我创建了自定义类，文件是 showBlock.h 和 showBlock.m，用于以编程方式加载 UIWebView，showBlock.m 的实现是

#import "showBlock.h"

@implementation showBlock;

@synthesize mainViewContObj;

- (void) showView {
    UIWebView *aWebView = [[UIWebView alloc] initWithFrame:CGRectMake(0, 0, 320, 50)];
    aWebView.autoresizesSubviews = YES;
    aWebView.autoresizingMask = (UIViewAutoresizingFlexibleHeight | UIViewAutoresizingFlexibleWidth);
    [aWebView setDelegate:[self mainViewContObj]];
    NSString *urlAddress = @"http://localhost/test/index.php";
    NSURL *url = [NSURL URLWithString:urlAddress];
    NSURLRequest *requestObj = [NSURLRequest requestWithURL:url];

    [aWebView loadRequest:requestObj];

    UIView *view = [[UIView alloc] initWithFrame:CGRectMake(0, 0, 0, 0)];
    [[[self mainViewContObj] view] addSubview:aWebView];

}
@end

它工作正常，正在加载包含 html 内容的 index.php 文件，但我想在 safari 浏览器中打开此 html 文件的链接，我需要为此做些什么？

最佳答案

需要在ShowBlock.m中添加下面的delegate方法实现

- (BOOL)webView:(UIWebView *)webView shouldStartLoadWithRequest:(NSURLRequest *)request
 navigationType:(UIWebViewNavigationType)navigationType {
    // This practically disables web navigation from the webView.
    if (navigationType == UIWebViewNavigationTypeLinkClicked) {
        [[UIApplication sharedApplication] openURL:[request URL]];
        return FALSE;
    }
    return TRUE;
}

关于IOS UIWebView : opening links in safari，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/13291836/