如何从复杂的 JSON 对象中获取值?
How do I get values from a complex JSON object?
有人可以告诉我如何打印出这些披萨店的名称吗?我的应用程序打印出预期的 "Status Code: 200"
。但是,我的控制台只显示空括号 []
。我怀疑我没有正确地从我的 JSON 对象中提取值。
我正在为我的 API 使用这个 link。
问题
如何从序列化的 JSON 对象中正确获取值?
相关代码:
// Response
if let httpResponse = response as? NSHTTPURLResponse where httpResponse.statusCode == 200, let data = data {
print("Status Code: \(httpResponse.statusCode)")
do {
let json = try NSJSONSerialization.JSONObjectWithData(data, options: .MutableContainers)
if let pizzaPlaces = json["response"] as? [[String: AnyObject]] {
for place in pizzaPlaces {
if let name = place ["name"] as? String {
self.PizzaClass.append(name)
}
}
}
} catch {
print("Error Serializing JSON Data: \(error)")
}
print(self.PizzaClass)
}
}).resume()
您需要将 NSJSONSerialization.JSONObjectWithData
结果转换为 [String:AnyObject]
。
let jsonObject = try NSJSONSerialization.JSONObjectWithData(returnedData, options: .MutableLeaves) as! [String: AnyObject]
一旦你有了这些,你需要做的就是注意你正在投射的内容。以下面的代码为例。如果我们想使用 jsonObject["response"]
获取我们的 response
对象,我们有什么样的数据结构?
"response": {
"venues": [{
//... continues
}]
}
在左边我们有 "response"
这是一个字符串,在右边我们有 {}
这是一个 AnyObject
。所以我们有 [String: AnyObject]
。您只需要逐个考虑要处理的对象。下面是一个工作示例,您可以将其粘贴到您的应用程序中。
完整的工作代码:
func getJson() {
let request = NSMutableURLRequest(URL: NSURL(string: "https://api.foursquare.com/v2/venues/search?client_id=0F5M0EYOOFYLBXUOKTFKL5JBRZQHAQF4HEM1AG5FDX5ABRME&client_secret=FCEG5DWOASDDYII4U3AAO4DQL2O3TCN3NRZBKK01GFMVB21G&v=20130815%20&ll=29.5961,-104.2243&query=burritos")!)
let session = NSURLSession.sharedSession()
request.HTTPMethod = "GET"
request.addValue("application/json", forHTTPHeaderField: "Accept")
let task = session.dataTaskWithRequest(request) { (data: NSData?, response: NSURLResponse?, error: NSError?) -> Void in
guard let testResponse = response as? NSHTTPURLResponse else {
print("\(response)")
return
}
guard let status = HTTPStatusCodes(rawValue: testResponse.statusCode) else {
print("failed to unwrap status")
return
}
print(status)
switch status {
case .Created:
print("ehem")
case .BadRequest:
print("bad request")
case .Ok:
print("ok")
guard let returnedData = data else {
print("no data was returned")
break
}
do {
let jsonObject = try NSJSONSerialization.JSONObjectWithData(returnedData, options: .MutableLeaves) as! [String: AnyObject]
guard let response = jsonObject["response"] as? [String: AnyObject] else { return }
guard let venues = response["venues"] as? [AnyObject] else { return }
guard let location = venues[0]["location"] as? [String:AnyObject] else { return }
guard let formattedAddress = location["formattedAddress"] else { return }
print("response: \n\n \(response)\n------")
print("venues : \n\n \(venues)\n-------")
print("location : \n\n \(location)\n------")
print("formatted address : \n \(formattedAddress)")
} catch let error {
print(error)
}
// update user interface
dispatch_sync(dispatch_get_main_queue()) {
print("update your interface on the main thread")
}
}
}
task.resume()
}
将它放在自己的文件中或 class 声明之外,
enum HTTPStatusCodes : Int {
case Created = 202
case Ok = 200
case BadRequest = 404
}
这并不是您要找的东西,但由于您是 Swift 的新手,请查看 Alamofire。它为您处理 JSON 序列化。当您需要链式调用时,PromiseKit 非常灵活。
Alamofire.request(.GET, url).responseJSON {response in
switch (response.result) {
case .Success(let value):
let pizzas = JSON(value).arrayValue
for place in pizzaPlaces {
if let name = place ["name"] as? String {
self.PizzaClass.append(name)
}
}
case .Failure(let error):
if let data = response.data, let dataString = String(data: data, encoding: NSUTF8StringEncoding) {
print("ERROR data: \(dataString)")
}
print("ERROR: \(error)")
}
}
有人可以告诉我如何打印出这些披萨店的名称吗?我的应用程序打印出预期的 "Status Code: 200"
。但是,我的控制台只显示空括号 []
。我怀疑我没有正确地从我的 JSON 对象中提取值。
我正在为我的 API 使用这个 link。
问题
如何从序列化的 JSON 对象中正确获取值?
相关代码:
// Response
if let httpResponse = response as? NSHTTPURLResponse where httpResponse.statusCode == 200, let data = data {
print("Status Code: \(httpResponse.statusCode)")
do {
let json = try NSJSONSerialization.JSONObjectWithData(data, options: .MutableContainers)
if let pizzaPlaces = json["response"] as? [[String: AnyObject]] {
for place in pizzaPlaces {
if let name = place ["name"] as? String {
self.PizzaClass.append(name)
}
}
}
} catch {
print("Error Serializing JSON Data: \(error)")
}
print(self.PizzaClass)
}
}).resume()
您需要将 NSJSONSerialization.JSONObjectWithData
结果转换为 [String:AnyObject]
。
let jsonObject = try NSJSONSerialization.JSONObjectWithData(returnedData, options: .MutableLeaves) as! [String: AnyObject]
一旦你有了这些,你需要做的就是注意你正在投射的内容。以下面的代码为例。如果我们想使用 jsonObject["response"]
获取我们的 response
对象,我们有什么样的数据结构?
"response": {
"venues": [{
//... continues
}]
}
在左边我们有 "response"
这是一个字符串,在右边我们有 {}
这是一个 AnyObject
。所以我们有 [String: AnyObject]
。您只需要逐个考虑要处理的对象。下面是一个工作示例,您可以将其粘贴到您的应用程序中。
完整的工作代码:
func getJson() {
let request = NSMutableURLRequest(URL: NSURL(string: "https://api.foursquare.com/v2/venues/search?client_id=0F5M0EYOOFYLBXUOKTFKL5JBRZQHAQF4HEM1AG5FDX5ABRME&client_secret=FCEG5DWOASDDYII4U3AAO4DQL2O3TCN3NRZBKK01GFMVB21G&v=20130815%20&ll=29.5961,-104.2243&query=burritos")!)
let session = NSURLSession.sharedSession()
request.HTTPMethod = "GET"
request.addValue("application/json", forHTTPHeaderField: "Accept")
let task = session.dataTaskWithRequest(request) { (data: NSData?, response: NSURLResponse?, error: NSError?) -> Void in
guard let testResponse = response as? NSHTTPURLResponse else {
print("\(response)")
return
}
guard let status = HTTPStatusCodes(rawValue: testResponse.statusCode) else {
print("failed to unwrap status")
return
}
print(status)
switch status {
case .Created:
print("ehem")
case .BadRequest:
print("bad request")
case .Ok:
print("ok")
guard let returnedData = data else {
print("no data was returned")
break
}
do {
let jsonObject = try NSJSONSerialization.JSONObjectWithData(returnedData, options: .MutableLeaves) as! [String: AnyObject]
guard let response = jsonObject["response"] as? [String: AnyObject] else { return }
guard let venues = response["venues"] as? [AnyObject] else { return }
guard let location = venues[0]["location"] as? [String:AnyObject] else { return }
guard let formattedAddress = location["formattedAddress"] else { return }
print("response: \n\n \(response)\n------")
print("venues : \n\n \(venues)\n-------")
print("location : \n\n \(location)\n------")
print("formatted address : \n \(formattedAddress)")
} catch let error {
print(error)
}
// update user interface
dispatch_sync(dispatch_get_main_queue()) {
print("update your interface on the main thread")
}
}
}
task.resume()
}
将它放在自己的文件中或 class 声明之外,
enum HTTPStatusCodes : Int {
case Created = 202
case Ok = 200
case BadRequest = 404
}
这并不是您要找的东西,但由于您是 Swift 的新手,请查看 Alamofire。它为您处理 JSON 序列化。当您需要链式调用时,PromiseKit 非常灵活。
Alamofire.request(.GET, url).responseJSON {response in
switch (response.result) {
case .Success(let value):
let pizzas = JSON(value).arrayValue
for place in pizzaPlaces {
if let name = place ["name"] as? String {
self.PizzaClass.append(name)
}
}
case .Failure(let error):
if let data = response.data, let dataString = String(data: data, encoding: NSUTF8StringEncoding) {
print("ERROR data: \(dataString)")
}
print("ERROR: \(error)")
}
}