How can i get back from an opened website in the webview?

If i opened a link on the webview component i can’t get back from this opened website to the given website at the start because there is no back button.
so i mean: if the starting page is google and i click on any of the shown links, i can’t get back to the google results.
How can i fix this?

You will need to handle this using the Component.onLocationChange event, and use formula to enable navigation buttons when the site is not yours.

which flow function do i need? And what do i have to write as formula?

You will need an IF condition, and you will need to use this as your formula:

IS_EQUAL(outputs[“Component onLocationChange”].event.url,“your site here”)

This compares the Webview value and your site. The first output on the IF block is what you want to do when it’s true. This could be displaying buttons.

I have changed the condition of the if-function to formula and got this error

I would recommend going through all the AppGyver tutorials. Once you do that, that error will be simple to solve. I can’t teach you everything here, unfortunately. There are some great resources on YouTube and Vimeo, however!

I didn’t got it… I now used the open web browser function instead of the webview. there i can close the page with an back button. But also the webview does not work correctly in the web preview. If i now click on an object with the open browser flow function, it opens a new tab of my browser.

In the web, it kinda makes sense that “Open browser” opens a new tab – it would be weird to implement e.g. Google within a web app.

Can you describe your use case in more detail so we could potentially offer alternative ideas on how to tackle it?

yes, that makes sense.