if statement - PHP "if" condition error -

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error in if condition. trying generate 5digit random no, , verify random no , textbox values ($_post['otp1']) equal move thank.php page else show popup error.

i did everything, if textbox values , $otp value equal showing popup message.

following 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 worng'); window.location.replace(\"otp.php\");</script>";     }     else     {         header('location: thank.php');      } } ?>

the problem otp code changed in form submit. have store otp code in hidden element , compare user's entered otp value.

<?php $otp = intval( "0" . rand(1,9) . rand(0,9) . rand(0,9) . rand(0,9) . rand(0,9) ); echo $otp; ?> <form action="test1.php" method="post"> <label>mobile :</label> <input type="text" name="mobile" /> <br /><br /> <label>otp :</label> <input type="hidden" name="otp" value="<?=$otp?>" />  <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'];   $otp = $_post['otp'];    if($otp_no != $otp) // condition not work   {       echo "<script>alert('your otp worng'); </script>";   }   else   {       echo "success";    } } ?>

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