Question

我一直在尝试创建一个注册表单，我已经尝试了几种方法，但最后我添加的是else if（$ status =＆＃34; OK＆＃34;）行。以下是我的代码的一部分，请帮助我仍然是编码新手;

<?php
$surnameErr = $firstnameErr = $usernameErr = $emailErr = $passwordErr = $password1Err = "";
$surname = $firstname = $username = $email = $password = $password1 = "";
$status = "OK";

if ($_SERVER["REQUEST_METHOD"] == "POST") {
    if (empty($_POST["surname"])) {
     $surnameErr = "Surname is required";
     $status = "NOTOK";
   } 
   else {
     $surname = test_input($_POST["surname"]);
        if (!preg_match("/^[a-zA-Z ]*$/",$surname)) {
        $surnameErr = "Only letters and white space allowed"; 
        $status = "NOTOK";
        }
   }

   if (empty($_POST["firstname"])) {
       $firstnameErr = "First Name is required";
       $status = "NOTOK";
   } 
   else {
     $firstname = test_input($_POST["firstname"]);
        if (!preg_match("/^[a-zA-Z ]*$/",$firstname)) {
            $firstnameErr = "Only letters and white space allowed";
            $status = "NOTOK";
        }
   }

   if (empty($_POST["username"])) {
     $usernameErr = "Username is required";
     $status = "NOTOK";
   } else {
     $username = test_input($_POST["username"]);
   }

   if (empty($_POST["email"])) {
     $emailErr = "Email is required";
     $status = "NOTOK";
   } else {
     $email = test_input($_POST["email"]);
     if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
        $emailErr = "Invalid Email format"; 
        $status = "NOTOK";
        }
   }

   if (empty($_POST["password"])) {
     $passwordErr = "Password is required";
     $status = "NOTOK";
   } 

   if (empty($_POST["password1"])) {
     $password1Err = "Repeat password to confirm";
     $status = "NOTOK";
   }

   if ($password != $password1){
       echo "<script>alert('Password does not match')</script>";
       $status = "NOTOK";
   }

    $connection = mysqli_connect("localhost", "root", "", "church") or die(mysqli_error());

    $checkuser = mysqli_query ($connection,"SELECT * FROM signup WHERE username = '$username'");
    if (mysqli_num_rows ($checkuser) > 0){
        echo "<script>alert('$username already exist in our database, please enter another one')</script>";
        $status = "NOTOK";
    }

    $checkemail = mysqli_query ($connection,"SELECT * FROM signup WHERE email = '$email'");
    if (mysqli_num_rows ($checkemail) > 0){
        echo "<script>alert('$email already exist in our database, please enter another one')</script>";
        $status = "NOTOK";
    }

    else if ($status = "OK"){
        $password=md5($password);
        $datainsert = mysqli_query ($connection, "INSERT INTO signup (surname, firstname, username, email, password) VALUES ('$surname', '$firstname','$username','$email','$password')") or die(mysqli_error());
            echo "<script>alert('Sign Up Successful!')</script>";
        }
   }   

function test_input($data) {
  $data = trim($data);
  $data = stripslashes($data);
  $data = htmlspecialchars($data);
  return $data;
}
?>

Answer 1

    else if ($status = "OK"){

此行表示“将$ status设置为”OK“的值并返回该值”。除非您将$ status设置为PHP认为为空的值（0，false，NULL，空字符串等），否则这将始终为true。

需要else if ($status == "OK"){才能正常工作。

另外，如果在测试的左侧放置常量或函数/方法调用，则可以避免此类错误。由于在正常执行期间无法为函数/方法调用或常量赋值，因此代码将失败并出现致命错误，因此调试非常容易。

    else if ("OK" = $status){ // Will fatal with a line number.  Fixable in 2 seconds flat

php - 验证后仍然提交数据返回false

1 个答案: