如何处理nil值变量 [英] how to handle the nil value variables

问题描述

我的模型如下.

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

在表格单元格中显示时，我正在做的事情如下.

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

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

如果我具有以下格式的数据，则应用程序现在崩溃

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

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

应用程序崩溃，说姓氏为空

The app is crashing saying lastName is nil

此检查为nil&的解决方案然后显示名称，但是我不想在运行时进行检查，因为 我必须检查所有变量(考虑到我有25个变量的模型) .下面是我可以做的.

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)"

解决方案2

我可以在模型本身中进行如下更新.

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!)"

但是我对此也有疑问.这样， 我必须再次创建N个变量.

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

如果Web服务中缺少该变量，还有其他替代方法可以分配默认值吗?

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

推荐答案

我建议使用以下两种方法之一:

I would recommend one of two options:

1. 将计算的属性添加到结构中以确定显示名称.
2. 手动解码，提供默认值. (如果需要，还可以添加显示名称属性)

我个人喜欢选项1.我认为它是最紧凑且最容易维护的.

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

选项1示例:

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: ""

选项2示例:

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: ""

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