Retaining drop down list box value of a form after posting
Retain a selected option safely
Use an allow-list for accepted option values. The same validated value can then be used to mark the matching option as selected.
$allowed = ['one', 'two', 'three'];
$selected = $_POST['class'] ?? '';
if (!in_array($selected, $allowed, true)) {
$selected = '';
}
echo '<option value="two"' . ($selected === 'two' ? ' selected' : '') . '>Two</option>';
From a drop down?> From a drop down list box or combo box visitor selects one option and submits the form. If the form validation fails then we have to show again the same form asking for changes in input data and submit again. In this process we have to keep our drop down box selected to the option previously selected by the user. For example in a travel information form, visitor select the week day of travel from a drop down list box having all days of the week and submits the form, then due to any reason the form validation fails and we ask the visitor to again fill the form, we must keep the previously selected day of the week of the visitor as selected only. So if the visitor has selected Monday then the form has to show day Monday as selected day of the week.
There are different ways to do this and one method where the data (or the option) are populated from table is already discussed in double drop down list box tutorial. There all options are matched with an if condition and accordingly the selected attribute is added. Note that there are many ways to achieve this and we will try to explore one way here.
A compact approach is to validate the submitted option against an allow-list and then add the
selected attribute while rendering the options. This keeps the sticky form on one page and avoids putting the selected value into a query string.
Here is the same-page POST version. The day is validated against an allow-list before it is used to restore the selected option.
Related: HTML forms and jQuery listbox.
$days = ['Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat', 'Sun'];
$t1 = $_POST['t1'] ?? 'Mon';
if (!in_array($t1, $days, true)) {
$t1 = 'Mon';
}
foreach ($days as $day) {
$selected = $t1 === $day ? ' selected' : '';
echo "<option value='$day'$selected>$day</option>";
}
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Subhendu Mohapatra
Author & Instructor at plus2net
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| smo1234 | 18-06-2012 |
| thanks this is working fine | |
| Jnanendra Veer | 05-09-2012 |
| thanks this is working fine | |
| Murugavel | 04-08-2014 |
| Working perfectly Dude.Thanks | |
