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parenclose where EoF is expected error

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Posted on by ahhollan

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I'm having trouble figuring out what is wrong with my formula. I keep getting the error Parenclose was found when EoF is expected. I'm sure it's that I haven't closed something off right. Below is my formula.

Set(varRecord,SubmitForm(Form3_1);If(!IsBlank(varRecord), Notify("Your request has been submitted! You may now close the window.", NotificationType.Success), Notify("Your request was not submitted", NotificationType.Error))

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Mr-Dang-MSFT Microsoft Employee on at

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Hi @ahhollan ,

I've reformatted your formula for readability below:

Set(varRecord,SubmitForm(Form3_1); If( !IsBlank(varRecord), Notify("Your request has been submitted! You may now close the window.", NotificationType.Success), Notify("Your request was not submitted", NotificationType.Error) )

It looks like you're missing a parenthesis to close the Set() formula.

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ahhollan 576 on at

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Now I get 'IsBlank has invalid argument' error. Any thoughts?

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WarrenBelz 155,290 Most Valuable Professional on at

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@ahhollan ,
Try - this is definitely valid syntax
Set( varRecord, SubmitForm(Form3_1) ); If( !IsBlank(varRecord), Notify( "Your request has been submitted! You may now close the window.", NotificationType.Success ), Notify( "Your request was not submitted", NotificationType.Error ) )

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ahhollan 576 on at

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get 'The function 'IsBlank' has some invalid arguments.

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Mr-Dang-MSFT Microsoft Employee on at

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Hi @ahhollan ,

If varRecord is indeed a full record which has many fields, it may not make sense to check if the full record itself is blank, but rather a specific field of the record.

So you can drill into varRecord.column_name and check if that specific column is blank. Change column_name to the column of your choice.

The resulting formula might look like this:

Set(varRecord,SubmitForm(Form3_1)); If( !IsBlank(varRecord.column_name), Notify("Your request has been submitted! You may now close the window.", NotificationType.Success), Notify("Your request was not submitted", NotificationType.Error) )

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WarrenBelz 155,290 Most Valuable Professional on at

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Hi @ahhollan ,
Back in my time zone - try
Set( varRecord, SubmitForm(Form3_1) ); If( !IsEmpty(varRecord), Notify( "Your request has been submitted! You may now close the window.", NotificationType.Success ), Notify( "Your request was not submitted", NotificationType.Error ) )

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WarrenBelz 155,290 Most Valuable Professional on at

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Hi @ahhollan ,
Just checking if you got the result you were looking for on this thread. Happy to help further if not.

Please click Accept as solution if my post helped you solve your issue. This will help others find it more readily. It also closes the item. If the content was useful in other ways, please consider giving it Thumbs Up.

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ahhollan 576 on at

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I ended up having to use your code in conjunction with a patch function. I did get it working.

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Mr-Dang-MSFT Microsoft Employee on at

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Hi @ahhollan ,

Upon using SubmitForm, you can reference the record that was created by something like:

EditForm1.LastSubmit

Where EditForm1 is the name of the form control and LastSubmit drills into the record that was created. If you want to look at a specific column, add another dot and the column name:

EditForm1.LastSubmit.column_name

So you won't need to create a variable to show the record that was created when using SubmitForm.

But for greater ease and taking advantage of the form control's properties, when using SubmitForm, you can place each notify action in the form's OnSuccess and OnFailure properties.

When a record is succesfully submitted, the OnSuccess notification will trigger.

When a record fails to be written, the form will show the fields that have an error and the Notify action you place in OnFailure will execute.

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