How to count the numbers of rows within a certain days - mysql

I would like to perform a query that counts the number of rows that have a close date within 60 days of open_date by store.
Sample: Table A
Store Open Date Close Date
A 2017-01-01 2017-01-31
B 2017-02-02 Null
A 2017-01-02 2018-01-21
Thanks in advance.

Use datediff()
SELECT count(*) FROM [Table A] WHERE datediff(Close_date,Open_date) <= 60;

Try this on of the queries below:
SELECT count(*) numberOfDays
FROM TableA
WHERE close_date>=CURRENT_DATE - INTERVAL 60 DAY;
SELECT count(*) numberOfDays
FROM TableA
WHERE close_date>=CURRENT_DATE - INTERVAL '60' DAY;
See it work on SQL Fiddle.

select store, open_date, count(*)
from tableA
where close_date between date_sub(open_date, interval 60 DAY) and date_add(open_date, interval 60 DAY)
group by store, open_date;
This will work but your starting date has to either be current date or the open_date (which is what your question was). Given that the open_date appears to be different for each row, your count is going to be ... interesting.
Also, count will always return something, so unless you use an if or case statement keying on 0 (zero) it will return a 0 not a null.

Related

In mysql how to get users that have no event X in past 14 days and neither in the future

Consider I have the following table and current date is 2022-09-01:
Requirement: I want to get all users that have no event_name like cbt care in the past 14 days and onwards into the future.
I have this query:
SELECT * FROM test_table
WHERE event_name LIKE "%cbt care%"
AND start_date <= DATE_SUB(NOW(), INTERVAL 14 DAY)
;
Which returns:
The issue is that user_id = x does have a cbt care event in 2022-09-10 which is 9 days ahead of current date (2022-09-01).
How to return only users satisfy requirement posted above?
SELECT user_id,
COUNT(CASE WHEN event_name LIKE '%cbt care%' AND start_date
> CURDATE() - INTERVAL 14 day THEN 1 END) AS count_recent
FROM test_table
GROUP BY user_id
HAVING count_recent = 0;
https://www.db-fiddle.com/f/64j7L1VZsVdLYqmcQ2NrvV/0
The CASE expression returns 1 for each row with the conditions you described (a specific event name and a start date after the date 14 days ago, which includes all of the future dates too). For rows that don't match that condition, the CASE returns NULL. There's an implicit ELSE NULL in any CASE expression.
COUNT(<expr>), like many set functions, ignores NULLs. It will only count the occurrences of non-NULL values. So if the count returns 0, then the CASE returned only NULLs, which means there are no recent or future 'cbt care' events for that user.
select id
,user_id
,event_name
,start_date
from (
select *
,count(case when abs(datediff(curdate(),start_date)) <= 14 and event_name like "%cbt care%" then 1 end) over (partition by user_id) as cw
from t
) t
where cw = 0
id
user_id
event_name
start_date
0
a
cbt care
2022-06-01 20:00:00
Fiddle

SQL Query by created 3 hours ago

Given a set of data with date_created stored like 2017-04-13 23:29:52, how would I construct an SQL query to select all items that were created within the last 3 hours?
I originally thought to do something like this:
SELECT
*,
MAX(date_created)
FROM items
GROUP BY date_created
but that would not be exactly what I want. I'm not sure how to go about this.
Use NOW() and INTERVAL in your WHERE clause
SELECT * FROM items WHERE date_created <= NOW() - INTERVAL 180 minute AND date_created >= NOW() - INTERVAL 210 minute
This one uses CURDATE, CURDATE and DATE_ADD:
SELECT *
FROM items
WHERE DATE(date_created) = CURDATE()
AND TIME(date_created) BETWEEN CURTIME() AND DATE_ADD(CURTIME(), INTERVAL -3 HOUR)
select * from items where
extract(hours from age(current_timestamp, date_created))>=3;
Extract keyword would extract hours difference from current timestamp and return only that is greater than or equal to 3.

