如何将JSON字符串转换为字典?

问题描述 投票:140回答:6

我想在我的swift项目中创建一个函数,将String转换为Dictionary json格式,但是我遇到了一个错误:

无法转换表达式的类型(@lvalue NSData,选项:IntegerLitralConvertible ...

这是我的代码:

func convertStringToDictionary (text:String) -> Dictionary<String,String> {

    var data :NSData = text.dataUsingEncoding(NSUTF8StringEncoding)!
    var json :Dictionary = NSJSONSerialization.JSONObjectWithData(data, options:0, error: nil)
    return json
} 

我在Objective-C中创建了这个函数:

- (NSDictionary*)convertStringToDictionary:(NSString*)string {
  NSError* error;
  //giving error as it takes dic, array,etc only. not custom object.
  NSData *data = [string dataUsingEncoding:NSUTF8StringEncoding];
  id json = [NSJSONSerialization JSONObjectWithData:data options:0 error:&error];
  return json;
}
json swift dictionary nsjsonserialization
6个回答
331
投票

警告:如果由于某种原因必须使用JSON字符串,则这是将JSON字符串转换为字典的便捷方法。但是如果你有可用的JSON数据,你应该改为work with the data,而根本不使用字符串。

斯威夫特3

func convertToDictionary(text: String) -> [String: Any]? {
    if let data = text.data(using: .utf8) {
        do {
            return try JSONSerialization.jsonObject(with: data, options: []) as? [String: Any]
        } catch {
            print(error.localizedDescription)
        }
    }
    return nil
}

let str = "{\"name\":\"James\"}"

let dict = convertToDictionary(text: str)

斯威夫特2

func convertStringToDictionary(text: String) -> [String:AnyObject]? {
    if let data = text.dataUsingEncoding(NSUTF8StringEncoding) {
        do {
            return try NSJSONSerialization.JSONObjectWithData(data, options: []) as? [String:AnyObject]
        } catch let error as NSError {
            print(error)
        }
    }
    return nil
}

let str = "{\"name\":\"James\"}"

let result = convertStringToDictionary(str)

Original Swift 1回答:

func convertStringToDictionary(text: String) -> [String:String]? {
    if let data = text.dataUsingEncoding(NSUTF8StringEncoding) {
        var error: NSError?
        let json = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.allZeros, error: &error) as? [String:String]
        if error != nil {
            println(error)
        }
        return json
    }
    return nil
}

let str = "{\"name\":\"James\"}"

let result = convertStringToDictionary(str) // ["name": "James"]

if let name = result?["name"] { // The `?` is here because our `convertStringToDictionary` function returns an Optional
    println(name) // "James"
}

在您的版本中,您没有将正确的参数传递给NSJSONSerialization并忘记转换结果。此外,最好检查可能的错误。最后一点:仅当您的值为String时,此方法才有效。如果它可能是另一种类型,最好像这样声明字典转换:

let json = NSJSONSerialization.JSONObjectWithData(data, options: NSJSONReadingOptions.allZeros, error: &error) as? [String:AnyObject]

当然,您还需要更改函数的返回类型:

func convertStringToDictionary(text: String) -> [String:AnyObject]? { ... }

50
投票

我已经更新了Eric D对Swift 2的回答:

func convertStringToDictionary(text: String) -> [String:AnyObject]? {
    if let data = text.dataUsingEncoding(NSUTF8StringEncoding) {
        do {
            let json = try NSJSONSerialization.JSONObjectWithData(data, options: .MutableContainers) as? [String:AnyObject]
            return json
        } catch {
            print("Something went wrong")
        }
    }
    return nil
}

22
投票

斯威夫特3:

if let data = text.data(using: String.Encoding.utf8) {
    do {
        let json = try JSONSerialization.jsonObject(with: data, options: .mutableContainers) as? [String:Any]
        print(json)
    } catch {
        print("Something went wrong")
    }
}

12
投票

使用Swift 3,JSONSerialization有一种叫做json​Object(with:​options:​)的方法。 json​Object(with:​options:​)有以下声明:

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

从给定的JSON数据返回Foundation对象。

当你使用json​Object(with:​options:​)时,你必须处理错误处理(trytry?try!)和类型转换(来自Any)。因此,您可以使用以下模式之一解决问题。


#1. Using a method that throws and returns a non-optional type

import Foundation

func convertToDictionary(from text: String) throws -> [String: String] {
    guard let data = text.data(using: .utf8) else { return [:] }
    let anyResult: Any = try JSONSerialization.jsonObject(with: data, options: [])
    return anyResult as? [String: String] ?? [:]
}

