I have a database with a table containing the history of the login times of all accounts. The table includes columns USERID and LOGIN_DATE. I want to find those users who have not logged in for over 60 days, so an SQL query that says
Find users who have a login date which was greater than 60 days ago, but have no entry for any date more recently than 60 days ago
Can anyone suggest how I would do this ?
Following can be a solution
select USERID, min(login_date) mind, max(login_date) maxd
from Logins
group by UserId
having max(login_date) < dateadd(d,-60,getdate())
You could use aggregation, and filter on users whose maximum login date is older than 60 days:
select userid
from mytable
group by userid
having max(login_date) < current_date - interval '60' day
You did not tell which database you are using, so this uses standard date arithmetics. You might need to adapt that to your actual database (all major databases have alternatives for this).
Now the question was tagged Oracle. The above would work; you might want to truncate the time portion of the date to check on entire days. And if you want to display the last login, just add the aggregate function to the select clause:
select userid, max(login_date) last_login_date
from mytable
group by userid
having max(login_date) < trunc(current_date) - interval '60' day
You can use aggregation:
select userid
from t
group by userid
having max(logindate) < trunc(sysdate) - interval '60' day;
Date/time functions are notoriously database-specific, so the exact syntax for the having clause might depend on your database.
Related
Hello I have table "os_txn.pay_link" and inside there are many columns.
What I want to do is that I want to count the row numbers by looking at "merchant_id" column for the current day.
So for example what I am looking for an output is that today one of "merchant_id" has
"8" rows. So I want to know the number of rows of the "merchant_id" column for current day.
I think I should use count(*) in view with select statement but couldnt succeed about syntax. So I am open your suggestions thank you.
If I understood you correctly, a simple option would be
select merchant_id, count(*)
from os_txn.pay_link
where date_column = trunc(sysdate)
group by merchant_id;
presuming that date_column contains date only (i.e. for today, 8th of October 2022, that's its value - no hours, minutes or seconds).
If date column contains time component, again - a simple option - would be
select merchant_id, count(*)
from os_txn.pay_link
where trunc(date_column) = trunc(sysdate)
group by merchant_id;
If there's an index on date_column, then such a code wouldn't use it (unless it is a function-based index) so you'd rather modify it to
where date_column >= trunc(sysdate)
and date_column < trunc(sysdate + 1)
If that's not it, do post sample data and desired result.
I'm looking through login logs (in Netezza) and trying to find users who have greater than a certain number of logins in any 1 hour time period (any consecutive 60 minute period, as opposed to strictly a clock hour) since December 1st. I've viewed the following posts, but most seem to address searching within a specific time range, not ANY given time period. Thanks.
https://dba.stackexchange.com/questions/137660/counting-number-of-occurences-in-a-time-period
https://dba.stackexchange.com/questions/67881/calculating-the-maximum-seen-so-far-for-each-point-in-time
Count records per hour within a time span
You could use the analytic function lag to look back in a sorted sequence of time stamps to see whether the record that came 19 entries earlier is within an hour difference:
with cte as (
select user_id,
login_time,
lag(login_time, 19) over (partition by user_id order by login_time) as lag_time
from userlog
order by user_id,
login_time
)
select user_id,
min(login_time) as login_time
from cte
where extract(epoch from (login_time - lag_time)) < 3600
group by user_id
The output will show the matching users with the first occurrence when they logged a twentieth time within an hour.
I think you might do something like that (I'll use a login table, with user, datetime as single column for the sake of simplicity):
with connections as (
select ua.user
, ua.datetime
from user_logons ua
where ua.datetime >= timestamp'2018-12-01 00:00:00'
)
select ua.user
, ua.datetime
, (select count(*)
from connections ut
where ut.user = ua.user
and ut.datetime between ua.datetime and (ua.datetime + 1 hour)
) as consecutive_logons
from connections ua
It is up to you to complete with your columns (user, datetime)
It is up to you to find the dateadd facilities (ua.datetime + 1 hour won't work); this is more or less dependent on the DB implementation, for example it is DATE_ADD in mySQL (https://www.w3schools.com/SQl/func_mysql_date_add.asp)
Due to the subquery (select count(*) ...), the whole query will not be the fastest because it is a corelative subquery - it needs to be reevaluated for each row.
