Question

我在网络服务器上有一个简单的PHP脚本，无法让我的iOS应用程序向其发送数据并将数据保存在php数据库中。检查php数据库时没有新评论。我检查了很多网站，似乎无法找到我做错了什么。我认为这是在PHP代码本身。如果我添加表单并在浏览器上访问该网站，我就能够使用它。

PHP代码（服务器，用户名，密码都正确）

<!DOCTYPE html>
<html>
<head>
    <title>Comment System</title>
</head>
<body>
<h1>Add a comment</h1>
</body>
<?php
        $name = $_POST['name'];
        $Judgename = "Judge Dredd";
        $IP= $_SERVER['REMOTE_ADDR'];
        $comment = $_POST['comment'];
        if($name){

                if($comment){
                    mysql_connect("Server","Username","Password");
                    mysql_select_db("comments");
                    mysql_query("INSERT INTO comments (id, name, comment, Judge, IP) VALUES ('','$name','$comment','$Judgename','$IP')");
                }
                else
                    echo "Empty Comment!";

        }
        else
            echo "Empty Name!";

?>
</html>

以下是我的Xcode项目中的代码。

- (void)phptesting{
    NSString *eventDescription = @"Hello2";
    NSString *EventCity = @"Hello3";
    NSString *myRequestString = [NSString stringWithFormat:@"&name=%@&comment=%@",eventDescription,EventCity];
    NSLog(@"%@", myRequestString);
    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");
    }
}

将iOS应用程序连接到PHP

0 个答案: