How can I get the retained customer from the previous year?
1. Customer Retention...the most recent trailing 12 months vs the previous 12 months...exclude new business...what % of customers are we still selling to? Ideally 100%...
explaining the above statement:-
I need the percentage of customers who were retained this year from the previous year depending on sales. for example- 1,2,3,4,5 (This year) are the customers doing business with us this year and 1,2,3,4,6 (previous Year) are the customers doing business in the previous year. so what I am looking for is the number of customers who have been retained from the previous year is 1,2,3,4 is (Four) divided by the total number of customers whom we are doing business within the previous year 1,2,3,4,6 is (Five). If I can get this somehow the percentage calculation is simple. So the percentage of customers who have been retained is (4/5)*100 = 80%. I want to calculate this as a measure for all the years.
Related
I have a table with the sales from last 2 years, and I want to compare the sales from this year with the same natural day last year. For example, Sunday 1st of April 2018 will be compared with Sunday 2nd April 2017.
In order to do that I have created the measure
sales_last_year = CALCULATE(Sales[Revenue]); SAMEPERIODLASTYEAR(DATEADD('Calendar'[Date];+1;DAY)))
And I have created another measure where I have the value from the same day last year:
Prueba_sales_last_year = CALCULATE(Sales[Revenue]); SAMEPERIODLASTYEAR('Calendar'[Date]))
The result is the following:
Sales last year
As you can see the sales per day shows 5.316€ and 3.546€, which is correct, but the total is 111.796 €, which is not correct. However, the measure with the formula without the natural day the sum of the two rows is correct. How could I solve this?
Thank you very much in advance
I just changed the order to calculate the date and it was solved.
sales_last_year = CALCULATE(Sales[Revenue]);DATEADD( SAMEPERIODLASTYEAR('Calendar'[Date]);+1;DAY))
I am trying to produce a query in SQLite where I can determine the average sales made each weekday in the year.
As an example, I'd say like to say
"The average sales for Monday are $400.50 in 2017"
I have a sales table - each row represents a sale you made. You can have multiple sales for the same day. Columns that would be of interest here:
Id, SalesTotal, DayCreated, MonthCreated, YearCreated, CreationDate, PeriodOfTheDay
Day/Month/Year are integers that represent the day/month/year of the week. DateCreated is a unix timestamp that represents the date/time it was created too (and is obviously equal to day/month/year).
PeriodOfTheDay is 0, or 1 (day, or night). You can have multiple records for a given day (typically you can have at most 2 but some people like to add all of their sales in individually, so you could have 5 or more for a day).
Where I am stuck
Because you can have two records on the same day (i.e. a day sales, and a night sales, or multiple of each) I can't just group by day of the week (i.e. group all records by Saturday).
This is because the number of sales you made does not equal the number of days you worked (i.e. I could have worked 10 saturdays, but had 30 sales, so grouping by 'saturday' would produce 30 sales since 30 records exist for saturday (some just happen to share the same day)
Furthermore, if I group by daycreated,monthcreated,yearcreated it works in the sense it produces x rows (where x is the number of days you worked) however that now means I need to return this resultset to the back end and do a row count. I'd rather do this in the query so I can take the sales and divide it by the number of days you worked.
Would anyone be able to assist?
Thanks!
UPDATE
I think I got it - I would love someone to tell me if I'm right:
SELECT COUNT(DISTINCT CAST(( julianday((datetime(CreationDate / 1000, 'unixepoch', 'localtime'))) ) / 7 AS INT))
FROM Sales
WHERE strftime('%w', datetime(CreationDate / 1000, 'unixepoch'), 'localtime') = '6'
AND YearCreated = 2017
This would produce the number for saturday, and then I'd just put this in as an inner query, dividing the sale total by this number of days.
Buddy,
You can group your query by getting the day of week and week number of day created or creation date.
In MSSQL
DATEPART(WEEK,'2017-08-14') // Will give you week 33
DATEPART(WEEKDAY,'2017-08-14') // Will give you day 2
In MYSQL
WEEK('2017-08-14') // Will give you week 33
DAYOFWEEK('2017-08-14') // Will give you day 2
See this figures..
Day of Week
1-Sunday, 2- Monday, 3-Tuesday, 4-Wednesday, 5-Thursday, 6-Saturday
Week Number
1 - 53 Weeks in a year
This will be the key so that you will have a separate Saturday's in every month.
Hope this can help in building your query.
I'm trying to create a query that can return the previous quarter balance for a series of records.
I have financial data for accountid (acctid), fiscal year (fyear), fiscal quarter (fquarter) and fiscal period (fperiod) that I'm summing and tracking through a series of other queries that I'm dropping in to a temporary table, that data includes the net change for the period (nperiod), net change for the quarter (nquarter) and net change for the year (nyear). Net change for the period is only the account transactions in the period, net change for the quarter is the cumulative total of the transactions that appear in periods 1-3, 4-6, 7-9 and 10-12 respectively (amount from previous periods are not calculated in proceeding quarters, ie the net change for the quarter resets to $0 for periods 4, 7 and 9) and the net change for the year is the total cumulative sum. I'm now trying to create a sql statement that returns the previous quarter end balance.
So for periods 1-3 I need to write a separate select statements to return the ending balance for the previous year, for periods 4-6 I want them ALL to return the net change for the quarter from period 3, for 7-9 I want to return the net change for the quarter from period 6 for all records and for period 10-12 I want to return net change for the quarter from period 9.
Can I get some assistance because I have a gigantic query that returns the max period per quarter, then the nquarter amount associated with that period and then trying to do a where exists, but something tells me there's a better way to do it.
Thanks!
I have a measure that gives me the month over month change in budget. The measure is defined as:
Bugdet Month over Month:=if(and(sum(budget[Value])<>0;[Budget, Previous Month]<>0);sum(budget[Value])-[Budget, Previous Month];0)
Budget, Previous Month:=CALCULATE(sum(budget[Value]);PREVIOUSMONTH(tDate[Date]))
When I show Bugdet Month over Month in a pivot table, it shows the correct numbers, but it shows 0 in the grand total. Is it possible to make some change somewhere to make it show the sum of all changes in the grand total?
Rough take is create a third measure:
Budget Month over Month Final:=SUMX ( VALUES(tdate[Date]), [Budget Month over Month])
This will independently iterate over every month in the given cell and calculated Budget Month over Month and then SUM them together. So for the month row values in your pivot table, this will be the equivalent of the current Budget Month over Month measure.
And for the Grand Total it will then show the sum of each individual month in the pivot.
I'm having trouble creating a subquery in Access to cough up what I need:
Input: SalespersonID, Amount, Date
I need to find what SalespersonID exceeds $2000 in a month (easy) AND what day in that month the running sum of Amount for that month exceeded $2000 (argh!).
I can groupby and get the first month any salesperson sum Amount > $2000 but I just can't figure out how to get the first date in that month when the month running sum of Amount>$22000
So you've got running-total in your tags - what did you try?
I would create a temporary table and generate the running total, and derive the month using the 'Month' function, and also derive the day of the month. If you sort by month and day each month's data will give you a running total that is what you want. This means you go month-by-month, and calculate the temp table for all salespersons.
It's not super elegant, but it'll give you what you want.