NSURL致命错误

Question

我尝试使链接在Swift中插入URL后出现致命错误。

我的代码：

let newName:String = closest.name.stringByReplacingOccurrencesOfString(" ", withString: "&", options: NSStringCompareOptions.LiteralSearch, range: nil) 
    print(closest.name) 
    print(newName) 
    let url:String = "waze://?q=\(newName)" 

    print(url) 

    let navAdd: NSURL? = NSURL(string:url)// here is the error 
    let wazeApp: NSURL? = NSURL(string: "http://itunes.apple.com/us/app/id323229106")! 
    print(navAdd) 
    if(true){ 
     UIApplication.sharedApplication().openURL(navAdd!) 
    }else{ 
     UIApplication.sharedApplication().openURL(wazeApp!) 
    } 

和错误是：

fatal error: unexpectedly found nil while unwrapping an Optional value

Answer 1

你解开一个可选的带出检查如果它不是零。

if let navAdd = NSURL(string:url) { 
    UIApplication.sharedApplication().openURL(navAdd) 
} else if let wazeApp = NSURL(string:"http://itunes.apple.com/us/app/id323229106"){ 
    UIApplication.sharedApplication().openURL(wazeApp) 
} else {print("Url not found")}