Hello, could you help me with the following question?
In my application, the user will enter a value. Based on this value, it will be checked if the value is between the minimum and the maximum pre-established. If it is between min and max, the value will receive a status of Y at the end, if it is a number less than min or greater than max, its result will be N. But in some cases, the minimum pre-established value is negative. When min is negative, any value entered, including those that technically will be outside the established MIN and MAX standard, is evaluated as Y.
Hi, I already use an IF expression in my app
IF (AND ([value_1]> = [min], [result_1] <= [max]), โYโ,
IF (AND ([value_2]> = [min], [value_2] <= [max]), โYโ, โNโ))
But how am I going to use this negative value issue? If any positive number is it, will it always be greater than it
IFS(), not IF().
Does not work. Would you have any examples to show me? I did the expression using IFS but it didnโt solve my problem
Please post a screenshot of the expression you tried.
Read the posted help article.
I used this and made the substitutions, it still doesnโt work
Where did you use it?
initial value
Was there another attempt at the expression since the last one you posted?
Yes, I made another attempt but it didnโt work. No matter what value I enter in the result, it always returns me as Y.
That was my last attempt.
IFS(
([value_1] < [min]),
โNโ,
([value_1] >= [min]),
โYโ,
([value_1] > [max]),
โNโ,
([value_1] <= [max]),
โYโ,
([value_2] < [min]),
โNโ,
([value_2] >= [min]),
โYโ,
([value_2] > [max]),
โNโ,
([value_2] <= [max]),
โYโ)
Good. Now re-incorporate the AND() expressions similar to how you had originally done. Remember you just have 3 conditions to check against, so you should only have 3 condition-value pairs in the IFS().
The doc posted above has several examples. Perhaps you should look at it.
You have 3 cases as shown above. Youโll have to check the last case first as it can override the logic for the other 2.
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