In my database I have a table with two columns. The first column contains dates and the second is a count variable. I was wondering if it is possible to compute the average counts for each weekday based on the dates and counts.
In the following a small example:
Table:
Date Count
02/01/2005 100
02/02/2005 200
02/03/2005 300
... ...
Output:
Days Average
Monday 120.5
Tuesday 200.2
Wednesday 300.5
You could a series of avg calls on case expression extracting the day's name:
SELECT AVG(CASE DAYOFWEEK(`date`) WHEN 2 THEN `count` ELSE NULL END) AS Monday,
AVG(CASE DAYOFWEEK(`date`) WHEN 3 THEN `count` ELSE NULL END) AS Tuesday,
AVG(CASE DAYOFWEEK(`date`) WHEN 4 THEN `count` ELSE NULL END) AS Wednesday,
AVG(CASE DAYOFWEEK(`date`) WHEN 5 THEN `count` ELSE NULL END) AS Thursday,
AVG(CASE DAYOFWEEK(`date`) WHEN 6 THEN `count` ELSE NULL END) AS Friday
FROM mytable
EDIT:
Given the updated expected output in the edited post, it's much easier to do - just group by the dayname:
SELECT DAYNAME(`date`), AVG(`count`)
FROM mytable
WHERE DAYOFWEEK(`date`) BETWEEN 2 AND 6
GROUP BY DAYNAME(`date`)
#Mureinik's answer also pivots the data sets. If you need the week days as rows, not columns (I'm not sure by your question), the query gets even easier (untested):
SELECT DAYNAME(`date`) AS day_of_week,
AVG(`count`) AS average
FROM yourtable
GROUP BY DAYOFWEEK(`date`)
ORDER BY DAYOFWEEK(`date`)
Related
I need to sum the number of visits in march and another column that show me the number of visits in april, how can I do it in one query?
I am running in Metabase and I never really used it before, I don't even know how to use the dashboard correctly so I am just querying as usual.
I want to be able to pull enough information to create a visitation over time graph.
Thanks in advance!
select
sum(case when month='MARCH' then 1 else 0 end) as VisitInMarch,
sum(case when month='APRIL' then 1 else 0 end) as VisitInApril
from your_table
or if some specific sum:
select
sum(case when month='MARCH' then [your_field] else 0 end) as VisitInMarch,
sum(case when month='APRIL' then [your_field] else 0 end) as VisitInApril
from your_table
Assuming each visit has a unique ID:
select
count( distinct case when month = 'March' then visit_ID else NULL end ) as March_Visits
, count( distinct case when month = 'April' then visit_ID else NULL end ) as April_Visits
from visits
Or if you want everything in one column, just group by month:
select
month
, count( distinct visit_ID ) as Visits
from visits
group by month
order by month
I have a MySQL DB as such:
Date Customer_ID
How can I turn it into:
Customer_ID | Count_Visits_Past_Week | Count_Visits_Past_Month | Count_Visits_Past_90days | Count_Total
Note : Count_Total =sum of the other three counts
Thanks
The first step is to determine the demarcation points for the specified date ranges.
There's several questions to answer here: did you want to compare just the DATE ('yyyy-mm-dd') and disregard any time component?
By "past week", does that mean within the last seven days, or does it mean so far since the previous Sunday, or does it mean the last last full week, from Sunday through Saturday.
For "past month", does that mean the previous whole month, from the first through the end of the month? Or does it mean that if the query is run on the 20th of the month, we want dates since the 20th of the previous month up until today? Or yesterday?
Once we know the points in time that begin and end each specified period, relative to today's date, we can build expressions that evaluate to those dates.
For example, "past week" could be represented as the most recent seven day period:
DATE(NOW())-INTERVAL 1 WEEK -thru- DATE(NOW())
And "past month" can be represented as the same "day of month" (e.g. 17th) of the immediately preceding month up until today:
DATE(NOW())-INTERVAL 1 MONTH -thru- DATE(NOW())
That's really the first step, to determine the begin and end dates of each specified period.
Once we have that, we can move on to building a query that gets a "count" of rows with a date column that falls within each period.
The "trick" is to use conditional tests in expressions in the SELECT list of the query. We'll use those conditional tests to return a 1 if the row is to be included in the "count", and return 0 or NULL if the row should be excluded.
