iPhone - How to have a custom browser be able to detect if back/forward buttons need to be enabled/disabled - iphone

Question

Hey guys! I have a tumblr app that I am developing, and I have a button that pushes a modal view that contains a webview and a toolbar on the bottom with refresh, back, forward and done buttons. The back forward and refresh buttons work, however I want to be able to tell if the webview can actually go back/forward and if not, disable the button..I have tried the code below, and the image nor the enable/disable changes.
- (IBAction)refreshPage {
[signUpWebView reload];
}
- (IBAction)goBack {
[signUpWebView goBack];
}
- (IBAction)goForward {
[signUpWebView goForward];
}
-(void)webViewDidFinishLoad:(UIWebView *)webView
{
// Enable or disable back
if ([signUpWebView canGoBack]) {
[backOnePage setEnabled:YES];
backOnePage.image = [UIImage imageNamed:#"backButton"];
} else {
[backOnePage setEnabled:NO];
backOnePage.image = [UIImage imageNamed:#"backButtonDisabled"];
}
// Enable or disable forward
if ([signUpWebView canGoForward]) {
[forwardOnePage setEnabled:YES];
forwardOnePage.image = [UIImage imageNamed:#"forwardButton"];
} else {
[forwardOnePage setEnabled:NO];
forwardOnePage.image = [UIImage imageNamed:#"forwardButtonDisabled"];
}
}
Any recommendations would be greatly appreciated!

Answer 1

I recommend binding the forward and backward button directly to the UIWebview.
Automatically enable and disable like this:
- (void)webViewDidStartLoad:(UIWebView *)mwebView {
backButton.enabled = (webView.canGoBack);
forwardButton.enabled = (webView.canGoForward);
}
- (void)webViewDidFinishLoad:(UIWebView *)webView {
backButton.enabled = (webView.canGoBack);
forwardButton.enabled = (webView.canGoForward);
}

Answer 2

When you first initialize your webView, try starting it with a request, like this:
[signUpWebView loadRequest:[NSURLRequest requestWithURL:[NSURL URLWithString:#"http://www.google.com"]]];
So you're initializing the webView with loadRequest: instead of loadHTMLString:, or something similar.