UIWebView 打开外部链接

Posted 2023-03-12

【中文标题】UIWebView 打开外部链接

【英文标题】：UIWebView opening external links

【发布时间】：2014-05-23 09:40:30

【问题描述】：

我有这个 uiwebview 代码，除了强制任何链接不使用 uiwebview 打开的部分（基本上我希望在浏览器中打开链接）之外，它在主要工作上有效。

谁能指出我正确的方向或看看我的代码有什么问题

JGViewController.h #导入

@interface JGViewController : UIViewController <UIWebViewDelegate> 
    IBOutlet UIWebView *customWebView;

@property (strong, nonatomic) IBOutlet UIWebView *customWebView;

@end

JGViewController.m

#import "JGViewController.h"

@interface JGViewController ()

@end

@implementation JGViewController
@synthesize customWebView;

//force any links to open in browser rather than in the uiwebview
-(BOOL) webView:(UIWebView *)customWebView shouldStartLoadWithRequest:(NSURLRequest*)request navigationType:(UIWebViewNavigationType)navigationType 
    NSURL* url = [request URL];
    if (UIWebViewNavigationTypeLinkClicked == navigationType) 
        [[UIApplication sharedApplication] openURL:url];
        return NO;
    

    return YES;


- (void)viewDidLoad

    [super viewDidLoad];

    // Do any additional setup after loading the view, typically from a nib.
    NSString *httpSource = @"http://www.google.co.uk";
    NSURL *fullUrl = [NSURL URLWithString:httpSource];
    NSURLRequest *httpRequest = [NSURLRequest requestWithURL:fullUrl];
    customWebView.scalesPageToFit = YES;

    customWebView.frame=self.view.bounds;

    [customWebView.scrollView setShowsHorizontalScrollIndicator:NO];
    customWebView.scrollView.scrollEnabled = NO;
    customWebView.scrollView.bounces = NO;

    //actually call the page into the uiwebview
    [customWebView loadRequest:httpRequest];



- (void)didReceiveMemoryWarning

    [super didReceiveMemoryWarning];
    // Dispose of any resources that can be recreated.


@end

【问题讨论】：

究竟有什么不适合你？您的代码似乎是正确的

【参考方案1】：

通过添加修复

self.webview.delegate = self;

查看DidLoad函数

【讨论】：

我相信它是通过 UI 设置的，因为你有 IBOutlet