Hi Folks,

I’m sure this is easy but I am stuck trying to figure out a formula that will return a different number based on step count.

Here’s what I want:

if #ZSC#<=4900,0

if #ZSC#>4900&&#ZSC<=6700,1

if #ZSC#>6700&&#ZSC<=8500,2

if #ZSC#>8500&&#ZSC<=10000,3

if #ZSC#>10000&&#ZSC<=11300,4

if #ZSC#>11300&&#ZSC<=13000,5

I thought about uploading pictures of the numbers but was wondering if a formula could do it.

Thanks!

Amy Beryl

Should just be

$#ZSC#<=4900?0: $

and

$#ZSC#>4900?0:$$#ZSC#<=6700?1:0$

1 Like

Are the 2 parts you showed above placed on the same element or on 2 separate elements?

If its on the same element, what is the conjunction or joiner? What replaces the “and”?

its up to you either way would work but i would just make one long code like this but i dont know how to test it lol

$#ZSC#<4900?0:$$#ZSC#<6700?1:$$#ZSC#<8500?2:$$#ZSC#<10000?3:$$#ZSC#<11300?4:$$#ZSC#<13000?5:0$

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Do the ranges have to be exactly the numbers you chose? If not you could slightly adjust them to fit on a straight line. In which case the following would give you around the same result

```
$#ZSC#<5070?0:(floor(0.0006253736*#ZSC#-3.170054))$
```

This would give you divisions at

5070 -v- 4900

6669 -v- 6700

8268 -v- 8500

9867 -v- 10000

11466 -v- 11300

13065 -v- 13000

Hi Amy,

Its sort of a combination of what they both said which is correct:

$#ZSC#<=4900?0:$$#ZSC#>4900&&#ZSC<=6700?1:$$#ZSC#>6700&&#ZSC<=8500?2:$ AND continue with the rest…

Enjoy~Sirhc

1 Like