Need Script for calculating workdays between 1st record, 2nd record, so on.
Ok, Thanks to I am able to calculate the working days since a previous record for a client.
with the following script
let t := this;
let xBase := ('Clark Leads'.Chronos order by 'Date + Time');
let xCurIndex := index(xBase, t);
if xCurIndex = 0 then
0
else
let xPrev := item(xBase, xCurIndex  1);
workdays(xPrev.date('Date + Time'), t.date('Date + Time'))  1
end
However, for accountability purposes, I have started another Sub Table that Provides Statistics for Avg (prefer median) working days for all client Chono Records. I have the overall average for the correct statuses I need using the following script:
do as server
avg((select Leads where 'Lead Student Type' = 2 and (Status = 1 or Status = 2 or Status = 12 or Status = 13)).'Days Since Last Contact')
end
Now I want to separate the overall Stats for the amount of working days between each contact. For instance, I want the avg (prefer median) of working days between the 1st contact (first chrono record for that client) and the second contact (second chrono record for that client). I will continue replicating this script all the way up to between their 4th and 5th contact.
I am keeping these stats on a subtable with the Parent Table being Leads.
Any help is greatly appreciated.
15 replies

Dave Irving said:
For instance, I want the avg (prefer median) of working days between the 1st contact (first chrono record for that client) and the second contact (second chrono record for that client).I'm unsure what you want here.
You have the number of days between chrono record for a parent client record. I don't know how you can find the average between 1st and 2nd contact of an individual client record since you only have 1 data set for 1st contact and one data set for 2nd.
Or are you wanting to find the average number of days between all 1st and 2nd contacts for all clients? This you can do since you will now have more than 1 number to average.
You have the number of days already figured out. You can just find the right filter then avg('Days Since Last Contact').

Yes, I need the avg for All Leads meeting a certain filter. But I need averages for each situation. The avg of days between 1st Chrono record and second, and so forth as described above. This, I need the script to be able to sort each individuals chronos records by date...avg the days between 1st ordered record by date and 2nd record by date. It's more complicated then just doing the average of days between all records, because it needs to split it up.

said:
This, I need the script to be able to sort each individuals chronos records by date...avg the days between 1st ordered record by date and 2nd record by date.You have the field Days Since Last Contact which finds the number of days between the current record in Chronos and the previous record when ordered by a date field. So if we can get the 2nd record in Chronos for each record in 'Clark Leads' then take the average of 'Days Since Last Contact' we will get what you want?
If that is true then this should do it:
let x := (select 'Clark Leads'); let y := for loop1 in x do item(loop1.Chronos order by 'Date + Time', 1) end; avg(y.'Days Since Last Contact')
Line 1, gets all the records in 'Clark Leads', if that is not the table name then change it to match your table name. We start here because we can use the relatonship to Chronos to find the 2nd record in Line 3.
Lines 2  4, is the for loop that will go through each record in 'Clark Leads'. Then it will take all related records in the Chronos relationship field order it by 'Date + Time' field and then find the 2nd record in the array of records. Since Ninox starts all list with a 0, the 2nd record is number 1.
Line 5, then finds the average of the field 'Days Since Lat Contact'.

For median you can try something like:
let mynumber := ((select 'My table') order by Number).Number; let mycnt := cnt(mynumber); let mymedian := if even(mycnt) then (item(mynumber, mycnt / 2  1) + item(mynumber, mycnt / 2)) / 2 else item(mynumber, mycnt / 2) end; mymedian
Since Ninox starts with 0 you have subtract 1 after you divide the count so you can get the correct value. For example if you have 4 numbers (2,5,8,12). The count is 4 and half of that is 2. If you ask for the value of the item 2 Ninox will return 8, since 2 is 0.

Dave Irving said:
However, still need to not include the 1st chrono record for all leads so we can remove the 0 values from the equation.This code might work. I think we still need to sort so we can make sure to drop the proper first record.
let curRec := this; let xStart := 'Start Date'; let xEnd := 'End Date'; let x := (select Leads where 'Lead Student Type' = 2 and 'Lead Entered Date' >= xStart and 'Lead Entered Date' <= xEnd); let y := for loop1 in range(1,count(x)) do item(x,loop1) end; avg(y.'Days Since Last Contact')
Since Ninox starts all arrays with 0, we create a range starting with 1 and we automagically have an array that drops the first instance.
Here is what it looks like in my test:
The first line is all the records in a table. The second line drops the first instance of the returned array.
Here is the code:
let x := (select Table1); let y := for loop1 in range(1, count(x)) do item(x, loop1) end; concat(x) + " " + concat(y)

said:
Here is the script for that "Days Since Last Contact" field. I replaced 0 with null, but that broke it.So if you want xCurIndex to be null then you have to think the opposite. Instead of looking for = 0 do the opposite and have no "else". Like:
let t := this; let xBase := ('Clark Leads'.Chronos order by 'Date + Time'); let xCurIndex := index(xBase, t); if xCurIndex != 0 then let xPrev := item(xBase, xCurIndex  1); workdays(xPrev.date('Date + Time'), t.date('Date + Time'))  1  cnt(select '20232024 Holiday Schedule' where Date >= xPrev.date('Date + Time') and Date <= t.date('Date + Time')) end
Zero is not null. Zero is a value, as you find out when you do an average. It is included.
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