AFNetworking 2.0 + PHP簡單HTTPRequest

Question

我試過在網上搜索，但找不到具體的答案我的問題。 我正在使用AFNetworking發送一個簡單的POST HTTP請求。在下面的下面的結果代碼：

$$da39a3ee5e6b4b0d3255bfef95601890afd8070##!! 
CREATE TABLE $$da39a3ee5e6b4b0d3255bfef95601890afd80709##!! 
(question VARCHAR(255), answer VARCHAR(255), choices VARCHAR(255), sentBy VARCHAR(255)); 
INSERT INTO games (gameId, player1, player2, turn) VALUES ('$$da39a3ee5e6b4b0d3255bfef95601890afd80709##!!', '', '',''); 

我使用SHA1（），因此$$da39a3ee5e6b4b0d3255bfef95601890afd80709
這裏是我的Objective-C代碼：

AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager]; 
NSDictionary *params = @{@"gameId" : datas[0], @"p1": datas[1], @"p2":datas[2], @"turn":datas[3] }; 
manager.requestSerializer = [AFHTTPRequestSerializer serializer]; 
manager.responseSerializer = [AFHTTPResponseSerializer serializer]; 
[manager POST:@"http://localhost/AnswerThisPHP/newgame.php" 
     parameters:params 
     success:^(AFHTTPRequestOperation *operation, id responseObject) { 
       NSLog(@"JSON: %@", responseObject); 
     } 
     failure:^(AFHTTPRequestOperation *operation, NSError *error) { 
      NSLog(@"Error: %@", error); 
     }]; 

我的PHP代碼：

$db = "db"; 
    $host = 'host'; 
    $username = "username"; 
    $password = 'password'; 
    $link = mysql_connect($host,$username,$password); 
    if (!$link) { 
     die('Could not connect: ' . mysql_error()); 
     echo ("Error - " . mysql_error()); 
    } 
//Seleccionamos la BBDD 
    @mysql_select_db($db); 
    //Creamos un array para almacenar los resultados 
    $p1 = $_GET['p1']; 
    $p2 = $_GET['p2']; 
    $gameID = $_GET['gameId']; 
    $gameIDSHA = "".$gameID."$$".sha1($gameID)."##".$p1."!!".$p2.""; 
    $turn = $_GET['turn']; 
    $querystring = "CREATE TABLE ".$gameIDSHA." (question VARCHAR(255), answer VARCHAR(255), choices VARCHAR(255), sentBy VARCHAR(255));"; 
    $querystring2 = "INSERT INTO games (gameId, player1, player2, turn) VALUES ('".$gameIDSHA."', '".$p1."', '".$p2."','".$turn."');"; 
    echo $gameIDSHA; 
    echo '<br/>'; 
    echo $querystring; 
    echo '<br />'; 
    echo $querystring2; 

    $insert = mysql_query($querystring); 
    $insert2 = mysql_query($querystring2); 

    echo $insert; 
    echo $insert2; 

Answer 1

您要在客戶端上發送POST，但您正在嘗試訪問它通過GET，使用$_POST而不是$_GET