Multiple select in an for loop is losing it's data when returning to the form

Question

Multiple select in an for loop is losing it's data when returning to the form

319 Views Asked by Muiter At 25 April 2025 at 04:23

I have an form with an multiple select in the while loop:

while ($row_i = mysql_fetch_array($res_i))
{
    $i++;

    // maak select name
    $name_bewerking_id = 'bewerking_id'.$i;
    ?>

    <tr valign="top">
        <td>
        <select name="<?php echo $name_bewerking_id ?>[]" multiple="multiple" size="2">
        <?php
            $sql = "SELECT id, bewerking FROM bewerkingen ORDER BY bewerking ASC";
            $res = mysql_query($sql,$con);
            while ($row = mysql_fetch_assoc($res))
            { ?>
                <option value="<?php echo $row['id']; ?>"><?php echo $row['bewerking']; ?></option>
        <?php } ?>
        </select>
        </td>
    </tr>
<?php
}

When the form is send:

$bewerking_id[$i] = array();
$bewerking_id[$i] = $_POST['name_bewerking_id'][$i];

if(isset($bewerking_id_temp[$i]))
{
    foreach($bewerking_id_temp[$i] as $temp[$i])
    {
        array_push($bewerking_id[$i], $temp[$i]);
    }
}

Returning to the form:

for ($i = 0; $i <= $aantal_regels_corr; $i++)
{
    // maak select name
    $name_bewerking_id = 'bewerking_id'.$i;
    ?>

    <tr valign="top">
        <td>
        <select name="<?php echo $name_bewerking_id ?>[]" multiple="multiple" size="2">
        <?php
            $sql = "SELECT id, bewerking FROM bewerkingen ORDER BY bewerking ASC";
            $res = mysql_query($sql,$con);
            while ($row = mysql_fetch_assoc($res))
            { ?>
                <option <?php if(isset($bewerking_id[$i]) && in_array($row['id'], $bewerking_id[$i])){ echo 'selected="selected"'; } ?> value="<?php echo $row['id']; ?>"><?php echo $row['bewerking']; ?></option>
        <?php } ?>
        </select>
        </td>
    </tr>
<?php
}

When returning to the form (when one of the other fields is not filled in) the chosen option(s) are lost and not selected again.

Where did I messed up?

Original Q&A

There are 2 best solutions below

**simopisto** · Answer 1

simopisto On 29 October 2011 at 12:27

You read POST data with this identifier 'name_bewerking_id'

but the select name is given by <?php echo $name_bewerking_id ?> with $name_bewerking_id = 'bewerking_id'.$i

**YTKColumba** · Answer 2

If assuming you are passing the form information between 3 different pages, you will need to use $_SESSION variables for that. Normal PHP variables cannot be passed between pages, only $_SESSION variables.

For example:

Page 1

<form ...>
    <input name="text1" type="text" />
    <input type="submit" />
</form>

Page 2

// must start the session before session variables can be used
start_session();

$inputTextBox1 = $_SESSION["textBox1"] = $_POST["text1"];

Page 3

<?php start_session(); ?>
<html>
    ...

    <form ...>
        <select>
            <?php while ... { ?>
                <option <?php if(!empty($_SESSION["textBox1"])) { echo "selected=\"selected\""; } ?>>Some Text</option>
            <?php } // End while ?>
        </select>
    </form>

    ...
</html>

Multiple select in an for loop is losing it's data when returning to the form

There are 2 best solutions below

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