Question

我觉得这不应该是这么复杂，但在这里我花了最后两天试图让这个工作。我试图将iOS应用程序中的两个变量发送到服务器上的php脚本。数据库添加一个条目，但2个信息字段为空。我很满意Objective C但不熟悉PHP。

PHP代码

<?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')");
?>

目标C代码

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");
    }

Answer 1

更新：我在更长时间后修复了问题。我拿出了

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

并将其替换为

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

将数据从iOS应用程序发送到PHP脚本

1 个答案: