mercredi 19 août 2015

PHP "if" condition error

Error in if condition. I am trying to generate 5digit random no, and verify that random no and textbox values ($_POST['otp1']) equal move to thank.php page else show popup error.

I did everything, if textbox values and $otp value is equal it showing popup message.

Following is the code

otp.php

<form action="otp.php" method="post">
<label>Mobile :</label>
<input type="text" name="mobile" /> <br /><br />
<label>OTP :</label>
<input type="text" name="otp1" /> <br /><br />
<input type="submit" name="send" value="Verifiy" />
</form>

<?php

$otp = intval( "0" . rand(1,9) . rand(0,9) . rand(0,9) . rand(0,9) . rand(0,9) );
echo $otp;

if(isset($_POST['send']))
{

    $mobile = $_POST['mobile'];
    $otp_no = $_POST['otp1'];

    if($otp_no != $otp) \\ Condition not work
    {
        echo "<script>alert('Your OTP is Worng'); window.location.replace(\"otp.php\");</script>";
    }
    else
    {
        header('Location: thank.php');

    }
}
?>




Aucun commentaire:

Enregistrer un commentaire