order NULLs are listed first, and if descending (DESC) they are listed last.
When a - (minus) sign is added before the column name, NULL become -NULL.
Since -NULL == NULL, adding DESC make all the rows sort by date in ascending order followed by NULLs at last.