[英]IOS UIWebView : opening links in safari
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,并加载带有html内容的index.php文件,但我想在safari浏览器中打开这个html文件的链接,我需要做些什么呢?
您需要在ShowBlock.m中添加下面的委托方法实现
- (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;
}
实现UIWebViewDelegate
协议并设置aWebView.delegate = self
。
然后实施
- (BOOL)webView:(UIWebView *)webView shouldStartLoadWithRequest:(NSURLRequest *)request navigationType:(UIWebViewNavigationType)navigationType;
单击链接时将调用此方法。 从请求中获取URL。
使用下面的代码在safari中打开一个链接:
[[UIApplication sharedApplication] openURL:[NSURL URLWithString: @\"http://www.google.com"]];
在UIWebView
委托中,定义webView:shouldStartLoadWithRequest
方法:
- (BOOL)webView:(UIWebView*)webView shouldStartLoadWithRequest:(NSURLRequest*)request navigationType:(UIWebViewNavigationType)navigationType {
if ([[request URL] checkCondition])
[[UIApplication sharedApplication] openURL:[request URL]];
return NO;
}
return YES;
}
checkCondition
是一种检查是否应该通过safari打开URL的方法(您可以根据域或其他方式进行检查)。 在最简单的情况下,始终调用openURL
(删除if
)
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