我想在sencha touch hybrid ios app中打开像'https://www.google.co.in/#q=adam+scott'这样的谷歌搜索链接。我尝试使用var ref = window.open(url,'_ blank','location = yes');但它没有加载页面,如果我将_blank更改为_system,它正在加载页面,但没有显示完成按钮以移动到上一页。
如果某人已经做过,请告诉我。
答案 0 :(得分:0)
我认为这就是你要找的东西:
navigator.app.loadUrl('https://www.google.co.in/#q=adam+scott',{openExternal:true});
答案 1 :(得分:0)
打开你的MainViewController.m类并在@end
之前添加这行代码- (BOOL) webView:(UIWebView*)theWebView shouldStartLoadWithRequest:(NSURLRequest*)request navigationType:(UIWebViewNavigationType)navigationType
{
NSURL *url = [request URL];
// Intercept the external http requests and forward to Safari.app
// Otherwise forward to the PhoneGap WebView
if ([[url scheme] isEqualToString:@"http"] || [[url scheme] isEqualToString:@"https"]){
[[UIApplication sharedApplication] openURL:url];
return NO;
}
else {
return [super webView:theWebView shouldStartLoadWithRequest:request navigationType:navigationType];
}
}