MySql - get days remaining

Users can sign up for a premium listing for a specified number of days, e.g. 30 days.
tblPremiumListings
user_id days created_date
---------------------------------
1 30 2013-05-21
2 60 2013-06-21
3 120 2012-06-21
How would I select records where there are still days remaining on a premium listing.
SELECT *
FROM tblPremiumListings
WHERE created_date + INTERVAL `days` DAY >= CURDATE()
It's easiest to read with INTERVAL
select *
from tblPremiumListings
where created_date + interval days day >= now();
But I would also change the table to instead of created_date and days instead store end_date. That way the query is
select *
from tblPremiumListings
where end_date >= now();
The benefit of doing like this is that you can put an index on end_date and quickly find all ended premium listings, with your original table you'll always have to do a full table scan to find the records with expired listing.
SELECT * FROM
tblPremiumListings
WHERE (DATEDIFF(NOW(), created_date) - days) <= 0
See if it solves your problem
One way to get the result:
SELECT t.user_id
, t.days
, t.created_date
FROM tblPremiumListings t
WHERE t.created_date + INTERVAL t.days DAY > DATE(NOW())
You may want a >= comparison operator (instead of >) depending on how you define days remaining.
NOTE: the access plan for this query will be full scan of all rows, since MySQL won't be able to do a range scan on an index. For large sets, having a column that can be indexed to satisfy the query may improve performance, e.g.
WHERE t.expire_date > DATE(NOW())
Try this one...
SELECT * FROM
tblPremiumListings
WHERE DATE_ADD(created_date, days)>DATE(NOW())

MySQL COUNT for days

I want to get the value of users visiting my page for 10 days in a chart. I need to COUNT() all the values from the last ten days.
The best layout would be
Day|COUNT(ip)
1 - 10
2 - 12
3 - 52
......
I hope you understand what I mean.
Can MySQL do this directly or need I to do this in PHP in 10 seperate querys?
Regards,
Moritz
Update with Tablestructure:
Id (Auto Increment)|Time (Unix Timestamp)|Ip|Referer
This should run fast for you
SELECT COUNT(ip) ipcount,dt FROM
(
SELECT ip,DATE(FROM_UNIXTIME(`Time`)) as dt FROM mytable
WHERE `Time` > TO_UNIXTIME(NOW() - INTERVAL 10 DAY)
) A GROUP BY dt;
Make sure you have an index on Time
ALTER TABLE mytable ADD INDEX TimeIndex (`Time`);
This will give you results with actual date values:
SELECT
COUNT(DISTINCT ip),
FROM_UNIXTIME(Time, '%m/%d/%Y') AS Day
FROM
tbl
WHERE
Time >= UNIX_TIMESTAMP(DATE_ADD(CURDATE(), INTERVAL -10 DAY))
GROUP BY
FROM_UNIXTIME(Time, '%m/%d/%Y')
try this:
SELECT CAST(DATE(FROM_UNIXTIME(`Time`)) AS CHAR) as dateoftime, COUNT(Ip) as cnt
FROM tablename
WHERE DATE(FROM_UNIXTIME(`Time`)) > DATE_SUB(current_timestamp, INTERVAL 10 DAY)
GROUP BY CAST(DATE(FROM_UNIXTIME(`Time`)) AS CHAR)

Display rows from MySQL where a datetime is within the next hour

I always have trouble with complicated SQL queries.
This is what I have
$query = '
SELECT id,
name,
info,
date_time
FROM acms_events
WHERE date_time = DATE_SUB(NOW(), INTERVAL 1 HOUR)
AND active = 1
ORDER BY date_time ASC
LIMIT 6
';
I want to get up to 6 rows that are upcoming within the hour. Is my query wrong? It does not seem to get events that are upcoming within the next hour when I test it.
What is the correct syntax for this?
I'm going to postulate that you're looking at a group of records that contain a range of DATETIME values, so you probably want something more like this:
SELECT id,
name,
info,
date_time
FROM acms_events
WHERE date_time < DATE_ADD(NOW(), INTERVAL 1 HOUR)
AND date_time >= NOW()
AND active = 1
ORDER BY date_time ASC
LIMIT 6
Otherwise, your query is looking for records with a date_time of exactly "now + 1 hour". I'm assuming all your dates aren't specific to that particular second. ;)
To clarify a bit, DATE_ADD() and DATE_SUB() return exact timestamps, so your query above roughly translates to something like SELECT ... WHERE date_time = '2010-04-14 23:10:05' ORDER BY ..., which I don't think is what you want.
WHERE date_time = DATE_SUB(NOW(), INTERVAL 1 HOUR)
means date_time equals exactly now minus one hour, which would result in any record exactly one hour old.
Why not use
WHERE TIMEDIFF(date_time, NOW()) < '01:00:00'
AND date_time > NOW()