I'm new to the iOS development
I have develop an application to open url in UIWebView
but that web site working fine in safari
web browser but when I taped a link coding as window.open()
it will load same webview
. here is my code
ViewController.h file
//
// ViewController.m
//
//
// Created by Code Kadiya on 5/18/17.
// Copyright © 2017 Code Kadiya. All rights reserved.
//
#import "ViewController.h"
@interface ViewController () <UIWebViewDelegate>
@end
@implementation ViewController
- (void)viewDidLoad {
[super viewDidLoad];
[self.webView setDelegate:self];
// Do any additional setup after loading the view, typically from a nib.
NSURL *websiteUrl = [NSURL URLWithString:@"http://some.url"];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:websiteUrl cachePolicy:NSURLRequestReturnCacheDataElseLoad timeoutInterval:15.0];
[_webView loadRequest:request];
}
- (BOOL) webView:(UIWebView *)webView shouldStartLoadWithRequest: (NSURLRequest *)request navigationType: (UIWebViewNavigationType)navigationType {
if (navigationType == UIWebViewNavigationTypeLinkClicked) {
NSURL *url = [request URL];
[[UIApplication sharedApplication] openURL: url options:@{} completionHandler:nil];
return NO;
}
return YES;
}
- (void)didReceiveMemoryWarning {
[super didReceiveMemoryWarning];
// Dispose of any resources that can be recreated.
}
@end
and I found this code from loading web application source
<a onclick="open_pdf('office')" target="_blank">
function open_pdf(type) {
window.open("http://some.url?type=" + type );
}
anyhow I need to open safari browser when I click that link but I cannot change the web application source code
I think
[[UIApplication sharedApplication] openURL: url options:@{} completionHandler:nil];
not executing for that type of links