Hello everyone, I’m a beginner, just starting to slowly understand the platform. Immediately I apologize for my English, as I use a translator. the question is, I’m making an application, and faced with the fact that you can’t do many screens, my application is tied to a website, and accordingly, it requires many screens for web viewing, since the side menu plays the role of a site menu, in general, you need help, you need to go to another and automatically launch a specific link in a web view, not one that plays the role of a home page, but a specific one, that is, an example, there is a screen two, it has five buttons, each button has its own link and when you click on certain button should The first screen opens with the link that was in the button. please help, I do not want to make a thousand screens, so that later the application would slow down
Instead of using multiple screens, you can use arrangement and set them to visible or invisible when needed. This topic might be some help.
Also, you might want to search for related topics before creating new ones.
You can do your whole app in one screen if you want. You say you have a lot of webpages? Then why don’t you just use one webviewer and call the webpages that the user request?
And for the links, you can use the WebViewer Go Back block like this:
If you can’t see the blocks, click on the link.
Post in your language. Kodular support also other language
@msr79526 Спасибо за разрешении общаться на русском языке)
Всем остальным спасибо за предложенные варианты решения проблемы, попробую каждый вариант
Your language doesn’t support kodular translator
Ok, then we will communicate using Google translator))
That last exchange was quite funny.
I don’t think it is working at all
In screen TWO you have 5 buttons with diferents link.
On click you can use the block “close screen with value”.
This block has a “return” property where you can put the link refered by the button.
In screen1 you should use block “when Screen1 Other Screen Closed”
This blocks has 2 properties: “otherScrenName, result” THEN:
IF “OtherScrenName” = “ScreenTwo”
webview.URL = “result” (this is the link you chose above)
Thanks to everyone for the proposed solutions to my problem, I figured out how to use vertical scrolling and web browsing, switching them with buttons, using component visibility, many thanks for the help