Solved

Clearing table values once user navigates away from page

Posted on 2011-03-04

PHP
AJAX

1 solution

468 Views

Last Modified: 2012-05-11

Hi Experts,

I have created a webpage that allows a user to enter table data and the table automatically refreshes using ajax.

However its only going to be used for proof of concept. So how can i clear any table entries a user has made once that user either leaves the website or refreshes the browser?

Many Thanks,
Dean

URL: http://www.the-lees.co.uk/site1/ajax_insert.php

Comment

Question by:deanlee17

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10 Comments

LVL 4

Expert Comment

by:Tomun

ID: 350374762011-03-04

Just don't save it in your database.

LVL 4

Expert Comment

by:LAMASE

ID: 350375312011-03-04

Since HTTP is stateless, you cannot "know" is a person is still on your page, or has his pc crashed or shutted duwn.
Normally sessions way of thinking is: "if the user comes back within x minutes, I keep his session active".
So what can you do is delete every entries not updated within the lapse you use for ajax updating. Keep in mind that the user can have temporary internet connection problems, and in this case you will think that the user gone away and delete his data...

An html trick can be using
<body onBeforeUnload="function()">

that is triggered when the user close the page. You can perform an ajax call (syncronous is better) that erases all the data, plus a timed script that deletes all data older than x minutes.

LVL 34

Expert Comment

by:Beverley Portlock

ID: 350376942011-03-04

Store the data in serailized format in the session rather than the database and then when the user closes the session the data is wiped automatically.

http://www.php.net/serialize
http://www.php.net/unserialize

Author Comment

by:deanlee17

ID: 350381922011-03-04

bportlock can you still print the data from a session so as its displayed as if its coming from a database? Click link in my original post to see what im talking about.

Thanks,
Dean

Author Comment

by:deanlee17

ID: 350382782011-03-04

Actually, thinking about it, is it possible to have a button that uses AJAX to empty the table without a page refresh?

LVL 14

Expert Comment

by:Scott Madeira

ID: 350391892011-03-04

If you want the ability to have a reset button on your form to clear the data then why don't you just use an HTML reset button:

<INPUT TYPE=RESET Value="Clear Form">

If that is all you want then you don't need AJAX to do it.

LVL 4

Expert Comment

by:Tomun

ID: 350393672011-03-04

smadeira: That would clear the form but not remove the things from the database that he's saved. Refreshing would bring it all back.

deanlee17:
Without seeing you php code I'm having to guess what your doing. I guess your ajax_data.php file saves the input to the database, then pulls it out again and echoes a <tr>...</tr> line with the data in it.

You could construct the table row directly from the form input without having to save it in the database at all. ajax_data.php seems to be broken at the moment. Show us the code if you want some help with it.

Author Comment

by:deanlee17

ID: 350394442011-03-04

Hi Tonun,

I have attached the code that constructs the page, ive now set the page back to the way it was....

http://www.the-lees.co.uk/site1/ajax_insert.php

Thanks,
Dean

<?
//connect to database
$dbhost = 'mysql01.myhostcp.com';
$dbuser = 'deans_user1';
$dbpass = 'qwerty123';
$dbname = 'deansghot5722com5233_deansdatabase';

$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbname);

//NOTE: MAKE SURE YOU DO YOUR OWN APPROPRIATE SERVERSIDE ERROR CHECKING HERE!!!
if(!empty($_POST) && isset($_POST))
{
    $emptyvariables = false;

    if(!empty($_POST)){
        foreach($_POST as $key => $val){
            if(empty($_POST[$key])){
                echo"Please insert $key<br />";
                $emptyvariables = true;
                }
        }
    }
    
    if($emptyvariables == false){
    


        //make variables safe to insert
        $id = mysql_real_escape_string($_POST['id']);
        $Forename = mysql_real_escape_string($_POST['Forename']);
        $Surname = mysql_real_escape_string($_POST['Surname']);
        $Arrival_Date = mysql_real_escape_string($_POST['Arrival_Date']);
        $Departure_Date = mysql_real_escape_string($_POST['Departure_Date']);
        $RoomNo = mysql_real_escape_string($_POST['RoomNo']);
        $Tel = mysql_real_escape_string($_POST['Tel']);

