Hi. I have been working on a survey all works great. Just wondering if I’m handing this the right way or not. After a code is entered and submitted I run it through a couple of checks. First is the code proper length (10 digits), is the code in the database and has this code been used. If the code passes all of these. I then start session and the survey can be completed. sample code below.
if($rowCheck >= "1") {
$row = mysql_fetch_array($query);
$status = $row['status'];
$washcode = $row['washCode'];
}
if($status == "1"){
die('<div align="center"><h1>Error the code you have entered has already been used.</h1><form action="'. $_SERVER['PHP_SELF'].'" method="POST">
<tr>
<th colspan="2">Please check your code and try again.<br />
<input type="text" name="code" autocomplete="off"><br />
<input type="submit" name="submitcode" value="Submit Code"></th>
</tr>
</form></div>');
}
else {
session_start();
$_SESSION['code'] = $code;
$_SESSION['redeemCode'] = $washcode;
echo '<div align="center"><b>Thank you for taking the time to complete our survey. Please click the link below to begin.<br /><a href="visitdate.php"><button type="button">Next</button></a></b></div>';
}
My main concern is that if the code does not pass this check I am currently using die(), and the rest of the page is not loaded leaving off the html footer and side nav. I am wanting to switch to echo’‘; to handle the error. Would this be alright? And can I get a little info on when it is best to use die(); instead of echo’';
I am working on converting all of my sql to mysqli when I started thinking about this.