Question

我试图在WebView中打开以下网址

http://1-dot-smartrefill-968.appspot.com/#/#mfucci@gmail_com

但无法制作NSURL的对象。

Answer 1

你的代码在哪里？你使用了webView吗？

我使用了这种方法：

[_webView loadRequest:[NSURLRequest requestWithURL:url]];

Answer 2

<强>步骤1

最初在.plist中添加NSAppTransportSecurity，请参阅this

<强>步骤-2

调用网址

NSString *urlAddress =  @"http://1-dot-smartrefill-968.appspot.com/#/#mfucci@gmail_com"; 
[self.yourwebviewName loadRequest:[NSURLRequest requestWithURL:[NSURL URLWithString:[NSURL URLWithString:[urlAddress stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]]]]];

Answer 3

尝试使用URL代替addingPercentEncoding对stringByAddingPercentEscapesUsingEncoding进行编码，因为它已被弃用。

在Swift中：

适用于Swift 3 ：

let strUrl = "http://1-dot-smartrefill-968.appspot.com/#/#mfucci@gmail_com"
let encodedUrl = strUrl.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)

对于Swift 2.3或更低版本：

 let encodedUrl = strUrl.stringByAddingPercentEncodingWithAllowedCharacters(.URLHostAllowedCharacterSet())

在Objective-C

NSString *strUrl = @"http://1-dot-smartrefill-968.appspot.com/#/#mfucci@gmail_com";
NSString *encodedUrl = [strUrl stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLHostAllowedCharacterSet]];

Answer 4

swift ：

$('#word').load()

objective-c

var url : NSString = "http://1-dot-smartrefill-968.appspot.com/#/#mfucci@gmail_com" 
var urlStr : NSString = url.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)! 
var searchURL : NSURL = NSURL(string: urlStr)! 
    webviewInstance.loadRequest(NSURLRequest(URL: searchURL))

如何在iOS中的WebView中打开URL？

4 个答案: