ios - 从模型的属性中获取可选值

Question

在我的代码中:

let userStatic:UserStaticSwift = UserStaticSwift.sharedInstance() // UserStaticSwift save the info

if userStatic.nor_businessName != "" {
     
    self.nor_name.text = userStatic.nor_businessName
    self.nor_vauth.isHidden = false
}else {
    self.nor_name.text = "un bind terant"
    self.nor_vauth.isHidden = true
}

结果出乎意料:

self.nor_name.text将设置为:"Optional(the real value)" ，但我想设置"the real value" .

但是userStatic的属性(property)是 optional ，我只想要值 the real value ，不是下面的:

(lldb) po userStatic.nor_businessName
"Optional(the real value)"

如何取出该值，而不是 Optional(value) ？

<小时/>

编辑-3

显示更多信息的附加信息:

在 @objc class UserStaticSwift:NSObject, NSCoding :

self.nor_status = UtilSwift.getIntFromAny(dic["status"] as Any)
self.nor_phone = UtilSwift.getStrFromAny(dic["phone"])
self.nor_name = UtilSwift.getStrFromAny(dic["name"])
self.nor_definition = UtilSwift.getStrFromAny(dic["definition"])
self.nor_log = UtilSwift.getStrFromAny(dic["logo"])
self.nor_businessName = UtilSwift.getStrFromAny(dic["businessName"]) //  attention, is there if I use the `dic["businessName"] as Any`, then will get the Optional value, so I pass the `dic["businessName"]`

在 UtilSwift :

class func getStrFromAny(_ any:Any?) -> String {
    
    if let nonNil = any, !(nonNil is NSNull) {
        
        return String(describing: nonNil)
    }
    return ""
}

<小时/>

Answer 1

试试这个(在可选元素处添加 !):

po userStatic.nor_businessName!

更新:

如果上述方法不适合您，那么您应该尝试以下方法:

guard let value = userStatic.nor_businessName  else {
print("fails");
return;
}

self.nor_name.text = value;