How to put array into session?

Question

This is assignment so not looking for anything perfectly safe and secure, just working.I have table in SQL database, I'm printing down all records, all records have unique reference number, for each row of the database I printed down I gave checkbox with value of the row's ref. number, when I submit them by "POST" everything is working and printing out:

if (!$_POST['checkbox']) {
    echo "Your basket is empty.";
} else {
    echo "<table border='0' id='games_table' cellspacing='1'>";
    echo "<tr id='basket_table_row'>";
    echo "<td colspan='3'>" . "Logged: " . $_SESSION['user'] . "</td>";
    echo "<td colspan ='2'>" . "OS used on this machine: " . "<script type='text/javascript'>document.write(yourOS())</script><noscript>Computers</noscript>" . "</td>";
    echo "</tr>";
    echo "<tr id='basket_table_row'>";
    echo "<td colspan='5'>" . "You put into the basket these games: " . "</td>";
    echo "</tr>";
    foreach ($_POST['checkbox'] as $value) {
        $_SESSION['basket']=array($value);

        $res=pg_query($conn,"select * from CSGames where refnumber='$value'");
        while ($a = pg_fetch_array ($res)) {
            echo "<tr id='games_table_row'>";
            echo "<td>" . $a["refnumber"] . "</td>";
            echo "<td>" . $a["title"] . "</td>";
            echo "<td>" . $a["platform"] . "</td>";
            echo "<td>" . $a["description"] . "</td>";
            echo "<td>" . $a["price"] . "</td>";
            echo "</tr>";   
        }
    }
    echo "</table>\n";
}

but only think which stays recorded in $_SESSION['basket'] is value of the last checkbox but I need all of them (60 or 70). what Am I doing wrong?

You need us to show some more. Your request doesnt really make sense. — Jan Dragsbaek
– Jan Dragsbaek, Commented Nov 24, 2011 at 15:09

Lao · Accepted Answer · 2011-11-24 15:10:21Z

2

You are overwriting te value of $_SESSION['basket'] at each iteration of the loop.

The last value is the only one stored.

answered Nov 24, 2011 at 15:10

Lao

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Jan Dragsbaek · Accepted Answer · 2011-11-24 15:11:02Z

1

Currently, you are only storing the last value, if you wish to store every value, you should add it like this:

$_SESSION['basket'][] = $value;

answered Nov 24, 2011 at 15:11

Jan Dragsbaek

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Julien · Accepted Answer · 2011-11-24 15:11:44Z

0

foreach($_POST['checkbox'] as $value){

    $_SESSION['basket']=array($value);

}

every iteration of your loop is overwriting the value in $_SESSION['basket'], hence you only seeing the last checkbox value.

answered Nov 24, 2011 at 15:11

Julien

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aorcsik · Accepted Answer · 2011-11-24 15:14:55Z

0

In every step of foreach, you create a new array in $_SESSION['basket'] with only one item, the current value of $value. It is corrected like this:

// ...
$_SESSION['basket'] = array();
foreach ($_POST['checkbox'] as $value) {
    $_SESSION['basket'][] = $value;
    // ...
}
// ...

answered Nov 24, 2011 at 15:14

aorcsik

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Comments

Leysam Rosario · Accepted Answer · 2011-11-24 15:19:05Z

0

You can directly do it like this

$_SESSION['basket'] = $_POST['checkbox'];

because the data of $_POST['checkbox'] is an array anyway. What you are doing is looping the $_POST['checkbox'] and saving each data as array to a session.

this $_SESSION['basket'] = $_POST['checkbox'];

and

 foreach ($_POST['checkbox'] as $value) {
        $_SESSION['basket'][]=$value;
 }

will have the same value.

What you have right only saves the last value of $_POST['checkbox']

answered Nov 24, 2011 at 15:19

Leysam Rosario

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