I prefer to use the SUM() aggregate function to get the "count". But it's also possible to use COUNT() aggregate. (If we use COUNT(), we need to use an expression that returns NULL when the row is to be excluded. I prefer to return a 1 or 0; I think it makes debugging easier.
Here's an outline of what a "count" query would look like.
SELECT t.Customer_Id
, SUM(IF( <some_condition> ,1,0) AS Count_something
, SUM(IF( <other_condition> ,1,0) AS Count_something_else
FROM mytable t
GROUP BY t.Customer_Id
When <some_condition> is true, we return a 1, otherwise we return 0.
To test the conditional expressions, it's often easiest to avoid doing the aggregation, and just return the individual rows:
That way, we can see which individual rows are going to be included in each "count".
For example:
SELECT t.Customer_ID
, t.date
, IF(t.date BETWEEN DATE(NOW())-INTERVAL 1 WEEK AND DATE(NOW()),1,0)
AS visit_past_week
, IF(t.date BETWEEN DATE(NOW())-INTERVAL 1 MONTH AND DATE(NOW()),1,0)
AS visit_past_month
FROM mytable t
ORDER BY t.date, t.Customer_Id
That query doesn't return the "count", it just returns the results of the expressions, which can be useful in testing. And of course we want to test the expressions that return the beginning and ending date of each period:
SELECT DATE(NOW()) - INTERVAL 1 WEEK AS past_week_begin
, DATE(NOW()) AS past_week_end
With this approach, the same row can be included in multiple "counts" with one query and one pass through the table.
Note that the expressions inside the SUM() aggregate in the query below are taking advantage of a convenient shorthand, an expression evaluated as a boolean will return 1 if TRUE, 0 if false, or a NULL.
To use the COUNT aggregate, we need to insure that the expression being aggregated returns a non-NULL when the row is to be "counted", and a NULL when the row is to be excluded from the count. So we use the convenient NULLIF function to return NULL if the value returned by the expression is a zero.
SELECT t.Customer_ID
, COUNT(NULLIF( t.date BETWEEN DATE(NOW())-INTERVAL 1 WEEK AND DATE(NOW()),0))
AS Count_Visits_Past_Week
, COUNT(NULLIF( t.date BETWEEN DATE(NOW())-INTERVAL 1 MONTH AND DATE(NOW()),0))
AS Count_Visits_Past_Month
, COUNT(NULLIF( t.date BETWEEN DATE(NOW())-INTERVAL 90 DAY AND DATE(NOW()),0))
AS Count_Visits_Past_90days
, COUNT(NULLIF( t.date BETWEEN DATE(NOW())-INTERVAL 1 WEEK AND DATE(NOW()),0))
+ COUNT(NULLIF( t.date BETWEEN DATE(NOW())-INTERVAL 1 MONTH AND DATE(NOW()),0))
+ COUNT(NULLIF( t.date BETWEEN DATE(NOW())-INTERVAL 90 DAY AND DATE(NOW()),0))
AS Count_Total
FROM mytable t
GROUP BY t.Customer_Id
NOTE: NULLIF(a,b) is a convenient shorthand for IF a IS NULL THEN return b ELSE return a
Returning the Count_Total is a bit odd, since it's got the potential to "count" the same row multiple times... but the value it returns should match the total of the individual counts.
I think this will give you what you want.
select customer_id,
sum(case when splitter = 'week' then num_visits else 0 end) as visits_this_week,
sum(case when splitter = 'month' then num_visits else 0 end) as visits_this_month,
sum(case when splitter = '90days' then num_visits else 0 end) as visits_last_90days,
sum(num_visits) as total
from (select customer_id, 'week' as splitter, count(*) as num_visits
from tbl
where extract(week from date) = extract(week from sysdate())
and extract(year from date) = extract(year from sysdate())
group by customer_id
union all
select customer_id, 'month' as splitter, count(*) as num_visits
from tbl
where extract(month from date) = extract(month from sysdate())
and extract(year from date) = extract(year from sysdate())
group by customer_id
union all
select customer_id, '90days' as splitter, count(*) as num_visits
from tbl
where date between date_sub(sysdate(), interval 90 day) and
sysdate()) x
group by customer_id
sql fiddle example: http://sqlfiddle.com/#!2/a762c/12/0
SELECT
EXTRACT( DAY FROM date) AS day,
EXTRACT( MONTH FROM date) AS month, who, sum ( wsd ) AS total FROM weekly
WHERE season = '08'
AND status = 1
AND who = 'NAME SURNAME'
GROUP BY day, month
day month who total
12 8 NAME SURNAME 18
I am getting totals of "wsd" column for every person in "who" column per day. If one person doesn't have records in table for a day, we can not see that name in the results as expected.
But i want to see that records too, with date and name with "0" in total column.
How can i do it with mysql only?
For this case, you can move the conditional logic from the where to the aggregation -- assuming that you have data for someone on every day:
SELECT EXTRACT( DAY FROM date) AS day, EXTRACT( MONTH FROM date) AS month, who,
sum(case when status = 1 AND who = 'NAME SURNAME' then wsd else 0 end) AS total
FROM weekly
WHERE season = '08'
GROUP BY day, month;
Did you try Coalesce or Isnull for inserting default values(in your case 0) ?
I have a table like this:
date day weather
2000-01-01 Monday Sunny
2000-01-02 Tuesday Rainy
. . .
I want to get number of rainy Mondays and sunny Mondays in one query like
day rainy_d sunny_d
Monday 2 5
How to accomplish that in Mysql and PostgreSQL?
select `Day`,
SUM(case when weather = 'Sunny' THEN 1 ELSE 0 end) as Sunny_D,
SUM(case when weather = 'Rainy' THEN 1 ELSE 0 end) as Rainy_D
FROM YOURTABLENAME
Where day = 'Monday'
Group by `Day`
Standard SQL, works in both:
SELECT
day,
SUM(CASE WHEN weather = 'Rainy' THEN 1 ELSE 0 END) AS rainy_d,
SUM(CASE WHEN weather = 'Sunny' THEN 1 ELSE 0 END) AS sunny_d
FROM yourtable
GROUP BY day
More concise version - MySQL only:
SELECT
day,
SUM(weather = 'Rainy') AS rainy_d,
SUM(weather = 'Sunny') AS sunny_d
FROM yourtable
GROUP BY day
More concise version - PostgreSQL only:
SELECT
day,
SUM((weather = 'Rainy')::int) AS rainy_d,
SUM((weather = 'Sunny')::int) AS sunny_d
FROM yourtable
GROUP BY day
The column day is probably redundant. I would delete it without substitution. The column date holds all the information. Then your query could look like this ..
In PostgreSQL:
SELECT to_char(date, 'Day') AS day
,COUNT(NULLIF(weather,'Sunny')) AS rainy_d
,COUNT(NULLIF(weather,'Rainy')) AS sunny_d
FROM tbl
GROUP BY 1;
In MySQL:
SELECT DAYNAME(date) AS day
... rest identical
The NULLIF() construct works for exactly two distinct (non-null) values in the column weather and is standard SQL. For more values use the alternatives provided by #Mark and #xQbert.
I have a table of transactions for purchases. Each transaction has a timestamp and purchase amount (in USD).
I'm trying to create some stats from this. I'd like to extract a row for each year that contains the sum for each month in the year. (I'd like months with no transaction to sum to 0 - not omitted.)
I know I could just do a plain SELECT of everything and process it in PHP, but I was wondering if it was at all possible to make MySQL do the work and extract the data like I want it?
What I'd like to see is rows like:
Year, Total_Jan, Total_Feb, ... Total_Dec, Total_Year
I am able to get the total per year, but I can't work out how to get the total per month into the same row.
SELECT
YEAR(dt) as the_year,
SUM(mc_gross) AS sum_total
FROM
transactions
GROUP BY
the_year
SELECT
YEAR(dt) as the_year,
SUM(CASE WHEN MONTH(dt) = 1 THEN mc_gross ELSE 0 END) AS Total_Jan,
SUM(CASE WHEN MONTH(dt) = 2 THEN mc_gross ELSE 0 END) AS Total_Feb,
...
SUM(CASE WHEN MONTH(dt) = 12 THEN mc_gross ELSE 0 END) AS Total_Dec
FROM
transactions
GROUP BY
the_year;