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amitk9244
amitk9244
Participating Frequently
Nov 7, 2022 6:08 AM UTC
Solved

How to round up large Number in appian

I want to roundup below result to ...upto two places like 3.13....how could i achieve that

please help

peter.lewis
peter.lewisAnswer
Employee
Nov 7, 2022 6:05 PM UTC

There's probably a more efficient way to do this, but here's at least an example that would work. I think you have to get your result into a state where it can easily be rounded, so I used a combination of log() and power():

a!localVariables(
  /* Original Variables */
  local!interest: 10,
  local!months: 12,
  
  /* Original Expression */
  local!base: 1+ local!interest,
  local!power: power(local!base, local!months),
  
  /* New Logic */
  local!numberOfDigits:floor(log(local!power, 10)),
  local!decimalOnly: local!power / power(10, local!numberOfDigits),
  local!rounded: round(local!decimalOnly, 2),
  
  /* Final Result */
  local!rounded * power(10, local!numberOfDigits)
)

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8 replies

mohank0008
mohank0008
Known Participant
Nov 7, 2022 6:24 AM UTC

roundup(7.36819e+12,2) gives the result of 7.37

amitk9244
amitk9244Author
Participating Frequently
Nov 7, 2022 6:27 AM UTC

Not working

sanchitg0002
sanchitg0002
Inspiring
Nov 7, 2022 6:30 AM UTC

You are trying to round up 3.138428e+12 (which is equivalent to 3138428000000) to 2 decimal points. How's that can be possible? roundup() will work when you will have a value containing decimal points. For example: roundup(3.1414, 2) gives 3.14

amitk9244
amitk9244Author
Participating Frequently
Nov 7, 2022 6:32 AM UTC

ok.. got it, but i want lesser value

sanchitg0002
sanchitg0002
Inspiring
Nov 7, 2022 6:37 AM UTC

What do you mean by lesser value?

Acc to your code, 11^12 will return 3.138428e+12 only

jeffw0532
New Participant
Nov 7, 2022 2:24 PM UTC

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csteward
csteward
Brainy
Nov 7, 2022 3:48 PM UTC

Also don't forget the fixed() function, just note it does return a text value of the number:

a!localVariables(
  local!base: 1+ri!interest,
  local!power: power(local!base,ri!months),
  /*local!round: round(local!power,3),*/

  fixed(local!power,2)
)

peter.lewis
peter.lewisAnswer
Employee
Nov 7, 2022 6:05 PM UTC

There's probably a more efficient way to do this, but here's at least an example that would work. I think you have to get your result into a state where it can easily be rounded, so I used a combination of log() and power():

a!localVariables(
  /* Original Variables */
  local!interest: 10,
  local!months: 12,
  
  /* Original Expression */
  local!base: 1+ local!interest,
  local!power: power(local!base, local!months),
  
  /* New Logic */
  local!numberOfDigits:floor(log(local!power, 10)),
  local!decimalOnly: local!power / power(10, local!numberOfDigits),
  local!rounded: round(local!decimalOnly, 2),
  
  /* Final Result */
  local!rounded * power(10, local!numberOfDigits)
)

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