警告:如果由于某种原因您必须使用 JSON 字符串,那么这是将 JSON 字符串转换为字典的便捷方法。但如果你有 JSONdata可用,你应该改为处理数据,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)
Swift 1 原始答案:
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
并忘记投射结果。另外,最好检查一下可能存在的错误。最后注意:只有当您的值是字符串时,这才有效。如果它可以是另一种类型,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]? { ... }