id modid userid timemodified FROM_UNIXTIME(timemodified,'%d-%m-%Y')
410 32 46 1438971403 03-08-2015
411 32 46 1438971403 03-08-2015
412 66 977 1438971403 07-08-2015
412 66 977 1438971403 07-08-2015
413 67 34 1438971423 07-08-2015
414 68 16 1438971424 07-08-2015
415 132 23 1438972154 07-08-2015
416 134 2 1438972465 08-08-2015
417 115 2 1438996430 08-08-2015
418 130 977 1438996869 08-08-2015
I got this query from framing the last 4weeks ago by calculating from today's date. Now, I want to show the users for 4 weeks individually like week1, week2, week3 & week4, either it could be column wise or row wise, which would be the best.
In detailed, from the above query, I need to separate data from week1 to week4,like
Week4 : No user
Week3 : 2 users (2,977)
Week2 : 4 users (16, 23, 34, 977)
Week1 : 1 user (46)
SET #unix_four_weeks_ago = UNIX_TIMESTAMP(curdate()) - 2419200;
SELECT *,FROM_UNIXTIME(timemodified,'%d-%m-%Y') FROM mod_users WHERE timemodified >= #unix_four_weeks_ago
My guess is that you want to split the user count per week according to the timemodified column. I would use the WEEK() function to do that.
The following SQL would add a weeknumber column to identify the week number:
SELECT WEEK(timemodified) weeknumber, dates.*
FROM dates
Then, if you want to get the distinct user count, you can simply use the following SQL:
SELECT WEEK(timemodified) weeknumber, COUNT(DISTINCT(userid)) users_count
FROM dates
GROUP BY weeknumber
You can also add a WHERE clause to get only certain weeks as you wish. So, to get the last 4 weeks from the 23-08-2015, I would do:
SELECT WEEK(timemodified) weeknumber, COUNT(DISTINCT(userid)) users_count
FROM dates
WHERE WEEK(timemodified) <= WEEK('2015-08-23')
AND WEEK(timemodified) > (WEEK('2015-08-23') - 4)
GROUP BY weeknumber
Let's hope I assumed correctly. :-)
Related
I´m trying to do some analysis in the following data
WeekDay Date Count
5 06/09/2018 20
6 07/09/2018 Null
7 08/09/2018 19
1 09/09/2018 16
2 10/09/2018 17
3 11/09/2018 24
4 12/09/2018 25
5 13/09/2018 24
6 14/09/2018 23
7 15/09/2018 23
1 16/09/2018 9
2 17/09/2018 23
3 18/09/2018 33
4 19/09/2018 22
5 20/09/2018 31
6 21/09/2018 17
7 22/09/2018 10
1 23/09/2018 12
2 24/09/2018 26
3 25/09/2018 29
4 26/09/2018 27
5 27/09/2018 24
6 28/09/2018 29
7 29/09/2018 27
1 30/09/2018 19
2 01/10/2018 26
3 02/10/2018 39
4 03/10/2018 32
5 04/10/2018 37
6 05/10/2018 Null
7 06/10/2018 26
1 07/10/2018 11
2 08/10/2018 32
3 09/10/2018 41
4 10/10/2018 37
5 11/10/2018 25
6 12/10/2018 20
The problem that I want to solve is: I want to create a table with the average of the 3 last same weekdays related to the day. But, when there is a NULL in the weekday, I want to ignore and do the average only with the remain numbers, not count NULL as an 0. I will give you an example here:
The date in this table is day/month/year :)
Ex: On day 12/10/2018, I need the average from
the days 05/10/2018; 28/09/2018; 21/09/2018. These are the last 3 same weekday(six) as 12/10/2018.
. Their values are Null; 29; 17. Then the result of this average must be 23, because I need to ignore the NULL, and not be 15,333.
How can I do this?
The count() function ignores nulls (i.e. does NOT increment if it encounters null) so I suggest you simply count the values then may contain the nulls you wish to ignore.
dow datecol value
6 21/09/2018 17
6 28/09/2018 29
6 05/10/2018 Null
e.g. sum(value) above = 46, and the count(value) = 2 so the average is 23.0 (and avg(value) will also return 23.0 as it also ignores nulls)
select
weekday
, `date`
, `count`
, (select (sum(`count`) * 1.0) / (count(`count`) * 1.0)
from atable as t2
where t2.weekday = t1.weekday
and t2.`date` < t1.`date
order by t2.`date` DESC
limit 3
) as average
from atable as t1
You could just use avg(count) in the query above, and get the same result.
ps. I do hope you do NOT use count as a column name! I also would suggest you do NOT use date as a column name either. i.e. Avoid using SQL terms as names.
SELECT WeekDay, AVG(Count)
FROM myTable
WHERE Count IS NOT NULL
GROUP BY WeekDay
Use IsNULL(Count,0) in your Select
SELECT WeekDay, AVG(IsNULL(Count,0))
FROM myTable
GROUP BY WeekDay
First off, you need to get the number of instances of that weekday in the data since you just need the last 3 same week days
create table table2
as
select
row_number() over(partition by weekday order by date desc) as rn
,weekday
,date
,count
from table
From here, you can get what you want. With you explanation, you don't need to filter out the NULL values for count. Just doing the avg() aggregation will simply ignore it.
select
weekday
,avg(count)
from table2
where rn in (1,2,3)
group by weekday
I have table following :
ClientNUM PIECES DID
NEWAGENC 10 5
NEWAGENC 25 5
WAY 30 4
CHCAH 20 2
AVC 21 2
i want the Result that sum the value for each client as below
CleintNUM Pieces DID
NEWAGENC 35 5
WAY 30 4
CHCAH 20 2
AVC 21 2
My query
SELECT
CLIENTNUM,
DID,
PIECES,
GETDATE() AS CURRENTDATE,
SUM(PIECES)
FROM Mytable
GROUP BY CLIENTNUM, DISPID, PIECES
So how can i do the sum for each CLIENTNUM in my query Means DISTINCT For each client Pieces like NEWAGENC has value 10 and in second row 25 so the pieces will be 10+ 25 = 35
Don't group by PIECES if you want to aggregate it
SELECT CLIENTNUM,
DID,
PIECES,
GETDATE() AS CURRENTDATE,
SUM(PIECES)
FROM Mytable
GROUP BY CLIENTNUM, DISPID
I'm stuck on this query. I need to do a group by date, card_id and only show the highest hits. I have this data:
date card_name card_id hits
29/02/2016 Paul Stanley 1345 12
29/02/2016 Phil Anselmo 1347 16
25/02/2016 Dave Mustaine 1349 10
25/02/2016 Ozzy 1351 17
23/02/2016 Jhonny Cash 1353 13
23/02/2016 Elvis 1355 15
20/02/2016 James Hethfield 1357 9
20/02/2016 Max Cavalera 1359 12
My query at the moment
SELECT DATE(card.create_date) `day`, `name`,card_model_id, count(1) hits
FROM card
Join card_model ON card.card_model_id = card_model.id
WHERE DATE(card.create_date) >= DATE(DATE_SUB(NOW(), INTERVAL 1 MONTH)) AND card_model.preview = 0
GROUP BY `day`, card_model_id
;
I want to group by date, card_id and filter the higher hits result showing only one row per date. As if I run a max(hits) with group by but I won't work
Like:
date card_name card_id hits
29/02/2016 Phil Anselmo 1347 16
25/02/2016 Ozzy 1351 17
23/02/2016 Elvis 1355 15
20/02/2016 Max Cavalera 1359 12
Any light on that will be appreciated. Thanks for reading.
Here is one way to do this. Based on your sample data (not the query):
select s.*
from sample s
where s.hits = (select max(s2.hits)
from sample s2
where date(s2.date) = date(s.date)
);
Your attempted query seems to have no relationship to the sample data, so it is unclear how to incorporate those tables (the attempted query has different columns and two tables).
I want a query that can give result with sum of last 7 day look back.
I want output date and sum of last 7 day look back impressions for each date
e.g. I have a table tblFactImps with below data:
dateFact impressions id
2015-07-01 4022 30
2015-07-02 4021 33
2015-07-03 4011 34
2015-07-04 4029 35
2015-07-05 1023 39
2015-07-06 3023 92
2015-07-07 8027 66
2015-07-08 2024 89
I need output with 2 columns:
dateFact impressions_last_7
query I got:
select dateFact, sum(if(datediff(curdate(), dateFact)<=7, impressions,0)) impressions_last_7 from tblFactImps group by dateFact;
Thanks!
If your fact table is not too big, then a correlated subquery is a simple way to do what you want:
select i.dateFact,
(select sum(i2.impressions)
from tblFactImps i2
where i2.dateFact >= i.dateFact - interval 6 day
) as impressions_last_7
from tblFactImps i;
You can achieve this by LEFT OUTER JOINing the table with itself on a date range, and summing the impressions grouped by date, as follows:
SELECT
t1.dateFact,
SUM(t2.impressions) AS impressions_last_7
FROM
tblFactImps t1
LEFT OUTER JOIN
tblFactImps t2
ON
t2.dateFact BETWEEN
DATE_SUB(t1.dateFact, INTERVAL 6 DAY)
AND t1.dateFact
GROUP BY
t1.dateFact;
This should give you a sliding 7-day sum for each date in your table.
Assuming your dateFact column is indexed, this query should also be relatively fast.
I have one table accounts. I have written following query
chk_account= mysql_query("SELECT GROUP_CONCAT( DISTINCT user_id ) AS userlist
FROM `accounts`
");
From this I get users id only. With this query I also want to fetch data price and date with column name price and created but I need only to select with lowest date
I have table structure like this:
id user_id price created
1 31 10 2013-04-09 17:30:15
2 32 20 2013-04-10 20:24:40
3 31 30 2013-04-11 04:44:25
4 33 40 2013-04-12 05:47:18
5 34 50 2013-04-13 19:54:15
6 34 50 2013-04-14 14:27:15
7 35 10 2013-04-15 13:54:45
8 35 60 2013-04-16 12:24:35
9 35 10 2013-04-17 20:34:10
I suspect that you want the earliest date and price for each user. You can do this using group_concat(), using a query such as:
select USER_ID,
substring_index(group_concat(price order by created), ',', 1) as price,
min(created)
from accounts a
group by user_id