        //query to insert data into table
        $sql = "
            INSERT INTO bookings
            SET
            id = '$id',
            Forename = '$Forename',
            Surname = '$Surname',
            Arrival_Date = '$Arrival_Date',
            Departure_Date = '$Departure_Date',
            RoomNo = '$RoomNo',
            Tel = '$Tel'";
            
            echo"$query";
        $result = mysql_query($sql);
        if(!$result)
        {
            echo "Failed to insert record<br>Please Ensure booking does not currently exist";
        }
        else
        {
            echo "Record inserted successfully";
        }
    }
}



?>


<table width="300" border="1">
	<tr>
		<td><b>ID</b></td>
		<td><b>Forename</b></td>
		<td><b>Surname</b></td>
		<td><b>Arrival_Date</b></td>
		<td><b>Departure_Date</b></td>
		<td><b>RoomNo</b></td>
		<td><b>Tel</b></td>

	</tr>
	<?
	//show data from tables
	$sql = "
		SELECT *
		FROM bookings
		ORDER BY ID ASC
		LIMIT 20";
	$result = mysql_query($sql);
	while($row = mysql_fetch_array($result))
	{
	//print data
	?>
	<tr>
		<td><? echo $row['id']; ?></td>
		<td><? echo $row['Forename']; ?></td>
		<td><? echo $row['Surname']; ?></td>
		<td><? echo $row['Arrival_Date']; ?></td>
		<td><? echo $row['Departure_Date']; ?></td>
		<td><? echo $row['RoomNo']; ?></td>
		<td><? echo $row['Tel']; ?></td>
	</tr>
	<?
	}
	?>
</table>

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LVL 4

Accepted Solution

by:Tomun

Tomun earned 500 total points

ID: 350398532011-03-04

Try it like this:

<?
//connect to database
$dbhost = 'mysql01.myhostcp.com';
$dbuser = 'deans_user1';
$dbpass = 'qwerty123';
$dbname = 'deansghot5722com5233_deansdatabase';

$conn = mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db($dbname);

//NOTE: MAKE SURE YOU DO YOUR OWN APPROPRIATE SERVERSIDE ERROR CHECKING HERE!!!
if(!empty($_POST) && isset($_POST))
{
    $emptyvariables = false;

    if(!empty($_POST)){
        foreach($_POST as $key => $val){
            if(empty($_POST[$key])){
                echo"Please insert $key<br />";
                $emptyvariables = true;
                }
        }
    }
    
    if($emptyvariables == false){
    


        //make variables safe to insert
        $id = mysql_real_escape_string($_POST['id']);
        $Forename = mysql_real_escape_string($_POST['Forename']);
        $Surname = mysql_real_escape_string($_POST['Surname']);
        $Arrival_Date = mysql_real_escape_string($_POST['Arrival_Date']);
        $Departure_Date = mysql_real_escape_string($_POST['Departure_Date']);
        $RoomNo = mysql_real_escape_string($_POST['RoomNo']);
        $Tel = mysql_real_escape_string($_POST['Tel']);

        //query to insert data into table
        $sql = "
            INSERT INTO bookings
            SET
            id = '$id',
            Forename = '$Forename',
            Surname = '$Surname',
            Arrival_Date = '$Arrival_Date',
            Departure_Date = '$Departure_Date',
            RoomNo = '$RoomNo',
            Tel = '$Tel'";
            
            echo"$query";
        /*$result = mysql_query($sql);
        if(!$result)
        {
            echo "Failed to insert record<br>Please Ensure booking does not currently exist";
        }
        else
        {*/
			echo "<td>$id</td><td>$Forename</td>$Surname<td>$Arrival_Date</td><td>$Departure_Date</td><td>$RoomNo</td><td>$Tel</td>";
        //}
    }
}

?>

Open in new window

For this to work you'll need to make two changes. First your ajax_insert.php should load the initial rows from the database and put them in the table as ajax_data.php will now only output the new row. Secondly instead of replacing the whole table just add the new row by calling jquery.form like this:

$(document).ready(function() {
    $('#myForm').ajaxForm({
      success: function(responseText) {
           $('#showdata tr:last').after('<tr>' + responseText + '</tr>');
      }
    });
  });

Open in new window

or something like that.

Author Comment

by:deanlee17

ID: 350401292011-03-04

Thanks very much

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