尝试使用来自注册表的PHP / HTML将数据输入到SQL表中

我尝试了很多不同的事情,并且进行了很多搜索,但都没有解决方案.我正在尝试使用html表单将数据提交到sql表.这是我的register.php文件的代码.$con = mysqli_connect(localhost, database_name, password datab...

我尝试了很多不同的事情,并且进行了很多搜索,但都没有解决方案.我正在尝试使用html表单将数据提交到sql表.

这是我的register.php文件的代码.

$con = mysqli_connect("localhost", "database_name", "password" "database_user");

if($con === false) {
    die("ERROR Could not Connect." . mysqli_connect_error());
}
$lasty= mysqli_real_escape_string($_POST['laz']);
$namez=mysqli_real_escape_string($_POST['namer']);
$emailAddr=mysqli_real_escape_string($_POST['emaila']);
$userName=mysqli_real_escape_string($_POST['usrn']);
$passwo=mysqli_real_escape_string($_POST['passw']);

$sqql = "INSERT INTO 'database_name' . table' (UserID, FirstName, LastName,   Email, UserName, Password) 
        VALUES (NULL, '$namez', '$lasty', '$emailAddr', '$userName', '$passwo')";


if (mysqli_query($con, $sqql)) {
    echo "Successfull";
} else {
    echo "Did not work!" . $con->error;
}

mysqli_close($con);

我的HTML文件是：

<form action="register.php" method="POST">
    First Name: <input type="text" name="namer"  placeholder="First Name"/> <br>
    Last Name: <input type='text' name='laz' /> <br>
    Email Address: <input type='text' name='emaila' /> <br>
    UserName: <input type='text' name='usrn' />
    Password: <input type='password' name='passw' />
    <input type='submit' id='button' value='Submit' name='login' />

</form>

我为那些奇怪的命名变量提前道歉,我担心其他文件会打扰我在这里试图做的事情.

解决方法:

$con = mysqli_connect("localhost", "database_name", "password" "database_user"); //open connection

if (mysqli_connect_errno()) { //if connection failed
    die("Connect failed: ", mysqli_connect_error());
    exit();
}

$lasty      = mysqli_real_escape_string($con, $_POST['laz']); //added $con needs two parameter (connection, input)
$namez      = mysqli_real_escape_string($con, $_POST['namer']);
$emailAddr  = mysqli_real_escape_string($con, $_POST['emaila']);
$UserName   = mysqli_real_escape_string($con, $_POST['usrn']);
$password   = mysqli_real_escape_string($con, $_POST['passw']);



$sqql = "INSERT INTO `table_name`(UserID, FirstName, LastName,   Email, UserName, Password) 
VALUES (NULL, '$namez', '$lasty', '$emailAddr', '$userName', '$passwo')";


if (mysqli_query($con, $sqql)) {
    echo "Row inserted";
}else{
    die("Error: ". mysqli_sqlstate($con));
}

mysqli_close($con);