如何将JSON字符串转换为字典？-Java 学习之路

118

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;
}

6 回答

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JSON data 可用，你应该改为work with the data，而根本不使用字符串 .

Swift 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)

Swift 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 answer:

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，您没有将适当的参数传递给 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]? { ... }

回复于 2024-04-29T02:35:08+08:00

303

我是've updated Eric D'的回答 Swift 2 ：

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
}

回复于 2024-04-29T02:35:08+08:00

Swift 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")
    }
}

回复于 2024-04-29T02:35:08+08:00

使用Swift 3， JSONSerialization 有一个名为jsonObject(with:options:)的方法 . jsonObject(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 .

当您使用 jsonObject(with:options:) 时，您必须处理错误处理（ try ， try? 或 try! ）并键入强制转换（来自 Any ） . 因此，您可以使用以下模式之一解决问题 .

＃1 . 使用抛出并返回非可选类型的方法

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 . 使用抛出并返回可选类型的方法

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 . 使用不抛出并返回非可选类型的方法

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 . 使用不抛出并返回可选类型的方法

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

回复于 2024-04-29T02:35:08+08:00

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
    }
}

回复于 2024-04-29T02:35:08+08:00

我找到了一个代码，将json字符串转换为NSDictionary或NSArray.just添加扩展名 SWIFT 3.0

HOW TO USE

let jsonData = (convertedJsonString as! String).parseJSONString

EXTENSTION

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
    }
}

}

回复于 2024-04-29T02:35:08+08:00

如何将JSON字符串转换为字典？

6 回答

＃1 . 使用抛出并返回非可选类型的方法

＃2 . 使用抛出并返回可选类型的方法

＃3 . 使用不抛出并返回非可选类型的方法

＃4 . 使用不抛出并返回可选类型的方法

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