I have an SQLite database which contains transactions, each of them having a price and a transDate.
I want to retrieve the sum of the transactions grouped by month. The retrieved records should be like the following:
Price month
230 2
500 3
400 4
it is always good while you group by MONTH it should check YEAR also
select SUM(transaction) as Price,
DATE_FORMAT(transDate, "%m-%Y") as 'month-year'
from transaction group by DATE_FORMAT(transDate, "%m-%Y");
FOR SQLITE
select SUM(transaction) as Price,
strftime("%m-%Y", transDate) as 'month-year'
from transaction group by strftime("%m-%Y", transDate);
You can group on the start of the month:
select date(DateColumn, 'start of month')
, sum(TransactionValueColumn)
from YourTable
group by
date(DateColumn, 'start of month')
Try the following:
SELECT SUM(price), strftime('%m', transDate) as month
FROM your_table
GROUP BY strftime('%m', transDate);
Use the corresponding page in SQLite documentation for future references.
SELECT
SUM(Price) as Price, strftime('%m', myDateCol) as Month
FROM
myTable
GROUP BY
strftime('%m', myDateCol)
This another form:
SELECT SUM(price) AS price,
STRFTIME('%Y-%m-01', created_at) as created_at
FROM records
GROUP BY STRFTIME('%Y-%m-01', created_at);
Another way is to substring the year and the month from the column and group by them. Assuming the format it's like 2021-05-27 12:58:00 you can substract the first 7 digits:
SELECT
substr(transDate, 1, 7) as YearMonth
SUM(price) AS price
FROM
records
GROUP BY
substr(transDate, 1, 7);
In Sqlite, if you are storing your date in unixepoch format, in seconds:
select count(myDate) as totalCount,
strftime('%Y-%m', myDate, 'unixepoch', 'localtime') as yearMonth
from myTable group by strftime('%Y-%m', myDate, 'unixepoch', 'localtime');
If you are storing the date in unixepoch format, in milliseconds, divide by 1000:
select count(myDate/1000) as totalCount,
strftime('%Y-%m, myDate/1000, 'unixepoch', 'localtime') as yearMonth
from myTable group by strftime('%Y-%m, myDate/1000, 'unixepoch', 'localtime');
Related
I have the following database on Google Big Query (SQL Standard) with the date and the revenue. Both in int format. I need to get the totals by month and year. I am not able to get the part of the date I am interested on. Basically the numbers of position 1 and 6 from the first column.
Revenue Database:
This is what I have tried but then I need to run this code for every month separately:
SELECT sum(revenue)
from revenue.table
where date between 20210601 and 20210630
Any clue on how to do this? Thanks!
If the value is in int format, then use arithmetic:
select floor(date / 100) as yyyymm, sum(revenue)
from revenue.table
group by yyyymm;
If it were stored properly as a date, then you would use the built-in date_trunc():
select date_trunc(date, month) as yyyymm, sum(revenue)
from revenue.table
group by yyyymm;
There are similar functions for related data types: timestamp_trunc() and datetime_trunc().
If your date is actually an INTEGER, then I would:
GROUP BY DIV(date/100)
I would go with below
select
date_trunc(parse_date('%Y%m%d', '' || date), month) month,
sum(revenue) revenue
from `revenue.table`
group by month
Try this:
SELECT substring(cast(Date as string),1,6) YearMonth,
sum(Revenue)
FROM `<Dataset_NAME>.<Table_NAME>`
group by YearMonth;
Adding my own solution in case it helps someone in the future:
SELECT substring(CAST(date AS STRING), 1, 6) AS month, sum(revenue) AS total_rev
FROM `revenue.table`
GROUP BY month
ORDER BY month DESC
I am trying to get the frequency of unique ID values for each month of the last year. However, I don't get the outcome.. including the error message "SELECT list expression references column user_id which is neither grouped nor aggregated".
How can I get the count of unique IDs in each month and them group them by month?
What I tried:
SELECT
user_id,
EXTRACT(MONTH FROM date) as month
FROM
TABLE
WHERE
date >= '2020-09-01'
GROUP BY
month
I want something like this:
month
count of unique user_id
1
300
2
200
...
...
12
250
You would use GROUP BY and COUNT(DISTINCT):
SELECT EXTRACT(MONTH FROM date) as month, COUNT(DISTINCT user_id)
FROM TABLE
WHERE date >= '2020-09-01'
GROUP BY 1;
I would advise you to include the year in the query. In BigQuery, this is simplest using DATE_TRUNC():
SELECT DATE_TRUNC(date, MONTH) as month, COUNT(DISTINCT user_id)
FROM TABLE
WHERE date >= '2020-09-01'
GROUP BY 1;
I have a table with 4 columns huge number of records. It has the following structure:
DATE_ENTERED EMP_NAME DATA ORIGINATED
01-JAN-20 A 545454 APPLE
I want to calculate no of records for every first day of every month in a year
is there any way can we fetch the data for every first day of month.
In oracle you can use TRUNC function on the date as follows:
SELECT TRUNC(DATE_ENTERED), COUNT(1) AS CNT
FROM YOUR_TABLE
WHERE TRUNC(DATE_ENTERED) = TRUNC(DATE_ENTERED, 'MON')
GROUP BY TRUNC(DATE_ENTERED, 'MON')
Please note that the TRUNC(DATE_ENTERED, 'MON') returns the first day of the month for DATE_ENTERED.
Cheers!!
SELECT Year, Month, COUNT(*)
FROM
(
SELECT
YEAR(DATE_ENTERED) Year
MONTH(DATE_ENTERED) Month
DAY(DATE_ENTERED) Day
FROM your_table
WHERE DAY(DATE_ENTERED) = 1
) A
GROUP BY Year, Month
Generally WHERE DAY(DATE_ENTERED) = 1 will get you the records only for dates at the start of each month. Thus using Year and Month function you can group them by in order to get a count for each year and each month
You mean something like
SELECT COUNT(*)
FROM Table
WHERE DAY(DATE_ENTERED) = 1 AND
YEAR(DATE_ENTERED) = Some_Year
GROUP BY DATE_ENTERED
You can also use DATE_ENTERED BETWEEN 'YYYY0101' and 'YYYY1231' (replace the YYYY with the year you want to retrieve data for) instead of YEAR(DATE_ENTERED) = Some_Year, if performance is an issue.
You can use something like this:
select * from your_table
where DAY(DATE_ENTERED) = 1
and DATE_ENTERED between '2020-01-01' and '2020-12-31'
for number of count use this:
select count(*) from your_table
where DAY(DATE_ENTERED)= 1
and DATE_ENTERED between '2020-01-01' and '2020-12-31'
UPDATE
select * from your_table where Extract(day FROM DATE_ENTERED) = 1 and DATE_ENTERED between '01-JAN-20 ' and '01-DEC-20 ';
this is how the data looks like:
For the list of records
select count(*) from your_table where Extract(day FROM DATE_ENTERED) = 1 and DATE_ENTERED between '01-JAN-20 ' and '01-DEC-20 ';
UPDATE-2
select EXTRACT(month from DATE_ENTERED) as Count,
to_char(to_date(DATE_ENTERED, 'DD-MM-YYYY'), 'Month') from your_table
where Extract(day FROM DATE_ENTERED) = 1 and DATE_ENTERED between '01-JAN-20
'and '01-DEC-20 ' group by EXTRACT(month from DATE_ENTERED),
to_char(to_date(DATE_ENTERED, 'DD-MM-YYYY'), 'Month');
Here is the output:
customer Date location
1 25Jan2018 texas
2 15Jan2018 texas
3 12Feb2018 Boston
4 19Mar2017 Boston.
I am trying to find out count of customers group by yearmon of Date column.Date column is of text data type
eg: In jan2018 ,the count is 2
I would do something like the following:
SELECT
date_part('year', formattedDate) as Year
,date_part('month', formattedDate) as Month
,count(*) as CustomerCountByYearMonth
FROM
(SELECT to_date(Date,'DDMonYYYY') as formattedDate from <table>) as tbl1
GROUP BY
date_part('year', formattedDate)
,date_part('month', formattedDate)
Any additional formatting for dates could be done on the inner query that will allow for adjustments in case some single digit days need to be padded or a month has four letters instead of three etc.
By converting to date type, you can properly order by date type and not alphabetical etc.
Optionally:
SELECT
Year
,Month
,count(*) as CustomerCountByYearMonth
FROM
(SELECT
date_part('year', to_date(Date,'DDMonYYYY')) as Year
,date_part('month', to_date(Date,'DDMonYYYY')) as Month
FROM <table>) as tbl1
GROUP BY
Year
,Month
You shouldn't store dates in a text column...
select substring(Date, length(Date)-6), count(*)
from tablename
group by substring(Date, length(Date)-6)
I thought #Jarlh asked a good question -- what about dates like January 1? Is it 01Jan2019 or 1Jan2019? If it can be either, perhaps a regex would work.
select
substring (date from '\d+(\D{3}\d{4})') as month,
count (distinct customer)
from t
group by month
The 'distinct customer' also presupposes you may have the same customer listed in the same month, but you only want to count it once. If that's not the case, just remove 'distinct.'
And, if you wanted the output in date format:
select
to_date (substring (date from '\d+(\D{3}\d{4})'), 'monyyyy') as month,
count (distinct customer)
from t
group by month
If it is a date column, you can truncate the date:
select date_trunc('month', date) as yyyymm, count(*)
from t
group by yyyymm
order by yyyymm;
I really read that the type was date. For a string, just use string functions:
select substr(date, 3, 7) as mmmyyyy, count(*)
from t
group by mmmyyyy;
Unfortunately, ordering doesn't work in this case. You should really be storing dates using the proper type.
I have a dataset in the format of: Date Amount such as:
Date Amount
2018-01 100
2018-02 200
2018-03 300
I want the sum year to date at the end of each month resulting in:
Date Amount
2018-01 100
2018-02 300 (100+200)
2018-03 600 (100+200+300)
How do I go about referencing the previous year-to-date sum in the previous row?
You can do this with a window function:
select date,
sum(amount) over (order by date) as so_far
from the_table
order by date;
The above is ANSI standard SQL
You need to use Running Total for this.
SELECT
Date,
SUM (Amount) OVER (ORDER BY Date ASC) AS RunningAmount
FROM Table1
Order by Date asc;
You can use OVER clause for that. Try as below.
Use PARTITION BY DATEPART(yy, Date) if you want to separate cumulative total for each year as below.
SELECT Date, Amount, SUM(Amount) OVER (PARTITION BY DATEPART(yy, Date) ORDER BY DATEPART(mm, Date)) AS CumulativeTotal
FROM Table
Or if you want to have cumulative total for all years then remove PARTITION BY DATEPART(yy, Date) part. Just order by is enough.
SELECT Date, Amount, SUM(Amount) OVER (ORDER BY Date) AS CumulativeTotal
FROM Table
Refer this for more information. Producing a moving average and cumulative total
Other option would to use a correlated subquery :
select t.*,
(select sum(t1.amount) from table t1 where t1.date <= t.date) as amount
from table t;
This would work almost in all DBMS.