我正在尝试从PHP解析信息,但我需要发送一个字典参数,所以我尝试的东西…我看到教程,例子,但我被困,所以我回到开始：(这是什么好的方法为了这样做？)
func asd(){
let urlPath = "http://xxxxx.PHP"
let url: NSURL = NSURL(string: urlPath)
let request = NSMutableuRLRequest(URL: url)
request.HTTPMethod = "GET"
var parm = ["id_xxxx": "900"] as Dictionary
//I THINK MY PROBLEM IT'S HERE! i dont kNow how to link parm with session,i try is with session.uploadTaskWithRequest(<#request: NSURLRequest?#>,fromData: <#NSData?#>) but doesn't work
let session = NSURLSession.sharedSession()
let task = session.dataTaskWithURL(url,completionHandler: {data,response,error -> Void in
println("Task completed")
if(error) {
// If there is an error in the web request,print it to the console
println(error.localizedDescription)
}
var err: NSError?
var jsonResult = NSJSONSerialization.JSONObjectWithData(data,options: NSJSONReadingOptions.MutableContainers,error: &err) as NSDictionary
if(err?) {
// If there is an error parsing JSON,print it to the console
println("JSON Error \(err!.localizedDescription)")
}
println(jsonResult.debugDescription)
let results: NSArray = jsonResult["x"] as NSArray
dispatch_async(dispatch_get_main_queue(),{
self.tableData = results
self.OfertaGridViewLista!.reloadData()
})
})
task.resume()
}
谢谢！
解决方法
GET数据需要是url的查询字符串的一部分.某些方法将接受POST / PUT请求的参数字典,但如果您使用GET方法,这些方法将不会为您添加字典到您的URL.
如果您希望将“GET”参数保留在“字典”中以保持清晰度或一致性,请考虑将以下方法添加到您的项目中：
func buildQueryString(fromDictionary parameters: [String:String]) -> String {
var urlVars:[String] = []
for (k,value) in parameters {
if let encodedValue = value.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLQueryAllowedCharacterSet()) {
urlVars.append(k + "=" + encodedValue)
}
}
return urlVars.isEmpty ? "" : "?" + urlVars.joinWithSeparator("&")
}
此方法将使用键/值对的字典,并返回可附加到您的网址的字符串.
例如,如果您的API请求允许多个请求方法(GET / POST / etc.),则只需要将此查询字符串附加到您的基本api网址以获取GET请求：
if (request.HTTPMethod == "GET") {
urlPath += buildQueryString(fromDictionary:parm)
}
如果您只是提交GET请求,则无需检查您将要使用哪种方法来发送数据.