Swift 4解码简单的根级别JSON值 [英] Swift 4 decode simple root level json value

问题描述

根据JSON标准 RFC 7159 ，这是有效的json：

According to the JSON standard RFC 7159, this is valid json:

22

如何使用swift4的可解码功能将其解码为Int？这不起作用

How do I decode this into an Int using swift4's decodable? This does not work

let twentyTwo = try? JSONDecoder().decode(Int.self, from: "22".data(using: .utf8)!)

推荐答案

它可以很好地与 JSONSerialization 和 .allowFragments
阅读选项。从文档：

allowFragments

指定解析器应允许不是其实例的顶级对象NSArray或NSDictionary。

Specifies that the parser should allow top-level objects that are not an instance of NSArray or NSDictionary.

示例：

let json = "22".data(using: .utf8)!

if let value = (try? JSONSerialization.jsonObject(with: json, options: .allowFragments)) as? Int {
    print(value) // 22
}

但是， JSONDecoder 没有此类选项，并且不接受不是数组或字典的顶级
对象。可以在
源代码中看到 decode（）方法调用
JSONSerialization.jsonObject（）时，没有任何选择：

However, JSONDecoder has no such option and does not accept top-level objects which are not arrays or dictionaries. One can see in the source code that the decode() method calls JSONSerialization.jsonObject() without any option:

open func decode<T : Decodable>(_ type: T.Type, from data: Data) throws -> T {
    let topLevel: Any
    do {
       topLevel = try JSONSerialization.jsonObject(with: data)
    } catch {
        throw DecodingError.dataCorrupted(DecodingError.Context(codingPath: [], debugDescription: "The given data was not valid JSON.", underlyingError: error))
    }

    // ...

    return value
}

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