如何将 JSON 字符串反序列化为 NSDictionary？（适用于 iOS 5+）

How do I deserialize a JSON string into an NSDictionary? (For iOS 5+)

提问by Andreas

In my iOS 5 app, I have an NSStringthat contains a JSON string. I would like to deserialize that JSON string representation into a native NSDictionaryobject.

在我的 iOS 5 应用程序中，我有一个NSString包含 JSON 字符串的。我想将该 JSON 字符串表示反序列化为本机NSDictionary对象。

 "{\"password\" : \"1234\",  \"user\" : \"andreas\"}"

I tried the following approach:

我尝试了以下方法：

NSDictionary *json = [NSJSONSerialization JSONObjectWithData:@"{\"2\":\"3\"}"
                                options:NSJSONReadingMutableContainers
                                  error:&e];  

But it throws the a runtime error. What am I doing wrong?

但它会引发运行时错误。我究竟做错了什么？

-[__NSCFConstantString bytes]: unrecognized selector sent to instance 0x1372c 
*** Terminating app due to uncaught exception 'NSInvalidArgumentException',
reason: '-[__NSCFConstantString bytes]: unrecognized selector sent to instance 0x1372c'

回答by Abizern

It looks like you are passing an NSStringparameter where you should be passing an NSDataparameter:

看起来您正在传递一个NSString应该传递NSData参数的参数：

NSError *jsonError;
NSData *objectData = [@"{\"2\":\"3\"}" dataUsingEncoding:NSUTF8StringEncoding];
NSDictionary *json = [NSJSONSerialization JSONObjectWithData:objectData
                                      options:NSJSONReadingMutableContainers 
                                        error:&jsonError];

回答by Desert Rose

NSData *data = [strChangetoJSON dataUsingEncoding:NSUTF8StringEncoding];
NSDictionary *jsonResponse = [NSJSONSerialization JSONObjectWithData:data
                                                             options:kNilOptions
                                                               error:&error];

For example you have a NSStringwith special characters in NSStringstrChangetoJSON. Then you can convert that string to JSON response using above code.

例如，您NSString在NSStringstrChangetoJSON 中有一个带有特殊字符的。然后您可以使用上述代码将该字符串转换为 JSON 响应。

回答by Hemang

I've made a category from @Abizern answer

我从@Abitern 回答中创建了一个类别

@implementation NSString (Extensions)
- (NSDictionary *) json_StringToDictionary {
    NSError *error;
    NSData *objectData = [self dataUsingEncoding:NSUTF8StringEncoding];
    NSDictionary *json = [NSJSONSerialization JSONObjectWithData:objectData options:NSJSONReadingMutableContainers error:&error];
    return (!json ? nil : json);
}
@end

Use it like this,

像这样使用它，

NSString *jsonString = @"{\"2\":\"3\"}";
NSLog(@"%@",[jsonString json_StringToDictionary]);

回答by Imanou Petit

With Swift 3 and Swift 4, Stringhas a method called data(using:allowLossyConversion:). data(using:allowLossyConversion:)has the following declaration:

在 Swift 3 和 Swift 4 中，String有一个名为data(using:allowLossyConversion:). data(using:allowLossyConversion:)有以下声明：

func data(using encoding: String.Encoding, allowLossyConversion: Bool = default) -> Data?

Returns a Data containing a representation of the String encoded using a given encoding.

返回包含使用给定编码编码的字符串表示的数据。

With Swift 4, String's data(using:allowLossyConversion:)can be used in conjunction with JSONDecoder's decode(_:from:)in order to deserialize a JSON string into a dictionary.

在 Swift 4 中，String'sdata(using:allowLossyConversion:)可以与JSONDecoder's结合使用decode(_:from:)，以便将 JSON 字符串反序列化为字典。

Furthermore, with Swift 3 and Swift 4, String's data(using:allowLossyConversion:)can also be used in conjunction with JSONSerialization's json?Object(with:?options:?)in order to deserialize a JSON string into a dictionary.

此外，在 Swift 3 和 Swift 4 中，String'sdata(using:allowLossyConversion:)也可以与JSONSerialization's结合使用json?Object(with:?options:?)，以便将 JSON 字符串反序列化为字典。

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#1. Swift 4 solution

#1. 斯威夫特 4 解决方案

With Swift 4, JSONDecoderhas a method called decode(_:from:). decode(_:from:)has the following declaration:

在 Swift 4 中，JSONDecoder有一个名为decode(_:from:). decode(_:from:)有以下声明：

func decode<T>(_ type: T.Type, from data: Data) throws -> T where T : Decodable

Decodes a top-level value of the given type from the given JSON representation.

从给定的 JSON 表示中解码给定类型的顶级值。

The Playground code below shows how to use data(using:allowLossyConversion:)and decode(_:from:)in order to get a Dictionaryfrom a JSON formatted String:

下面的 Playground 代码显示了如何使用data(using:allowLossyConversion:)并从 JSON 格式中decode(_:from:)获取 a ：DictionaryString

let jsonString = """
{"password" : "1234",  "user" : "andreas"}
"""

if let data = jsonString.data(using: String.Encoding.utf8) {
    do {
        let decoder = JSONDecoder()
        let jsonDictionary = try decoder.decode(Dictionary<String, String>.self, from: data)
        print(jsonDictionary) // prints: ["user": "andreas", "password": "1234"]
    } catch {
        // Handle error
        print(error)
    }
}

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#2. Swift 3 and Swift 4 solution

#2. Swift 3 和 Swift 4 解决方案

With Swift 3 and Swift 4, JSONSerializationhas a method called json?Object(with:?options:?). json?Object(with:?options:?)has the following declaration:

在 Swift 3 和 Swift 4 中，JSONSerialization有一个名为json?Object(with:?options:?). json?Object(with:?options:?)有以下声明：

class func jsonObject(with data: Data, options opt: JSONSerialization.ReadingOptions = []) throws -> Any

Returns a Foundation object from given JSON data.

从给定的 JSON 数据返回一个 Foundation 对象。

The Playground code below shows how to use data(using:allowLossyConversion:)and json?Object(with:?options:?)in order to get a Dictionaryfrom a JSON formatted String:

下面的 Playground 代码显示了如何使用data(using:allowLossyConversion:)并从 JSON 格式中json?Object(with:?options:?)获取 a ：DictionaryString

import Foundation

let jsonString = "{\"password\" : \"1234\",  \"user\" : \"andreas\"}"

if let data = jsonString.data(using: String.Encoding.utf8) {
    do {
        let jsonDictionary = try JSONSerialization.jsonObject(with: data, options: []) as? [String : String]
        print(String(describing: jsonDictionary)) // prints: Optional(["user": "andreas", "password": "1234"])
    } catch {
        // Handle error
        print(error)
    }
}

回答by IOS Singh

Using Abizerncode for swift 2.2

使用Abilern代码进行 swift 2.2

let objectData = responseString!.dataUsingEncoding(NSUTF8StringEncoding)
let json = try NSJSONSerialization.JSONObjectWithData(objectData!, options: NSJSONReadingOptions.MutableContainers)