GoTo URL in UIActionSheet

[newbieToo]

I'm new to sdk development and am sure it's very simple, but can't seem to get it...what I would like to do is go to a specific URL after clicking on the "GoToURL1" button in a UIActionSheet.

This is the code I have so far:

----

UIActionSheet *actionSheet = [[UIActionSheet alloc]
initWithTitle:@"Title"
delegate:self
cancelButtonTitle:@"Cancel"
destructiveButtonTitle:@"Button1"
otherButtonTitles:@"GoToURL1", @"GoToURL2", @"Open in Safari", nil];
[actionSheet showInView:self.view];
[actionSheet release];
}

---

Any help would be greatly appreciated!

[ftm]

Try something like this:

-(void)actionSheet:(UIActionSheet *)actionSheet clickedButtonAtIndex:(NSInteger)buttonIndex
{
	if (buttonIndex == 0)
	{
		[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://yourUrlHere"]];
	}
}

[newbieToo]

Originally Posted by ftm

Try something like this:

-(void)actionSheet:(UIActionSheet *)actionSheet clickedButtonAtIndex:(NSInteger)buttonIndex
{
	if (buttonIndex == 0)
	{
		[[UIApplication sharedApplication] openURL:[NSURL URLWithString:@"http://yourUrlHere"]];
	}
}

Thanks for taking the time to answer! It does take me to the URL, but how can i have the URL open in the current app without opening up in Safari?

To elaborate on my app, it's a web browser that allows users to click on a button to be taken to a particular url. I would like them to stay in my app unless they choose to open it up in Safari.

[ftm]

If you want them to stay in your App, you need to put a UIWebView component in your app. If the UIWebView is visible when the button is pressed, just set the URL of the UIWebView to the URL of your page. If it is not visible, make it visible, then set the URL.

[S.LakshmiKanth Reddy]

Originally Posted by ftm

If you want them to stay in your App, you need to put a UIWebView component in your app. If the UIWebView is visible when the button is pressed, just set the URL of the UIWebView to the URL of your page. If it is not visible, make it visible, then set the URL.

If i use openURL for sending SMS and mails then it is terminating my application, how to reopen my application, if anyone knows how to do it please help me out.

Thank you,

Lakshmikanth.

[ftm]

Using openURL will always exit the App and take you to Safari. There is not a way to automatically get back in to your application.

If you want the user to view a web page and return to your App, you will have use a UIWebView component in your app instead. This will allow viewing of web pages without leaving the application.

[newbieToo]

Originally Posted by ftm

Using openURL will always exit the App and take you to Safari. There is not a way to automatically get back in to your application.

If you want the user to view a web page and return to your App, you will have use a UIWebView component in your app instead. This will allow viewing of web pages without leaving the application.

If you open a new UIWebView (to keep the user in the app), can you also keep the previous view to quickly move back to the original webpage without reloading, similar to how you can open multiple web sites in Safari?

[jsd]

Sure, you can keep as many UIWebViews around as you want... just be sure to handle the low memory warning by tossing the ones that aren't in use at the moment.

[S.LakshmiKanth Reddy]

Originally Posted by ftm

Using openURL will always exit the App and take you to Safari. There is not a way to automatically get back in to your application.

If you want the user to view a web page and return to your App, you will have use a UIWebView component in your app instead. This will allow viewing of web pages without leaving the application.

Yes, i agree with you, we can do that only if we are willing to open a web page, but here i am calling the sms service like "sms:17737452345", this will redirect to the sms application.

[TTBAKIATWOAM]

Originally Posted by S.LakshmiKanth Reddy

Yes, i agree with you, we can do that only if we are willing to open a web page, but here i am calling the sms service like "sms:17737452345", this will redirect to the sms application.

What about on a button press if I want to call a script on a web site but I do not want to display the page I am calling and I do not want to leave the app.

Example...

I want a text entry field that you can enter information into then hit submit. with out leaving the app i need that text to be placed in the database Via a PHP script task But after they hit submit the user should not see a thing. except maybe a success message from the server

...

I want to call urls with out opening them How do i do that?

-TTBAKIATWOAM

[jsd]

Originally Posted by TTBAKIATWOAM

What about on a button press if I want to call a script on a web site but I do not want to display the page I am calling and I do not want to leave the app.

Example...

I want a text entry field that you can enter information into then hit submit. with out leaving the app i need that text to be placed in the database Via a PHP script task But after they hit submit the user should not see a thing. except maybe a success message from the server

...

I want to call urls with out opening them How do i do that?

-TTBAKIATWOAM

You can use NSURLRequest to talk to a web server without leaving your app or displaying any UI elements. Personally I prefer the third-party ASIHTTPRequest library, as it handles a lot of the boring detail work for you. ASIHTTPRequest Documentation - All-Seeing Interactive