Syntax Troubleshooting for Show_If Formula

[mykrobinson]

I am trying to craft an expression to dynamically show/hide a field based on the status of another column. Essentially what I am trying to say is, if a certain field is set to "signed off" to so the long text field for "sign off notes."

The expression passes in the expression assistant, but once I get back to the data columns view, it shows an error and breaks the app and shows an invalid constaint.

Need some help understanding what i've done wrong here, it's a bit confusing since the expression seems to pass.
EDITED: wrong image posted initially

[mykrobinson · Accepted solution]

UPDATE: I think i fixed it. The expression passed in the assistant, but it was wrong in the context of the SHOW expression, because SHOW_IF is assumed. I did not need to add an IF statement, i just needed to put the condition i wanted to return true.

So i just changed the expression to   [Status]="Signed Off"

[Joseph_Seddik]

It is not about the use of IF statement which you can surely use. You might have a more complex condition and you might need to use IF() or any other statement as long as it returns a Yes/No value.

The problem with your expression is it does NOT return a Yes/No value, rather a text value. This is because you have quoted the return values, forcing them to be of text type. Therefore, instead of "True" and "False", you should have written TRUE and FALSE without the quotes, and your expression should work equally.