I have this query i use to get statistics of blogs in our own tracking system.
I use union select over 2 tables as we daily aggregate data in 1 table and keeps todays data in another table.
I want to have the last 10 months of traffic show.. This query does that, but of there is no traffic in a specific month that row is not in the result.
I have previously used a calendar table in mysql to join against to at avoid that, but im simply not skilled enoght to rewrite this query to join against that calendar table.
The calendart table has 1 field called "datefield" which i date format YYY-MM-DD
This is the current query i use
SELECT FORMAT(SUM(`count`),0) as `count`, DATE(`date`) as `date`
FROM
(
SELECT count(distinct(uniq_id)) as `count`, `timestamp` as `date`
FROM tracking
WHERE `timestamp` > now() - INTERVAL 1 DAY AND target_bid = 92
group by `datestamp`
UNION ALL
select sum(`count`),`datestamp` as `date`
from aggregate_visits
where `datestamp` > now() - interval 10 month
and target_bid = 92
group by `datestamp`
) a
GROUP BY MONTH(date)
Something like this?
select sum(COALESCE(t.`count`,0)),s.date as `date`
from DateTable s
LEFT JOIN (SELECT * FROM aggregate_visits
where `datestamp` > now() - interval 10 month
and target_bid = 92) t
ON(s.date = t.datestamp)
group by s.date
Related
I have a column in my sql table called loggedTime which is a datetime field and I want to select between two dates startDate and endDate along with the interval may be 5 minutes, 10 minutes, 1 hour etc. I tried to write the SQL query but it says You have syntax error next interval, I am not sure what wrong with my query. If I remove INTERVAL 5 MINUTE my query works fine but I want to pass the Interval along with the date so it will select all rows between two dates and also with interval
Here is SQL
SELECT * FROM mytable WHERE loggedTime BETWEEN '2021-06-01' and '2021-06-03' INTERVAL 5 MINUTE
If you have any unique consecutively increasing column like id, then you can use an INNER JOIN as done followingly:
SELECT *
FROM mytable a
INNER JOIN mytable b
ON a.ID = b.ID + 1
WHERE TIMESTAMPDIFF(minute, a.timestamp, b.timestamp) = 5;
If you do not have that column in your table then use this code :
SELECT *
FROM (SELECT mt.*,
TIMESTAMPDIFF(minute, #prevTS, `loggedTime`) AS timeinterval,
#prevTS:=mt.`loggedTime`
FROM mytable mt,
(SELECT #prevTS := (SELECT MIN(`loggedTime`)
FROM yourTable)) vars
ORDER BY ID)subquery_alias
WHERE loggedTime BETWEEN '2021-06-01' AND '2021-06-03'
AND timeinterval = 5
Check this thread as reference too.
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;
RDBMS: MySQL
The time column(s) datatype is of datetime
For every hour of the 24 hour day I need to retrieve the number of rows in which their start_time matches the hour OR the end_time is great than or equal to the hour.
Below is the current query I have which returns the data I need but only based off of one hour. I can loop through and do 24 separate queries for each hour of the day but I would love to have this in one query.
SELECT COUNT(*) as total_online
FROM broadcasts
WHERE DATE(start_time) = '2018-01-01' AND (HOUR(start_time) = '0' OR
HOUR(end_time) >= '0')
Is there a better way of querying the data I need? Perhaps by using group by somehow? Thank you.
Not exactly sure if i am following, but try something like this:
select datepart(hh, getdate()) , count(*)
from broadcasts
where datepart(hh, starttime) <=datepart(hh, endtime)
and cast(starttime as date)=cast(getdate() as date) and cast(endtime as date)=cast(getdate() as date)
group by datepart(hh, getdate())
Join with a subquery that returns all the hour numbers:
SELECT h.hour_num, COUNT(*) AS total_online
FROM (SELECT 0 AS hour_num UNION SELECT 1 UNION SELECT 2 ... UNION SELECT 23) AS h
JOIN broadcasts AS b ON HOUR(b.start_time) = h.hour_num OR HOUR(b.end_time) >= h.hour_num
WHERE DATE(b.start_time) = '2018-01-01'
GROUP BY h.hour_num
I have a table, that pretty much looks like this:
users (id INT, masterId INT, date DATETIME)
Every user has exactly one master. But masters can have n users.
Now I want to find out how many users each master has. I'm doing that this way:
SELECT `masterId`, COUNT(`id`) AS `total` FROM `users` GROUP BY `masterId` ORDER BY `total` DESC
But now I also want to know how many new users a master has since the last 14 days. I could do it with this query:
SELECT `masterId`, COUNT(`id`) AS `last14days` FROM `users` WHERE `date` > DATE_SUB(NOW(), INTERVAL 14 DAY) GROUP BY `masterId` ORDER BY `total` DESC
Now the question: Could I somehow get this information with one query, instead of using 2 queries?
You can use conditional aggregation to do this by only counting rows for with the condition is true. In standard SQL this would be done using a case expression inside the aggregate function:
SELECT
masterId,
COUNT(id) AS total,
SUM(CASE WHEN date > DATE_SUB(NOW(), INTERVAL 14 DAY) THEN 1 ELSE 0 END) AS last14days
FROM users
GROUP BY masterId
ORDER BY total DESC
Sample SQL Fiddle
I have a table
id user Visitor timestamp
13 username abc 2014-01-16 15:01:44
I have to 'Count' total visitors for a 'User' for last seven days group by date(not timestamp)
SELECT count(*) from tableA WHERE user=username GROUPBY __How to do it__ LIMIT for last seven day from today.
If any day no visitor came so, no row would be there so it should show 0.
What would be correct QUERY?
There is no need to GROUP BY resultset, you need to count visits for a week (with unspecified user). Try this:
SELECT
COUNT(*)
FROM
`table`
WHERE
`timestamp` >= (NOW() - INTERVAL 7 DAY);
If you need to track visits for a specified user, then try this:
SELECT
DATE(`timestamp`) as `date`,
COUNT(*) as `count`
FROM
`table`
WHERE
(`timestamp` >= (NOW() - INTERVAL 7 DAY))
AND
(`user` = 'username')
GROUP BY
`date`;
MySQL DATE() function reference.
Try this:
SELECT DATE(a.timestamp), COUNT(*)
FROM tableA a
WHERE a.user='username' AND DATEDIFF(NOW(), DATE(a.timestamp)) <= 7
GROUP BY DATE(a.timestamp);
i think it's work :)
SELECT Count(*)
from table A
WHERE user = username AND DATEDIFF(NOW(),timestamp)<=7