如果单元素解码失败，Swift JSONDecode解码数组也会失败

一种选择是使用包装器类型，尝试对给定值进行解码。nil如果不成功，则存储：

struct FailableDecodable

let base: Base?

init(from decoder: Decoder) throws {
    let container = try decoder.singleValueContainer()
    self.base = try? container.decode(Base.self)
}

} 然后，我们可以解码这些数组，并GroceryProduct在Base占位符中填写以下内容：

import Foundation

let json = """ [ { "name": "Banana", "points": 200, "description": "A banana grown in Ecuador." }, { "name": "Orange" } ] """.data(using: .utf8)!

struct GroceryProduct : Codable { var name: String var points: Int var description: String? }

let products = try JSONDecoder() .decode([FailableDecodable ].self, from: json) .compactMap { $0.base } // .flatMap in Swift 4.0

print(products)

// [ // GroceryProduct( // name: "Banana", points: 200, // description: Optional("A banana grown in Ecuador.") // ) // ] 然后，我们.compactMap { $0.base }用于过滤掉nil元素（那些在解码时引发错误的元素）。

这将创建一个的中间数组[FailableDecodable ]，这不应该成为问题。但是，如果您希望避免这种情况，则可以始终创建另一个包装器类型，该包装器类型将对来自无键容器的每个元素进行解码和解包：

struct FailableCodableArray : Codable {

var elements: [Element]

init(from decoder: Decoder) throws {

    var container = try decoder.unkeyedContainer()

    var elements = [Element]()
    if let count = container.count {
        elements.reserveCapacity(count)
    }

    while !container.isAtEnd {
        if let element = try container
            .decode(FailableDecodable<Element>.self).base {

            elements.append(element)
        }
    }

    self.elements = elements
}

func encode(to encoder: Encoder) throws {
    var container = encoder.singleValueContainer()
    try container.encode(elements)
}

} 然后，您将解码为：

let products = try JSONDecoder() .decode(FailableCodableArray .self, from: json) .elements

print(products)

// [ // GroceryProduct( // name: "Banana", points: 200, // description: Optional("A banana grown in Ecuador.") // ) // ]