问题

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

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

I tried the following approach:

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

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

-[__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1:

It looks like you are passing an NSString parameter where you should be passing an NSData parameter:

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

回答2:

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

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

回答3:

I've made a category from @Abizern answer

@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,

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

回答4:

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

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

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

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.

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

#1. Swift 4 solution

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

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.

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

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)
    }
}

#2. Swift 3 and Swift 4 solution

With Swift 3 and Swift 4, JSONSerialization has a method called jsonObject(with:options:). jsonObject(with:options:) has the following declaration:

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

Returns a Foundation object from given JSON data.

The Playground code below shows how to use data(using:allowLossyConversion:) and jsonObject(with:options:) in order to get a Dictionary from a JSON formatted String:

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5:

Using Abizern code for swift 2.2

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

来源：https://stackoverflow.com/questions/8606444/how-do-i-deserialize-a-json-string-into-an-nsdictionary-for-ios-5

标签

ios