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Power Platform Community / Forums / Power Apps / Notify function isn't ...
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Notify function isn't working using the result of others TextInputs

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Posted on by 65

Hi, im try to set a notification using the Notify() function in a TextInput, but isn't working and i don't know why.

 

In the OnChange property of the TextInput15_1 im using the next code :

 

 

 

If(Value(TextInput15_1.Text)>12;Notify("Value is greater than 12";NotificationType.Error;10))

 

 

 

And in the Default property by adding the TextInput6 and TextInput15. Please see the next images

 

Crissceron_2-1659637794237.png

 

Crissceron_1-1659637613276.png

 

Then, when i play the App and type a values in the TextInput6 and TextInput15 with a result greater than 12 nothing happens.

No notification appears.

 

Can someone tell me why isn't working or how can i do it?.

Thanks! 

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I have the same question (0)
  • Akash17 Profile Picture
    549 Super User 2024 Season 1 on at

    Hi @Crissceron 

    If you go to TextInput15_1(Result textfield) and you just Manually enter any value > greater than 12 into your TextInput15_1 textfield and click on any part of the screen(anywhere on screen) you will get notify this is how OnChange property works

  • Crissceron Profile Picture
    65 on at

    Hi @Akash17 

    I know that with a Manually value is gonna work... but i need that the Notification appear with the result of a formula.. (something like if i type a value in the TextIpunt6 and in the TextInput15.. then just with the result of the TextInput15_1 the notification appear... without typing anything in the TextInput15_1).

     

  • Verified answer
    Crissceron Profile Picture
    65 on at

    If someone is searching for a solution to a similar problem... this is the solution.

     

    - Add a toggle and in the Default property set the code as:

     

    If(Value(TextInput15.Text)<>12;true;false)

     

    - Then in the OnChange property of the Toggle set the next code:

     

    If(Toggle1.Value=true;Notify("Value is not 12";NotificationType.Error;1000))

     

    - Then in the Visible property of the toggle set as false.

     

    Thats make that when you type values into the TextInput6 and TextInput15 the toggle is gonna check the value of the result in TextInput15 and if the value is not 12, then the toggle is true and the notification will appear. Otherwise nothing happens.

     

     

     

     

  • Akash17 Profile Picture
    549 Super User 2024 Season 1 on at

    Great!
    You got the solution...

    I thought you are looking for the solution with texinput field control itself........ 😀

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