CODEWITHSUNDEEP
oracle-databaseoracle-apex
Dean Newstead
Dean Newstead

Reputation: 35

ORACLE APEX - Set up a Button to populate a textfield

I have a button which I would like to take the Input text field called P7_INPUT and Divide (/) it by 2000 and pop it into the P7_OUTPUT

Users enters say 3000 into P7_INPUT field and pressed the Button then in the P7_OUTPUT is will show the value of 1.5

enter image description here

Upvotes: 0

Views: 1790

Answers (3)

Jeffrey Kemp
Jeffrey Kemp

Reputation: 60262

Since you require the action to occur dynamically when the user clicks a button, I would create a Dynamic Action on button click with a Set Value action for P7_OUTPUT set to a JavaScript expression:

parseFloat($v("P7_INPUT)) / 2000

This avoids a server round-trip.

Upvotes: 1

Imran
Imran

Reputation: 169

Right-click on the :P7_INPUT and select Create Dynamic Action. Set the following properties:

Event: Key Release
Selection Type: Items(s)
Item: P7_INPUT

Right-click the dynamic action and select either Create TRUE Action. Set the following properties:

Action: Set Value
Set Type: PL/SQL expression 
PL/SQL expression : NVL(:P7_INPUT,0) / 2000
Items to Submit: P7_INPUT
Affected Elements > Selection Type: Item(s)
Affected Elements > Item(s): P7_OUTPUT

Upvotes: 3

Littlefoot
Littlefoot

Reputation: 142733

You don't even need a button.

  • create P7_INPUT as a text item; set its "Submit when enter pressed" property to YES
  • create P7_OUTPUT

  • create a process which will contain a PL/SQL code:

    :P7_OUTPUT := :P7_INPUT / 2000;

That's all; when you enter e.g. 3000 into the input item and press enter, process will do the calculation and display the result in the output item.

If it must be with a button, create it and let it submit the page. Process will look the same, you'd just put P7_BUTTON into process' "Server side condition's" "When button pressed" property.

Upvotes: 1

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