Sir I am making an Browser application. There are 5 screen in app, the 1 to 4 screen shows multiple web pages with home URL of Google. In the Fifth screen there are buttons name Tab1, Tab2, Tab3, Tab4. In my app I want that when user click on tab1 it opens the screen1 with the last search on screen1 but it opens the home URL. I don’t know how to set up this function in blocks , Please Please help me Sir.
Variables are not passed between screens in Classic. So you need to do an Screen initialize reload variables, and you can keep them in a TinyDB
You have to be careful how you close and open screens or you will have other issues.
Most important. You probably don’t need four screens. Look at how to hide and show elements based o actions. Most apps could be created on one screen if done properly.
Sir can you sent me the picture of Blocks placement to solve this problem. I don’t know how to do the 1 point issue.
Take a look at this thread. It handles a few of the issues
try this solution
Note : You have to change the tag [ i.e ct_1 ] according to the Screen. i.e for Screen 1 tag is ct_1. Similarly for the Screen 2 tag will be ct_2 and so on.
Clock Properties :
- Uncheck Timer Always Fires
- Uncheck Timer Enabled
- Timer Interval (ms) : 1000
Hope this solution will help you out!!
You have to make sure you CLOSE the screens you open once you are done with them, otherwise you run out of memory or have other issues.
see the general tips, especially tip 1
what about using only one screen to avoid the hassle?