嗯,我在做同样的事情。我不得不运行一个冻结我的UI的同步ajax请求。所以这是我如何修复它:
__block NSString *message;
dispatch_queue_t q = dispatch_queue_create("sign up Q", NULL);
dispatch_async(q, ^{
NSString *function = [[NSString alloc] initWithFormat: @"signup(\'%@\',\'%@\',\'%@\')",self.email.text,self.password.text,self.name.text];
dispatch_async(dispatch_get_main_queue(), ^{
NSString *result = [self.webView stringByEvaluatingJavaScriptFromString:function];
NSLog(@"%@",result);
if ([result isEqualToString:@"1"]) {
message = [NSString stringWithFormat:@"Welcome %@",self.name.text];
[self.activityIndicator stopAnimating];
[UIApplication sharedApplication].networkActivityIndicatorVisible = NO;
}
else {
message = [NSString stringWithFormat:@"%@ is a registered user",self.name.text];
[self.activityIndicator stopAnimating];
[UIApplication sharedApplication].networkActivityIndicatorVisible = NO;
}
UIAlertView *alertView = [[UIAlertView alloc] initWithTitle:@"Message" message:message delegate:self cancelButtonTitle:@"Okay" otherButtonTitles: nil];
[alertView show];
});
});
逻辑很简单。去一个新的线程,并从内部,派遣到主队列,然后做JS工作,一切工作像我的魅力...
这将有助于? – 2012-02-08 17:19:07