我正在尝试根据 this 访问 home_timeline 请求示例。
但是,我不断收到以下错误:
'URLRequestWithMethod' with an argument list of type '(String, URL: String, parameters: NSArray, error: inout NSError?)'
func getHomeTimeLine(){
var clientError:NSError?
let params = []
let request = Twitter.sharedInstance().APIClient.URLRequestWithMethod(
"GET",
URL: "https://api.twitter.com/1.1/statuses/home_timeline.json",
parameters: params,
error: &clientError)
if request != nil {
Twitter.sharedInstance().APIClient.sendTwitterRequest(request) {
(response, data, connectionError) -> Void in
if (connectionError == nil) {
var jsonError : NSError?
let json : AnyObject? =
NSJSONSerialization.JSONObjectWithData(data,
options: nil,
error: &jsonError)
}
else {
println("Error: \(connectionError)")
}
}
}
else {
println("Error: \(clientError)")
}
}
提前致谢。
最佳答案
将参数定义为字典并使用它。
let params: Dictionary = Dictionary()
func getHomeTimeLine() {
var clientError:NSError?
let params: Dictionary = Dictionary<String, String>()
let request: NSURLRequest! = Twitter.sharedInstance().APIClient.URLRequestWithMethod(
"GET",
URL: "https://api.twitter.com/1.1/statuses/home_timeline.json",
parameters: params,
error: &clientError)
if request != nil {
Twitter.sharedInstance().APIClient.sendTwitterRequest(request!) {
(response, data, connectionError) -> Void in
if (connectionError == nil) {
var jsonError : NSError?
let json : AnyObject? =
NSJSONSerialization.JSONObjectWithData(data!,
options: nil,
error: &jsonError)
// check for json data
if (json != nil) {
println("response = \(json)")
} else {
println("error loading json data = \(jsonError)")
}
}
else {
println("Error: \(connectionError)")
}
}
}
else {
println("Error: \(clientError)")
}
}
关于ios - 使用 Swift 的 Twitter API 无效 URLRequestWithMethod,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30832991/