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detect when F5 or submit

hi,

Please assist how can we stop or prevent the page from adding the same record that was previously added when the user hits F5.
PHP mysql -a sample code will help.

<?php

if(@$_POST['submit']=="add")
{
$myQuery="insert into mytable (name) values( " . @$_POST['name'] . " )";
mysql_query($myQuery) or die(mysql_error());
}

<form name ="create" method="post">
<Input name="name" type="Text" <?php $_POST['name'] ?>>
<input name "submit type="submit" value="add">
</form>

Thanks.
0
FrankPorter
Asked:
FrankPorter
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3 Solutions
 
RoonaanCommented:
You can redirect to a special thank you page:

if(@$_POST['submit']=="add")
{
  $myQuery="insert into mytable (name) values( " . mysql_real_escape_string(@$_POST['name']) . " )";
  mysql_query($myQuery) or die(mysql_error());
  header('Location:thankyou.php');
  exit();
}

-r-
0
 
FrankPorterAuthor Commented:
after redirecting to the thank you page, can we redirect again the the add page?
is there a standard way to do this rather than redirecting here and there?

Thanks.
0
 
RoonaanCommented:
You can redirect to the "add" page as well.
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FrankPorterAuthor Commented:
Is this the standard way of dealing F5 refresh scenario ? or there's a better way of handling..

Thanks.
0
 
RoonaanCommented:
It is a solution to the F5 "the page you are trying to load was submitted using POST Data" warning.

Alternatives to duplicate prevention also account for testing if data has been recently added to database, or tracking variables in sessions.

-r-
0
 
Chris GralikeSpecialistCommented:
an other method might be assigning a session value to detect an F5 stroke

<?php

$show_form = false;

if(isset($_POST['submit']) ){
    session_start();
   
    if(isset($_SESSION['submitted']) ){
                $show_form = true;
                echo "Submitted!";
    }else{
                $myQuery="insert into mytable (name) values( " . mysql_real_escape_string(@$_POST['name']) . " )";
                if(mysql_query($myQuery) ){
                       $_SESSION['submitted'] = 'true';
                }else{
                      if(isset($_SESSION['count']) ){
                               $_SESSION['count'] = "1";
                      }else{
                               $_SESSION['count'] ++;
                      }
                       
                      $_SESSION['submitted'] = 'false';
                     
                      if($_SESSION['count'] < 3){
                           header('location'.$_SERVER['PHP_SELF']);
                      }else{
                           die('3 Retries Failed, please contact the administrator');
                      }
                }
    }
}else{
      if($show_form){
            print(' <table><form etc ');
      }else{
            echo "Enable Session VARS for this script to work!";
      }
}

?>


Well there are enough ways about it i guess, so good luck! and Regards,
0
 
Chris GralikeSpecialistCommented:
$show_form = false;         << This one should be "true"; :$

Sorry for that ;)
0
 
Bruce PieterseTechnical LeadCommented:
An simplier way of checking if the data exists, is by checking in the database like Roonan said earlier. You can try this:

<?php

if(@$_POST['submit']=="add") {
      $checkDB = "SELECT name FROM mytable WHERE name = '". $_POST['name'] ."'";
      $resCheckDB = mysql_db_query($checkDB);
      $recordValid = mysql_num_rows($resCheckDB);
      if($recordValid >= 1) {
            echo "Your Name is already in the Database! ";
      }
else{
      $myQuery="insert into mytable (name) values( " . @$_POST['name'] . " )";
      mysql_query($myQuery) or die(mysql_error());
      }
}

<form name ="create" method="post">
<Input name="name" type="Text" <?php $_POST['name'] ?>>
<input name "submit type="submit" value="add">
</form>
Hope that helps,
0

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