從iOS應用程序發送數據到PHP腳本

我覺得這不應該是這麼複雜，但在這之後我花了兩天的時間試圖讓它工作。我試圖從我的iOS應用程序發送兩個變量到我在服務器上的一個php腳本。該數據庫添加一個條目，但2個信息字段爲空。我沒有客觀的C但是PHP新手。從iOS應用程序發送數據到PHP腳本

PHP代碼

<?php 
     $name = $_POST['name']; 
     $comment = $_POST['comment']; 
        mysql_connect("Server","Username","Password"); 
        mysql_select_db("comments"); 
        mysql_query("INSERT INTO comments (id, name, comment) VALUES ('','$name','$comment')"); 
?>

目標C代碼

NSString *name = @"Name1"; 
    NSString *comment = @"Comment1"; 
    NSString *myRequestString = [NSString stringWithFormat:@"&name=%@&comment=%@",name,comment]; 
    NSData *myRequestData =[myRequestString dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES]; 
    NSString *postLength = [NSString stringWithFormat:@"%d",[myRequestData length]]; 
    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL: [NSURL URLWithString: @"http://www.notguiltyapp.com/harris.php"]]; 

    [request setHTTPMethod: @"POST"]; 

    [request setValue:postLength forHTTPHeaderField:@"Content-Length"]; 
    [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"content-type"]; 

    // Set Request Body 
    [request setHTTPBody: myRequestData]; 
    NSURLConnection *conn = [[NSURLConnection alloc]initWithRequest:request delegate:self]; 

    if(conn) 
    { 
     NSLog(@"Connection Successful"); 
    } 
    else 
    { 
     NSLog(@"Connection could not be made"); 
    }

來源

2014-07-15 user3157910

取代它沒有工作。你爲什麼要爲每個變量添加引號？ – user3157910

更新：我有更多的時間後，解決了這一問題。我拿出

NSURLConnection *conn = [[NSURLConnection alloc]initWithRequest:request delegate:self];

與

NSData *returnData=[NSURLConnection sendSynchronousRequest: request returningResponse:&response error:&err];

來源

2014-08-03 23:55:24 user3157910

從iOS應用程序發送數據到PHP腳本

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