Hi,
I have a tableview which shows a webview on clicking some row in the table, which in turn picks up the data from an sqlite. if a user closes the app by pressing home key while viewing a description in webview and reopens it after sometimes, I should be making the user to see the same screen. how to show the same view again ? What is the efficient way ?
The easiest way is to write a string to a file in your documents directory that holds your state information in some easily parsed format. For example "screen_name:database_id". Or build a JSON string and parse it with JSON.
Everytime you switch screens, rewrite the file. When you open next time, read the file, parse it, and show the appropriate screen in your app.
Well I think the easiest way is to store the state of the application in NSUserDefaults. There is a delegate method on UIApplication called:
- (void)applicationWillTerminate:(UIApplication *)application
This delegate method gets called when the user quits the app. This is the time where you can save the state of your application off to the NSUserDefaults. But be aware that you cannot do time intensive stuff there. If you do, you get killed by the OS.
In your case why not simply store the row the user picked in NSUserDefaults and then check in
-(void)applicationDidFinishLaunching:(UIApplication *)application
if there is a saved row and restore the screen approbiately.