Question

我是OOP的新手。我想在mySql查询中使用类函数的值。但是我在检索值时遇到了困难。也许我的方法不正确。

以下是我的代码和结果。如您所见，回显的值为空。

（如果附加的代码不够，请告诉我。）

代码：

if(isset($_POST['hidden']))

{
    $validator = new FormValidator();

    $fname =$validator->addValidation("fname","req","Please fill in First Name");

    $email= $validator->addValidation("email","email","The input for Email should be a valid email value");

    $lname= $validator->addValidation("lname","req","Please fill in Last Name");

    $pass= $validator->addValidation("pass","req","Please fill in Password");

    $con_pass= $validator->addValidation("confirmpass","req","Please fill in Confirm Password");

    $sname= $validator->addValidation("sname","req","Please fill in Screen Name");



    if($validator->ValidateForm())

    {

        echo $insert_query ="insert into registeration set fname = '".$fname."', lname = '".$lname."', email = '".$email."', pass = '".$pass."', cpass = '".$con_pass."', sname = '".$sname."' ";

         mysql_query($insert_query);

结果：

insert into registeration set fname = '', lname = '', email = '', pass = '', cpass = '', sname = ''

Answer 1

如果您要更新现有记录，

您正在使用的查询更合适

     if($validator->ValidateForm())
{

           $fname = $_POST['fname'];  // name of your input field 
           $lname = $_POST['lname'];
           $email = $_POST['email'];
           $pass = $_POST['pass'];
           $con_pass = $_POST['con_pass'];
           $sname = $_POST['sname'];


    $insert_query = mysql_query("INSERT INTO registeration (fname,lname,email,pass,cpass,sname) 
    VALUES ('$fname','$lname','$email','$pass','$con_pass','$sname')") or die(mysql_error());
}

（已编辑）和不工作的原因是因为您的变量等于您调用的方法实例，您需要使它们等于输入字段。

希望这可以帮到你

通过课堂形式验证

1 个答案: