how to handle the nil value variables

问题:

I have model as below.

struc Info: Decodable {
    var firstName: String?
    var lastName: String?
}

While displaying in tableview cell, what I am doing is as below.

personName.text = "\(personArray[indexPath.row].firstName!) \(personArray[indexPath.row].lastName!)"

Now the app is crashing if I have data in below format

[
    {
        "firstName" : "F 1",
        "lastName" : "L 1"
    },
    {
        "firstName" : "F 2"
    },
    {
        "lastName" : "L 3"
    }
]

The app is crashing saying lastName is nil

---

Solution 1

The solution for this check for nil & then show name, however I don't want to do the check at run time because that I have to check this for all variables (considering I have model of 25 variables). Below is what I could have done.

var firstName = ""
if (personArray[indexPath.row].firstName == nil) {
    firstName = ""
} else {
    firstName = personArray[indexPath.row].firstName!
}

var lastName = ""
if (personArray[indexPath.row].lastName == nil) {
    lastName = ""
} else {
    lastName = personArray[indexPath.row].lastName!
}

personName.text = "\(firstName) \(lastName)"

Solution 2

I can do the update in the model itself as below.

struc Info: Decodable {
    var firstName: String?
    var lastName: String?

    var firstName2 : String? {
    get {
        if (self.firstName==nil) {
            return ""
        }
        return firstName
    }

    var lastName2 : String? {
    get {
        if (self.lastName==nil) {
            return ""
        }
        return lastName
    }
}

personName.text = "\(personArray[indexPath.row].firstName2!) \(personArray[indexPath.row].lastName2!)"

However I have problem with this also. This way, again I have to create N number of variables again.

Is there any other alternate way where default value will get assigned if that variable is missing in the webservice?

回答1:

I would recommend one of two options:

1. Add computed property to the struct to determine the display name.
2. Manually decode, providing default values. (And also add a display name property if you want)

Personally, I like option 1. I think it's the most compact and also the easiest to maintain.

Option 1 Example:

struct Info1: Decodable {
    var firstName: String?
    var lastName: String?

    var displayName: String {
        return [self.firstName, self.lastName]
            .compactMap { $0 } // Ignore 'nil'
            .joined(separator: " ") // Combine with a space
    }
}

print(Info1(firstName: "John", lastName: "Smith").displayName)
// Output: "John Smith"

print(Info1(firstName: "John", lastName: nil).displayName)
// Output: "John"

print(Info1(firstName: nil, lastName: "Smith").displayName)
// Output: "Smith"

print(Info1(firstName: nil, lastName: nil).displayName)
// Output: ""

Option 2 Example:

struct Info2: Decodable {
    var firstName: String
    var lastName: String

    enum CodingKeys: String, CodingKey {
        case firstName, lastName
    }

    init(from decoder: Decoder) throws {
        let container = try decoder.container(keyedBy: CodingKeys.self)

        self.firstName = try container.decodeIfPresent(String.self, forKey: .firstName) ?? ""
        self.lastName = try container.decodeIfPresent(String.self, forKey: .lastName) ?? ""
    }

    // Optional:
    var displayName: String {
        return [self.firstName, self.lastName]
            .compactMap { $0.isEmpty ? nil : $0 } // Ignore empty strings
            .joined(separator: " ") // Combine with a space
    }

    // TEST:
    init(from dict: [String: Any]) {
        let data = try! JSONSerialization.data(withJSONObject: dict, options: .prettyPrinted)
        self = try! JSONDecoder().decode(Info2.self, from: data)
    }
}

print(Info2(from: ["firstName": "John", "lastName": "Smith"]).displayName)
// Output: "John Smith"

print(Info2(from: ["lastName": "Smith"]).displayName)
// Output: "Smith"

print(Info2(from: ["firstName": "John"]).displayName)
// Output: "John"

print(Info2(from: [String: Any]()).displayName)
// Output: ""