Matching Display of Digital Switch Button to Actual Position

I completed a garage door project that determines the status of the door and open/close it via Blynk. Works great and will publish. My wife (yea… I know…blame the wife…it’s never me :wink:) has a tendency to click things on her iPhone by accident. To prevent accidental garage door openings or closings I put a ‘lockout’ on a digital switch button. So, to open/close the door you must first press on your phone the unlock button (a digital push button in switch mode) that activates (closes) a relay and then another digital button (in push mode) to activate another relay that will open/close the door. When the door starts to open (or close) my code opens the lockout relay to reactivate the lock.

Here’s the problem, abeit a minor one. After opening/closing the door the display on the lockout button on my Blynk app does not match up with the actual position of the relay. I’ve tried Blynk.syncAll() with no success. It seems as if the commands available are focused of the button being a virtual pin
Any ideas ?

If by your statement you mean that you are using direct pin control from the widget, then yes… what you see is what you get.

If you wish to have colour and/or label changes, or even button/slider state or status feedback, based on other inputs, then you need to be using virtual pins, Blynk.setProperty(vPin, “property”, value) and Blynk.virtualWrite(vPin, value) commands, and write your code accordingly.

Post up your code so we can assist more.

Thanks. That is what I thought was needed. I was hoping there was a command that I missed in the Docs.