用法:

let string1 = "{\"City\":\"Paris\"}"
do {
    let dictionary = try convertToDictionary(from: string1)
    print(dictionary) // prints: ["City": "Paris"]
} catch {
    print(error)
}
let string2 = "{\"Quantity\":100}"
do {
    let dictionary = try convertToDictionary(from: string2)
    print(dictionary) // prints [:]
} catch {
    print(error)
}
let string3 = "{\"Object\"}"
do {
    let dictionary = try convertToDictionary(from: string3)
    print(dictionary)
} catch {
    print(error) // prints: Error Domain=NSCocoaErrorDomain Code=3840 "No value for key in object around character 9." UserInfo={NSDebugDescription=No value for key in object around character 9.}
}

#2. Using a method that throws and returns an optional type

import Foundation

func convertToDictionary(from text: String) throws -> [String: String]? {
    guard let data = text.data(using: .utf8) else { return [:] }
    let anyResult: Any = try JSONSerialization.jsonObject(with: data, options: [])
    return anyResult as? [String: String]
}

用法:

let string1 = "{\"City\":\"Paris\"}"
do {
    let dictionary = try convertToDictionary(from: string1)
    print(String(describing: dictionary)) // prints: Optional(["City": "Paris"])
} catch {
    print(error)
}
let string2 = "{\"Quantity\":100}"
do {
    let dictionary = try convertToDictionary(from: string2)
    print(String(describing: dictionary)) // prints nil
} catch {
    print(error)
}
let string3 = "{\"Object\"}"
do {
    let dictionary = try convertToDictionary(from: string3)
    print(String(describing: dictionary))
} catch {
    print(error) // prints: Error Domain=NSCocoaErrorDomain Code=3840 "No value for key in object around character 9." UserInfo={NSDebugDescription=No value for key in object around character 9.}
}

#3. Using a method that does not throw and returns a non-optional type

import Foundation

func convertToDictionary(from text: String) -> [String: String] {
    guard let data = text.data(using: .utf8) else { return [:] }
    let anyResult: Any? = try? JSONSerialization.jsonObject(with: data, options: [])
    return anyResult as? [String: String] ?? [:]
}

用法:

let string1 = "{\"City\":\"Paris\"}"
let dictionary1 = convertToDictionary(from: string1)
print(dictionary1) // prints: ["City": "Paris"]
let string2 = "{\"Quantity\":100}"
let dictionary2 = convertToDictionary(from: string2)
print(dictionary2) // prints: [:]
let string3 = "{\"Object\"}"
let dictionary3 = convertToDictionary(from: string3)
print(dictionary3) // prints: [:]

#4. Using a method that does not throw and returns an optional type

import Foundation

func convertToDictionary(from text: String) -> [String: String]? {
    guard let data = text.data(using: .utf8) else { return nil }
    let anyResult = try? JSONSerialization.jsonObject(with: data, options: [])
    return anyResult as? [String: String]
}

用法:

let string1 = "{\"City\":\"Paris\"}"
let dictionary1 = convertToDictionary(from: string1)
print(String(describing: dictionary1)) // prints: Optional(["City": "Paris"])
let string2 = "{\"Quantity\":100}"
let dictionary2 = convertToDictionary(from: string2)
print(String(describing: dictionary2)) // prints: nil
let string3 = "{\"Object\"}"
let dictionary3 = convertToDictionary(from: string3)
print(String(describing: dictionary3)) // prints: nil

2
投票

我找到了一个代码,它将json字符串转换为NSDictionary或NSArray。只需添加扩展名SWIFT 3.0

如何使用

let jsonData = (convertedJsonString as! String).parseJSONString

延期

 extension String
{
var parseJSONString: AnyObject?
{
    let data = self.data(using: String.Encoding.utf8, allowLossyConversion: false)
    if let jsonData = data
    {
        // Will return an object or nil if JSON decoding fails
        do
        {
            let message = try JSONSerialization.jsonObject(with: jsonData, options:.mutableContainers)
            if let jsonResult = message as? NSMutableArray {
                return jsonResult //Will return the json array output
            } else if let jsonResult = message as? NSMutableDictionary {
                return jsonResult //Will return the json dictionary output
            } else {
                return nil
            }
        }
        catch let error as NSError
        {
            print("An error occurred: \(error)")
            return nil
        }
    }
    else
    {
        // Lossless conversion of the string was not possible
        return nil
    }
}

}


2
投票

斯威夫特4

extension String {
    func convertToDictionary() -> [String: Any]? {
        if let data = self.data(using: .utf8) {
            do {
                return try JSONSerialization.jsonObject(with: data, options: []) as? [String: Any]
            } catch {
                print(error.localizedDescription)
            }
        }
        return nil
    }
}
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