CODEWITHSUNDEEP
phphtmlmysql
Diarmuid Naughton
Diarmuid Naughton

Reputation: 27

php script not working in Internet explorer and Firefox

I have a button on a webpage that allows users to add a video on that page to their list of favourites. behind this button is a form and some php. The PHP code uses a session variable to retrieve the username. This information is used to get the relevant user id from the database and store its value in a variable. Using the input value from the form it was possible to retrieve the tuple from the videos database table that related to the video in question and store the values of the video title and URL attributes in variables. The code then checks if the user has already added the video as a “favourite”. The favourites database entity is checked for tuples containing both the user id and video id. If both are contained in a single row of the database table the user has already added the video and is notified of this. Otherwise, the user id, video id, video title and URL are inserted into the favourites database entity and the user is informed that the video has been added this all works fine in chrome or safari but does nothing in ie or firefox. The database is updated and message is displayed only in Chrome and safari. I've attached the code, please note the session has already been started in earlier code on the webpage. Any assistance would be greatly appreciated.

    <div id="addfav">
    <form action="python.php" method="post">
        <input name="add" src="images/add.png" type="image"
        value="3">
    </form>
    <?php 
        $user=$_SESSION['user'];

        if ( isset( $_POST['add'] ) )
        {
            $vid = $_POST["add"];
            $sql = "SELECT * FROM `users` WHERE username = '$user'";
            $result = mysql_query($sql) or die(mysql_error());
            $row = mysql_fetch_array($result);
            $uid= $row['user_id'];

            $sql = "SELECT * FROM `Video` WHERE Video_id = '$vid'";
            $result = mysql_query($sql) or die(mysql_error());
            $row = mysql_fetch_array($result);
            $url=$row['URL'];
            $title=$row['Title'];

            $check = mysql_query("SELECT * FROM `favourites` WHERE Uid = '$uid' AND vid_id = '$vid'") or die (mysql_error());
            $r = mysql_num_rows($check);

            if ($r>=1)
            {
                echo "already added to favourites";

                echo '<script type="text/javascript">window.alert("Already added to favourites")</script>';
                //'<span style="color: red;" />Already added to favourites </span>' ;   
            }

            else 
            {  
                mysql_query("INSERT INTO `favourites` (`Uid`,
                `vid_id`,`url`,`title`) VALUES ('$uid',
                '$vid','$url','$title')")or die(mysql_error());
                echo "Added to favourites"; 
            }
        }
    ?>
</div>

Upvotes: 1

Views: 2009

Answers (3)

Viktor S.
Viktor S.

Reputation: 12815

When user clicks on <input type="image" /> browser will pass coordinates of a click. Chrome will send three values:

add.x = x_coord
add.y = y_coord
add = input_value (3 in your case)

Note that in php you can access add.x/add.y value with $_POST["add_x"]/$_POST["add_y"] (see dot replaced with underscore)

At the same time, IE will not pass third value. That is why your if ( isset( $_POST['add'] ) ) will never return true. Option is to put video id value into some hidden field and use its name in that if. You can easily check that behavior by doing var_dump($_POST); in php

PS:

You should never use values received in request without them being sanitized in sql queries. Right now code below is opened to sql injections:

$sql = "SELECT * FROM `Video` WHERE Video_id = '$vid'";

You should, at least, user mysql_real_escape_string function before value is inserted into a query:

$sql = "SELECT * FROM `Video` WHERE Video_id = '".mysql_real_escape_string($vid)."'";

And take a look at warning message on top of php manual page linked above: mysql_* functions are deprecated and you should better use PDO or mysqli extension.

Upvotes: 0

Javid
Javid

Reputation: 2935

PHP is run server-side. This means that regardless of what browser you're using, it works as expected. The problem is definitely from HTML codes you've written if IE and Firefox can connect to your website without any problem. I think the problem is inside your form tag because I think it's not standard you can either use a GET method inform that your form is submitted, or use a hidden input indicating it.

P.S. I think your code has security issues. (SQL Injection)

Upvotes: 0

JoDev
JoDev

Reputation: 6873

(Just a debug idea) Try to change your input image to a hidden element like this : 

<form action="python.php" method="post">
     <!-- I don't remove this, to keep the image shown-->
     <input name="addimg" src="images/add.png" type="image"  value="3">

     <input type='hidden' name='add' value='3' />
 </form>

Does it works now?

Upvotes: 4

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