Block of code php appear after a form [duplicate] - php

Question

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PHP code is not being executed, but the code shows in the browser source code
(35 answers)
Closed 6 years ago.
I have a problem for a connection page i'm trying to set up, after the connection form, the conditionnal block of code that should verify the info just appear instead of executing. Everything appear after , even the semi-colon and parenthesis that should end the if. What did I do wrong?
<?php
// if ($id!=0) {erreur(ERR_IS_CO)};
if (!isset($_POST['mail']))
{
echo "<form method="post" action="connexion.php">
<fieldset>
<legend>Connexion</legend>
<p>
<label for="mail">Mail :</label><input name="mail" type="text" id="mail" /><br />
<label for="password">Mot de Passe :</label><input type="password" name="password" id="password" />
</p>
</fieldset>
<p><input type="submit" value="Connexion" /></p></form>" ;
}
else
{
$message;
if (empty($_POST['mail']) || empty($_POST['password']) ) //empty space
{
$message = "<p>une erreur s\'est produite pendant votre identification. Vous devez remplir tous les champs</p>
<p>Cliquez ici pour revenir</p>";
}
else //password checker
{
if ( md5($_POST['password']) == 'student') // Student
{
$_SESSION['mail'] = $_POST['mail'];
$_SESSION['id'] = "student";
$message = "<p>Bienvenue student
vous êtes maintenant connecté!</p>" //.$data['Nom']., need to fetch name
<p>Cliquez ici pour revenir à la page d accueil</p>;
}
else // Access denied
{
$message = "<p>Une erreur s\est produite pendant votre identification.<br /> Le mot de passe ou le pseudo
entré n\est pas correct.</p><p>Cliquez ici
pour revenir à la page précédente <br />";
}
$query->CloseCursor();
}
echo $message;
}
?>

Answer 1

Syntax error is reason of showing blank page, use Netbeans or Sublime software helps you and you need concatination or use single quotes inside double quotes
Example
echo "<form method='post' action='connexion.php'>
and also enable your errors to see reason of blank page