I want my query to return number of users registered in a particular day between current date and 7 days before. Instead of days, in date-time format it shows it in integer format. Example, for date 20/12/2012 its showing as 0, 21/12/2012 as 1 and so on
My Query:
select date(update_timestamp) between date_sub(sysdate(), interval 7 day) and sysdate() as date, count(*) users
from registered
group by date(update_timestamp)
Desired Output
DATE | USERS
20/12/2012 | 5
21/12/2012 | 6
Output i'm getting
DATE | USERS
0 | 5
1 | 6
What is wrong with my sql query? Also, i need to know the alternate way of fetching last 7 days data
You have put the condition in the select statement, so it will be evaluated and returned, instead of limiting the result. Put the condition in the where statement:
select date(update_timestamp) as date, count(*) users
from registered
where date(update_timestamp) between date_sub(sysdate(), interval 7 day) and sysdate()
group by date(update_timestamp)
To get all the dates in the interval even if there are no users, you need to join in another source of data:
select date_sub(sysdate(), interval d.offset day) date, count(r.update_timestamp) users
from registered r
inner join (
select 0 as offset union all
select 1 union all
select 2 union all
select 3 union all
select 4 union all
select 5 union all
select 6 union all
select 7
) d on date(r.update_timestamp) = date_sub(sysdate(), interval d.offset day)
group by date_sub(sysdate(), interval d.offset day)
I think you want a WHERE clause:
select date(update_timestamp) Date, count(*) users
from registered
where date(update_timestamp) between date_sub(sysdate(), interval 7 day) and sysdate()
group by date(update_timestamp)
try this query ::
where clause is the filtering part, For the date type logic you should use your current day - 7 days logic in that part of the query.
select date(update_timestamp) ,count(*) users
from registered
where date(update_timestamp)
between
date_sub(sysdate(), interval 7 day) and sysdate()
group by date(update_timestamp)
Related
The following query returns the visitors and pageviews of last 7 days. However, if there are no results (let's say it is a fresh account), nothing is returned.
How to edit this in order to return 0 in days that there are no entries?
SELECT Date(timestamp) AS day,
Count(DISTINCT hash) AS visitors,
Count(*) AS pageviews
FROM behaviour
WHERE company_id = 1
AND timestamp >= Subdate(Curdate(), 7)
GROUP BY day
Assuming that you always have at least one record in the table for each of the last 7 days (regardless of the company_id), then you can use conditional aggregation as follows:
select
date(timestamp) as day,
count(distinct case when company_id = 1 then hash end) as visitors,
sum(company_id = 1) as pageviews
from behaviour
where timestamp >= curdate() - interval 7 day
group by day
Note that I changed you query to use standard date arithmetics, which I find easier to understand that date functions.
Otherwise, you would need to move the condition on the date from the where clause to the aggregate functions:
select
date(timestamp) as day,
count(distinct case when timestamp >= curdate() - interval 7 day and company_id = 1 then hash end) as visitors,
sum(timestamp >= curdate() - interval 7 day and company_id = 1) as pageviews
from behaviour
group by day
If your table is big, this can be expensive so I would not recommend that.
Alternatively, you can generate a derived table of dates and left join it with your original query:
select
curdate - interval x.n day day,
count(distinct b.hash) visitors,
count(b.hash) page_views
from (
select 1 n union all select 2 union all select 3 union all select 4
union all select 5 union all select 6 union all select 7
) x
left join behavior b
on b.company_id = 1
and b.timestamp >= curdate() - interval x.n day
and b.timestamp < curdate() - interval (x.n - 1) day
group by x.n
Use a query that returns all the dates from today minus 7 days to today and left join the table behaviour:
SELECT t.timestamp AS day,
Count(DISTINCT b.hash) AS visitors,
Count(b.timestamp) AS pageviews
FROM (
SELECT Subdate(Curdate(), 7) timestamp UNION ALL SELECT Subdate(Curdate(), 6) UNION ALL
SELECT Subdate(Curdate(), 5) UNION ALL SELECT Subdate(Curdate(), 4) UNION ALL SELECT Subdate(Curdate(), 3) UNION ALL
SELECT Subdate(Curdate(), 2) UNION ALL SELECT Subdate(Curdate(), 1) UNION ALL SELECT Curdate()
) t LEFT JOIN behaviour b
ON Date(b.timestamp) = t.timestamp AND b.company_id = 1
GROUP BY day
Use IFNULL:
IFNULL(expr1, 0)
From the documentation:
If expr1 is not NULL, IFNULL() returns expr1; otherwise it returns expr2. IFNULL() returns >a numeric or string value, depending on the context in which it is used.
You can use next trick:
First, get query that return 1 dummy row: SELECT 1;
Next use LEFT JOIN to connect summary row(s) without condition. This join will return values in case data exists on NULL values in other case.
Last select from joined queries onle what we need and convert NULL's to ZERO's
using IFNULL dunction.
SELECT
IFNULL(b.day,0) AS DAY,
IFNULL(b.visitors,0) AS visitors,
IFNULL(b.pageviews,0) AS pageviews
FROM (
SELECT 1
) a
LEFT JOIN (
SELECT DATE(TIMESTAMP) AS DAY,
COUNT(DISTINCT HASH) AS visitors,
COUNT(*) AS pageviews
FROM behaviour
WHERE company_id = 1
AND TIMESTAMP >= SUBDATE(CURDATE(), 7)
GROUP BY DAY
) b ON 1 = 1;
I have the entries of monthly dues like,
Table name : month_dues,
Columns:
customer_id,
due_date,
due_amount
These table have lot of due entries.
customer_id due_date due_amount
--------------------------------------
1 2018-12-01 100
1 2019-01-01 100
1 2019-02-01 100
1 2019-10-01 100
so, how to select pending due months from these record?
in my table customer 1 not paid dues for these months,
2019-03,2019-04,2019-05,2019-06,2019-07,2019-08, 2019-09
customer pay the due every month so
For select paid dues from table using,
SELECT customer_id, due_date, due_amount FROM month_dues where customer_id='1' where due_date>='2019-01-01' and due_date<='2019-10-18';
How to get pending due month and year using this table?
Which means, find month and year not in this record.
IF you are searching between the two dates
YOu can use this
select * from month_dues
where due_date between '2012-03-11 00:00:00' and '2012-05-11
23:59:00' && customer_id='1'
order by month_dues desc;
If you want to search the date(due_date) lower than today
SELECT * FROM month_dues WHERE due_date < CURDATE();
And you are refeering to the record that is not belong to the query you can find some NOT IN query
Like
`SELECT * FROM month_dues NOT IN ( select * from month_dues
where due_date between '2012-03-11 00:00:00' and '2012-05-11
23:59:00' && customer_id='1'
order by month_dues desc;
)`
So basically you need to validate your table against some kind of calendar, here is a simple solution that only works for the current year, maybe you can use it as a start or someone else could improve it
SELECT m.MONTH
FROM (SELECT 1 AS MONTH
UNION SELECT 2 AS MONTH
UNION SELECT 3 AS MONTH
UNION SELECT 4 AS MONTH
UNION SELECT 5 AS MONTH
UNION SELECT 6 AS MONTH
UNION SELECT 7 AS MONTH
UNION SELECT 8 AS MONTH
UNION SELECT 9 AS MONTH
UNION SELECT 10 AS MONTH
UNION SELECT 11 AS MONTH
UNION SELECT 12 AS MONTH) as m
WHERE m.MONTH NOT IN (SELECT MONTH(due_date)
FROM due_months
WHERE customer_id = 1
AND YEAR(due_date) = YEAR(CURDATE()))
AND m.MONTH < MONTH(CURDATE()) -- needs to be improved as well
I have a database table visitors with three columns:
id | Name | checkin_date |
1 | Reg | 2018-04-20T08:28:54.446Z |
2 | Meg | 2018-04-21T08:28:54.446Z |
3 | Ted | 2018-04-21T08:28:54.446Z |
4 | Bin | 2018-04-23T08:28:54.446Z |
There are several records such as these.
I want to fetch the count of records per each day for only the past 7 days. Right now i was able to fetch the count of visitors per day for all the dates using :
select count(id) as no_of_users
, DATE_FORMAT(checkin_date, '%d %b, %Y') as date
from visitors
GROUP
BY DATE(checkin_date)
But this displays the count of users per each day of all the records. How to get the records of only past 7 days.
select count(id) as no_of_users, DATE_FORMAT(checkin_date, '%d %b, %Y') as date from visitors
where checkin_date >= DATE(NOW()) - INTERVAL 7 DAY
GROUP BY DATE(checkin_date)
in the where is where you want to do the date field >= last 7 days
From your question.
You need to create a calendar table, then LEFT JOIN on the calendar table.
SELECT DATE(t.dt),count(t1.id) cnt
FROM
(
SELECT NOW() dt
UNION ALL
SELECT NOW() - INTERVAL 1 DAY
UNION ALL
SELECT NOW() - INTERVAL 2 DAY
UNION ALL
SELECT NOW() - INTERVAL 3 DAY
UNION ALL
SELECT NOW() - INTERVAL 4 DAY
UNION ALL
SELECT NOW() - INTERVAL 5 DAY
UNION ALL
SELECT NOW() - INTERVAL 6 DAY
UNION ALL
SELECT NOW() - INTERVAL 7 DAY
) t LEFT JOIN T t1
ON DATE(t.dt) = DATE(t1.checkin_date)
group by t1.name,DATE(t.dt)
sqlfiddle:http://sqlfiddle.com/#!9/59f49b/5
select id, count(id) as TOTAL, min (checkin_date) as no_of_users
from visitors
where checkin_date between '<Start Date>' and '<End Date>'
GROUP
BY Id,checkin_date
I am having trouble with a query. This is taken from a similar query where i count number of rows per month.
I want to count all rows for each day of the last week and display the day name and a count. If there are no rows for that day, display zero.
I know the UNIONS won't work but i don't know what to replace it with.
At the moment it get the last 7 days but the day name is NULL
SELECT DAYNAME(STR_TO_DATE(Days.ID, '%a')) AS `day`, COUNT(`returns`.list_date) AS `total`
FROM
(
SELECT 1 as ID UNION SELECT 2 as ID UNION SELECT 3 as ID UNION SELECT 4 as ID
UNION
SELECT 5 as ID UNION SELECT 6 as ID UNION SELECT 7 as ID
) as Days
WHERE (list_date >= DATE_SUB(NOW(), INTERVAL 1 WEEK))
GROUP BY Days.id
UPDATE:
I have created a SQL fiddle showing the code output from #Gordon Linoff answer below which doesn't get the counted rows
http://sqlfiddle.com/#!9/969463/1
One method for doing what you want is a LEFT JOIN:
SELECT DAYNAME(DATE_SUB(CURDATE(), INTERVAL Days.n DAY)) AS `day`,
COUNT(r.list_date) AS `total`
FROM (SELECT 1 as n UNION ALL SELECT 2 as n UNION ALL
SELECT 3 as n UNION ALL SELECT 4 as n UNION ALL
SELECT 5 as n UNION ALL SELECT 6 as n UNION ALL
SELECT 7 as n
) Days LEFT JOIN
returns r
ON r.list_date = DATE_SUB(CURDATE(), INTERVAL Days.n DAY))
GROUP BY Days.n
ORDER BY Days.n;
Some notes:
Use UNION ALL instead of UNION, unless you have a good reason for incurring the overhead of removing duplicates.
This assumes that returns.list_date is actually a date, because it uses = rather than >=.
The use of now() and >= is a bit confusing, because now() has a time component, which you generally want to ignore.
I have a query which returns the total of users who registered for each day. Problem is if a day had no one register it doesn't return any value, it just skips it. I would rather it returned zero
this is my query so far
SELECT count(*) total FROM users WHERE created_at < NOW() AND created_at >
DATE_SUB(NOW(), INTERVAL 7 DAY) AND owner_id = ? GROUP BY DAY(created_at)
ORDER BY created_at DESC
Edit
i grouped the data so i would get a count for each day- As for the date range, i wanted the total users registered for the previous seven days
A variation on the theme "build your on 7 day calendar inline":
SELECT D, count(created_at) AS total FROM
(SELECT DATE_SUB(NOW(), INTERVAL D DAY) AS D
FROM
(SELECT 0 as D
UNION SELECT 1
UNION SELECT 2
UNION SELECT 3
UNION SELECT 4
UNION SELECT 5
UNION SELECT 6
) AS D
) AS D
LEFT JOIN users ON date(created_at) = date(D)
WHERE owner_id = ? or owner_id is null
GROUP BY D
ORDER BY D DESC
I don't have your table structure at hand, so that would need adjustment probably. In the same order of idea, you will see I use NOW() as a reference date. But that's easily adjustable. Anyway that's the spirit...
See for a live demo http://sqlfiddle.com/#!2/ab5cf/11
If you had a table that held all of your days you could do a left join from there to your users table.
SELECT SUM(CASE WHEN U.Id IS NOT NULL THEN 1 ELSE 0 END)
FROM DimDate D
LEFT JOIN Users U ON CONVERT(DATE,U.Created_at) = D.DateValue
WHERE YourCriteria
GROUP BY YourGroupBy
The tricky bit is that you group by the date field in your data, which might have 'holes' in it, and thus miss records for that date.
A way to solve it is by filling a table with all dates for the past 10 and next 100 years or so, and to (outer)join that to your data. Then you will have one record for each day (or week or whatever) for sure.
I had to do this only for MS SqlServer, so how to fill a date table (or perhaps you can do it dynamically) is for someone else to answer.
A bit long winded, but I think this will work...
SELECT count(users.created_at) total FROM
(SELECT DATE_SUB(CURDATE(),INTERVAL 6 DAY) as cdate UNION ALL
SELECT DATE_SUB(CURDATE(),INTERVAL 5 DAY) UNION ALL
SELECT DATE_SUB(CURDATE(),INTERVAL 4 DAY) UNION ALL
SELECT DATE_SUB(CURDATE(),INTERVAL 3 DAY) UNION ALL
SELECT DATE_SUB(CURDATE(),INTERVAL 2 DAY) UNION ALL
SELECT DATE_SUB(CURDATE(),INTERVAL 1 DAY) UNION ALL
SELECT CURDATE()) t1 left join users
ON date(created_at)=t1.cdate
WHERE owner_id = ? or owner_id is null
GROUP BY t1.cdate
ORDER BY t1.cdate DESC
It differs from your query slightly in that it works on dates rather than date times which your query is doing. From your description I have assumed you mean to use whole days and therefore have used dates.