Swift3 call可以抛出，但它没有标记为"；试一试；并且错误不会被处理

重写func viewDidLoad（）{ super.viewDidLoad（）

数据（contentsOf:url！as url）可能引发异常，您需要尝试调用。

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通过使用let optData=try？…如果引发异常，您将拥有有效的数据对象或为零

谢谢，您节省了我的时间。guard let jsonresponse=try？JSON（数据：数据）else{failure（“error”）return}

    //get the values from sql/Json
    let url = URL(string: "https://example.com/dropdownmenu/phpGet.php")
    let data = Data(contentsOf: url! as URL)
    var tmpValues = try! JSONSerialization.jsonObject(with: data as Data, options: JSONSerialization.ReadingOptions.mutableContainers) as! NSArray
    tmpValues = tmpValues.reversed() as NSArray
    reloadInputViews()

    for candidate in tmpValues {
    if let cdict = candidate as? NSDictionary {

            //fullName is the column name in sql/json

            let names = cdict["fullName"]
            self.values.append(names! as AnyObject)

        }
    }

}

let optData = try? Data(contentsOf: url! as URL)
guard let data = optData else {
    return
}