A syntax error occurs

Code Help PHP
1

Question:
Hello, I encountered an error like this in PHP, how should I deal with it?

Parse error: syntax error, unexpected token "if" in /home/runner/php/ad/1/2/3/4/server.php on line 10

The code will be this.

<!DOCTYPE html>
<html lang="ja">
<head>
  <meta charset="UTF-8">
  <title>Server console</title>
</head>
<body>
  <?php
  $logincookie = $_COOKIE
  if ($logincookie == namae) {

  } elseif ($logincookie == null) {
    
  } else {
      echo "Sorry, some error prevented access. Please log in again.";
  }
  echo"<h1>Welcome to the Server console.</h1>";
  echo "<a href="."logout.html".">Logout</a>";
   echo "<a href="."https://e887b4f9-57f7-4ba1-b25f-5a1633f91c06-00-5n6j13l7d88u.sisko.replit.dev/ad/1/2/3/4/5/6/os-% 20release.php"."> <br> os-
    release"
  ?>
</body>
</html>

What is wrong?
Repl link:

code snippet

screenshot

screenshot (47)
screenshot (47)1365Ă—644 10.6 KB

2

Seems like it doesn’t know how to handle the if statement on line 10, double check your syntax and formatting perhaps?

3

There are two mistakes in your code

You are missing an end semicolon on line 9
your variable $logincookie is expecting a string value of “namae” and needs to be enclosed with quotation marks

Below is the corrected code

<!DOCTYPE html>
<html lang="ja">
<head>
  <meta charset="UTF-8">
  <title>Server console</title>
</head>
<body>
  <?php
  $logincookie = $_COOKIE;
  if ($logincookie == "namae") {

  } elseif ($logincookie == null) {

  } else {
      echo "Sorry, some error prevented access. Please log in again.";
  }
  echo"<h1>Welcome to the Server console.</h1>";
  echo "<a href="."logout.html".">Logout</a>";
   echo "<a href="."https://e887b4f9-57f7-4ba1-b25f-5a1633f91c06-00-5n6j13l7d88u.sisko.replit.dev/ad/1/2/3/4/5/6/os-% 20release.php"."> <br> os-
    release"
  ?>
</body>
</html>
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