如何从共享类到ViewController获取JSON响应数据?
Posted
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了如何从共享类到ViewController获取JSON响应数据?相关的知识,希望对你有一定的参考价值。
我没有使用Alamofire,所以我想在SharedClass中使用JSON post方法,我想将我的api名称和所有参数发送到该函数。最后,我希望得到回复。我试过了,但它不起作用。如果不正确,请更正我或如果有其他选择,请建议我。
我在SharedClass中的代码
func postRequestFunction(apiName:String , parameters:String ) -> [String:Any] {
var localURL = "hostname/public/index.php/v/***?"
localURL = localURL.replacingOccurrences(of: "***", with: apiName)
var request = URLRequest(url: URL(string: localURL)!)
request.setValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")
request.httpMethod = "POST"
print("shared URL : (request)")
request.httpBody = parameters.data(using: .utf8)
var returnRes:[String:Any] = [:]
let task = URLSession.shared.dataTask(with: request) { data, response, error in guard let data = data, error == nil else { // check for fundamental networking error
print(error!)
// print("error=(String(describing: error))")
print("localizedDescription : (String(describing: error?.localizedDescription))")
return
}
if let httpStatus = response as? HTTPURLResponse, httpStatus.statusCode != 200 { // check for http errors
print("statusCode should be 200, but is (httpStatus.statusCode)")
print("response = (String(describing: response))")
}
do {
returnRes = try JSONSerialization.jsonObject(with: data, options: []) as! [String : Any]
print(returnRes)
} catch let error as NSError {
print(error)
}
}
task.resume()
return returnRes
}
在我的视图控制器类中,我的代码是。我在这里呼唤功能
func getProjectDetails() {
let response = SharedClass.sharedInstance.postRequestFunction(apiName: "API Name", parameters: parameters)
print(response)
let res = response["Response"] as! [String:Any]
let status = res["status"] as! String
if status == "SUCCESS" {
//I will handle response here
} else {
let message = res["message"] as! String
//Call alert function
SharedClass.sharedInstance.alert(view: self, title: "", message: message)
}
}
答案
这是我的解决方案:
class APIManager {
private init () {}
static let shared = APIManager()
func postRequestFunction(apiName: String , parameters: String, onCompletion: @escaping (_ success: Bool, _ error: Error?, _ result: [String: Any]?)->()) {
var localURL = "hostname/public/index.php/v/***?"
localURL = localURL.replacingOccurrences(of: "***", with: apiName)
var request = URLRequest(url: URL(string: localURL)!)
request.setValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")
request.httpMethod = "POST"
print("shared URL : (request)")
request.httpBody = parameters.data(using: .utf8)
var returnRes:[String:Any] = [:]
let task = URLSession.shared.dataTask(with: request) { data, response, error in
if let error = error {
onCompletion(false, error, nil)
} else {
guard let data = data else {
onCompletion(false, error, nil)
return
}
if let httpStatus = response as? HTTPURLResponse, httpStatus.statusCode == 200 {
do {
returnRes = try JSONSerialization.jsonObject(with: data, options: []) as! [String : Any]
onCompletion(true, nil, returnRes)
} catch let error as NSError {
onCompletion(false, error, nil)
}
} else {
onCompletion(false, error, nil)
}
}
}
task.resume()
}
}
func getProjectDetails() {
/* Notes:
** onCompletion Block Parameters:
success - This indicates whether the API called successfully or not.
error - This indicates errors from either API calling failed, JSON parsing, or httpStatus is not 200.
result - This indicates the JSON parsed result.
** APIManager:
I have renamed your SharedClass to APIManager for better readibility.
** sharedInstance:
I have renamed sharedInstance to shared for better readibility.
*/
APIManager.shared.postRequestFunction(apiName: "API Name", parameters: "parameters") { (success, error, result) in
if success {
if let res = result?["Response"] as? [String: Any] {
if let status = res["status"] as? String {
if status == "SUCCESS" {
//You can handle response here.
} else {
let message = res["message"] as! String
//Call alert function.
}
}
}
} else {
print(error?.localizedDescription)
}
}
}
另一答案
您忘记了Service的异步范例,您可以在Closure中返回您的API响应,如下所示
func postRequestFunction(apiName:String , parameters:String, returnRes: @escaping ([String: Any]) -> () ) {
var localURL = "hostname/public/index.php/v/***?"
localURL = localURL.replacingOccurrences(of: "***", with: apiName)
var request = URLRequest(url: URL(string: localURL)!)
request.setValue("application/x-www-form-urlencoded", forHTTPHeaderField: "Content-Type")
request.httpMethod = "POST"
print("shared URL : (request)")
request.httpBody = parameters.data(using: .utf8)
let task = URLSession.shared.dataTask(with: request) { data, response, error in guard let data = data, error == nil else {
// check for fundamental networking error
return
}
if let httpStatus = response as? HTTPURLResponse, httpStatus.statusCode != 200 { // check for http errors
print("statusCode should be 200, but is (httpStatus.statusCode)")
print("response = (String(describing: response))")
}
do {
if let response = try JSONSerialization.jsonObject(with: data, options: []) as? [String : Any] {
returnRes(response)
}
} catch let error as NSError {
print(error)
}
}
task.resume()
}
并使用如下
postRequestFunction(apiName: "yourUrl", parameters: "Param") { (response) in
print(response)
}
以上是关于如何从共享类到ViewController获取JSON响应数据?的主要内容,如果未能解决你的问题,请参考以下文章
目标 C:如何从当前 ViewController 后面的 ViewController 获取导航控制器
有办法在两个 viewController 之间共享 UI 元素吗?
如何从 AppDelegate 上的 TabBarController 获取 ViewController?
如何使用 Swift 从 ViewController 获取 TextView 的文本值?