CODEWITHSUNDEEP
php
user3542048
user3542048

Reputation: 43

Avoiding adding data multiple times by Refreshing the page

I am developing a fun script where user will be guessing images by putting points for each image which the game owner will be displaying. If he guesses right, his points will be doubled. If he fails to guess, the points will be added to game owner.

In this, After submitting the form, the images which user has guessed and the images game owner displays, will be shown in the page. After that the calculations will happen.

I have properly done all the steps, but if the user reloads the page from browser refresh button, the calculations happens again and again . Not getting how to avoid this.

I can not use

header("location: reditect_page.php");
exit();

As the user should see the data.

Can somebody suggest me to resolve this?

Here is the my submission page

if(isset($_POST['guess']))
{
//user entered/guessed images
$img1 = $_POST['img1'];
$img2 = $_POST['img2'];
$img3 = $_POST['img3'];

$img1_pnt = $_POST['point1'];
$img2_pnt = $_POST['point2'];
$img3_pnt = $_POST['point3'];

$sn1 = mysqli_query($con, "SELECT * from images where id=".$img1."");
$sn2 = mysqli_fetch_array($sn1);

$mn1 = mysqli_query($con, "SELECT * from images where id=".$img2."");
$mn2 = mysqli_fetch_array($mn1);

$is1 = mysqli_query($con, "SELECT * from images where id=".$img2."");
$is2 = mysqli_fetch_array($is1);

//game Owner's image
$k1 = "SELECT * FROM days_data WHERE id=".$gid." AND lead=".$_SESSION['lead']." ORDER BY id DESC LIMIT 1";
$k2 = mysqli_query($con, $k1);
$k3 = mysqli_fetch_array($k2);
$var1 = $k3['img1'];
$var2 = $k3['img2'];
$var3 = $k3['img3'];

$vararray = array($var1, $var2, $var3);

$s1 = mysqli_query($con, "select * from images where id = ".$var1."");
$s2 = mysqli_query($con, "select * from images where id = ".$var2."");
$s3 = mysqli_query($con, "select * from images where id = ".$var3."");

$q1 = mysqli_fetch_array($s1);
$q2 = mysqli_fetch_array($s2);
$q3 = mysqli_fetch_array($s3); 

//Display User guessed images
echo '<div class="row gutter30" style="margin-top:20px;">
 <h2>Guessed Data</h2>
        <div class="col-xs-4">
        <img src="img/'.$mn2['image'].'" class="image" /> 
</div>
        <div class="col-xs-4">
        <img src="img/'.$sn2['image'].'" class="image" /> 
</div>

        <div class="col-xs-4">
        <img src="img/'.$is2['image'].'" class="image" /> 
</div>
</div>

//Game Owner's Images
<div class="row gutter30" style="margin-top:20px;">
 <h2>RESULT</h2>
        <div class="col-xs-4">
        <img src="img/'.$q1['image'].'" class="image" /> 
</div>
        <div class="col-xs-4">
        <img src="img/'.$q2['image'].'" class="image" /> 
</div>

        <div class="col-xs-4">
        <img src="img/'.$q3['image'].'" class="image" /> 
</div>

//Here i will do the calculations of points
if(($_POST['img1_pnt'])!='')

{
 if(in_array($_POST['img1'], $vararray))
{
 // Calculations part
}
}
}

Upvotes: 1

Views: 664

Answers (3)

A. Kiyoshi
A. Kiyoshi

Reputation: 61

I hope this is what you are looking for:

If I understand you correctly, if the page is refresh, the data is display twice, if it refreshes 3 times, a similar data is display 3 times.

This works for me.

  1. Put your form in index.php
  2. put your action code and your display in another file.g. action.php
  3. In your action.php file, make sure you end the file with this code:
  4. Make sure you include your action file in your index.php file like include 'action.php';
>  echo '<META HTTP-EQUIV=REFRESH CONTENT="1;
> '.$_SERVER["PHP_SELF"].'">';

or any code similar. Since you are currently in your index.php file, the code above ask to refresh your action.php file which does not contain your form. Hope this would work.

Upvotes: 0

Vivek
Vivek

Reputation: 2755

You can use session to keep track of the user attempt count. Then use this value to prevent the user from attempting multiple times.

In your code where you start the session, add the following code to start tracking attempts.

session_start();
if (!isset($_SESSION['attempt_count'])) {
  $_SESSION['attempt_count'] = 0;
}

Then you need to increment the attempt count for each attempt made by the user and prevent multiple submissions. For this, replace the following code.

if(isset($_POST['guess']))
{
  //user entered/guessed images
  $img1 = $_POST['img1'];
  $img2 = $_POST['img2'];
  $img3 = $_POST['img3'];

With the code below.

if(isset($_POST['guess']) && $_SESSION['attempt_count']==0)
{
  $_SESSION['attempt_count']++;

  //user entered/guessed images
  $img1 = $_POST['img1'];
  $img2 = $_POST['img2'];
  $img3 = $_POST['img3'];

Upvotes: 1

TarangP
TarangP

Reputation: 2738

Put this code in starting of your php code outside isset submit button

if(empty($_SESSION['key']) && !isset($_SESSION['key'])){
  $randomkey = rand(0,99999);
  $mykey = $_SESSION['key'] = $randomkey
}

It set Session a unique random key. now in form field take one hidden field

<input type="hidden" name="key" value="<?=$mykey?>">

Now On Submit

  1. First Check Current $mykey Matches With Hidden Variable
  2. After Performing Certain Action Unset $mykey

So it prevent user adding data multiple times by Refreshing the page.

I hope it Helps... :-)

Upvotes: 1

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