The with is simply to compute a subset of user_logons to minimize its cost. This might not be useful, however this will lessen the complexity of the query.
You might have better performance using a stored function or a language driven (eg: java, php, ...) function.
I want to display the amount of data by month and year. This is an example of displaying data by date:
select count(*) from db.trx where trxdate = to_date('2018-04-23','yyyy-mm-dd')
When I try to display the amount of data by month and year, no query results appear. Is there something wrong with the query?
The query:
select count(*) from db.trx where trxdate = to_date('2018-04','yyyy-mm')
You need to apply the function to trxdate. Using your logic:
SELECT Count(*)
FROM olap.trxh2hpdam
WHERE To_char(trxdate, 'YYYY-MM') = '2018-04';
However, I strongly recommend that you use direct date comparisons:
WHERE trxdate >= date '2018-04-01'
AND
trxdate < date '2018-05-01'
This will allow the database to use an index on trxdate.
There are a couple of ways of accomplishing what you're trying to do. Which one works for you will depend on your database design (for example, the indexes you've created). One way might be this:
SELECT COUNT(*) FROM olap.trxh2hpdam
WHERE TRUNC(trxdate, 'MONTH') = DATE'2018-04-01';
This will round the date down to the first of the month (and, of course, remove any time portion). Then you simply compare it to the first of the month for which you want the data. However, unless you have an index on TRUNC(trxdate, 'MONTH'), this may not be the best course of action; if trxdate is indexed, you'll want to use:
SELECT COUNT(*) FROM olap.trxh2hpdam
WHERE trxdate >= DATE'2018-04-01'
AND trxdate < DATE'2018-05-01';
There are a number of functions at your disposal in Oracle (e.g. ADD_MONTHS()) in the event that the date you use in your query is supposed to be dynamic rather than static.
Just FYI, there is no reason not to use ANSI date literals when trying to retrieve data by day as well. I'm not sure your original query is a good example of getting data for a particular day, since the Oracle DATE datatype does at least potentially include a time:
SELECT COUNT(*) FROM olap.trxh2hpdam
WHERE trxdate >= DATE'2018-04-23'
AND trxdate < DATE'2018-04-24';
or:
SELECT COUNT(*) FROM olap.trxh2hpdam
WHERE TRUNC(trxdate) = DATE'2018-04-23';
EDIT
In case the month and year are dynamic, I would build a date from them (e.g., TO_DATE('<year>-<month>-01', 'YYYY-MM-DD')) and then use the following query:
SELECT COUNT(*) FROM olap.trxh2hpdam
WHERE trxdate >= TO_DATE('<year>-<month>-01', 'YYYY-MM-DD')
AND trxdate < ADD_MONTHS( TO_DATE('<year>-<month>-01', 'YYYY-MM-DD'), 1 );
Hope this helps.
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 );
What's the best way to select only those rows from the table that have been created in last 7 days?
There are dozens of time and date functions in MySQL and I'm a little bit confused about what's the easiest way to do this.
For the sake of this question, assume that you have a table called "my_table" and it contains a row "created_at" which is a DATETIME.
SELECT * FROM my_table WHERE ...
What would you fill in the WHERE clause?
WHERE DATEDIFF(NOW(), created_at) <= 7;
I like it because it reads: "Where the Difference in Date between Now and when it was created is at most 7 (days)." in my own head
...WHERE created_at >= Date_Add(now(), INTERVAL -7 DAY)
This is my preferred way because it's so...clear. But ADDDATE is fine too (and you can use the INTERVAL form with that for clarity as well; its default is days so you see people leaving it off). You don't want to do a calculation on created_at and compare it to now() because that requires the computation on created_at on each row (assuming MySQL doesn't optimise it out), whereas modifying now() and comparing to an unmodified created_at means MySQL does that bit once and uses the result when comparing against rows, not to mention indexes.
...... WHERE created_at >= DATE_SUB(CURRENT_DATE, INTERVAL 7 DAY)
hopefully that will help
WHERE ADDDATE(datefield, 7) > NOW();
SELECT * FROM my_table
WHERE DATE(created_at) >= SUBDATE(DATE(NOW()), 7)
SELECT * FROM my_table WHERE my_table.datefield > sysdate - 7