我「米試圖將數據發佈,從目標C服務器,我試圖讓在它返回一個JSON
。目標C從Snapchat/Snaphax API獲取JSON
我看Snaphax的API PHP
和Snaphaxpy API,並試圖從PHP改寫爲目的C. 的代碼鏈接: https://github.com/tlack/snaphax https://github.com/jasonanovak/snaphaxpy/blob/master/snaphaxpy.py
我還特別看:http://adamcaudill.com/2012/06/16/snapchat-api-and-security/ 但顯然這是過時的
我的代碼是:
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:@"https://feelinsonice.appspot.com/ph/login"]];
[request setHTTPMethod:@"POST"];
[request addValue:@"******testusername******" forHTTPHeaderField:@"username"];
[request addValue:@"*********" forHTTPHeaderField:@"password"];
[request addValue:@"M02cnQ51Ji97vwT4" forHTTPHeaderField:@"blob_enc_key"];
[request addValue:@"false" forHTTPHeaderField:@"debug"];
[request addValue:@"iEk21fuwZApXlz93750dmW22pw389dPwOk" forHTTPHeaderField:@"secret"];
[request addValue:@"m198sOkJEn37DjqZ32lpRu76xmw288xSQ9" forHTTPHeaderField:@"static_token"];
[request addValue:@"Snaphax 4.0.1 (iPad; iPhone OS 6.0; en_US)" forHTTPHeaderField:@"user_agent"];
[request addValue:@"930cf95a6731dc986ef3bceef6abbaf420e94d8d197dca87b9b47314d8c51b3b" forHTTPHeaderField:@"req_token"];
[request addValue:@"1355776346532" forHTTPHeaderField:@"timestamp"];
NSData *data = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil];
NSString *dataString = [[NSString alloc] initWithData:data encoding:NSASCIIStringEncoding];
NSLog(@"%@", dataString);
我沒有將真正的username/password
。但是,怎麼會這樣,因爲我從字面上複製了我用新語言發現並實現的所有內容......
我是不是正確地發佈數據?我嘗試使用ASIHTTPRequest
,但我無法得到那個工作...
根據經驗的任何建議或想法?
您是否使用NSURLConnection初始化了您的請求? – Engnyl
您尚未指定任何POST數據。使用「application/x-www-form-urlencoded」作爲內容類型的原始代碼是什麼? – CouchDeveloper