php multiple select drop down

0
votes

here is my mysql and php code layout:

I have 3 tables

  • tableA stores unique "person" information
  • tableB stores unique "places" information
  • tableC stores not unique information about a person and places they have "beenTo".

here is how i layed out my form: -one big form to insert into "person" tableA; "beenTo" tableC in the form, a person mulitple selects "places" which get inserted into "beenTo"

my question is, when i am editing a "person" how do i display what the user has already selected to appear on my multiple select options drop down menu?

my drop down menu at the moment query "places" table and displays it in a multiple select drop down menu. its easier when a person have beenTo one place, the problem arrises when there is more than one "beenTo" places?

3
phpmysql
I'm confused ... can you post some code somewhere (pastebin or here)? - Nathan Loding
i don't have a code. basically, i have 3 tables that i need to select to display one person information to edit. eg. if i want to edit person id=111 i click on that id and i populate that person info in my form fields. i have a multiple select in my form fields. which again, i query from the db to display all the available options. the 3rd table select comes when i am trying to match whata person has already selected with my options values. again, its easier when a person has only one options selected in the db, the problem comes when i am trying to display all the person's multiple options. - NULL

3 Answers

2
votes

Foreach option, check if they have beenTo it. Then add the selected="selected" attribute to the tag if true.

Example:

<select multiple="multiple">
    <option selected="selected">Rome</option>
    <option>France</option>
    <option selected="selected">Underpants</option>
</select>

And in PHP this might look like:

$beenTo = array("Rome","Underpants");
$places = array("Rome","France","Underpants");
?> <select multiple="multiple"> <?php
foreach($places as $place) {
    echo "<option";
    $found = false;
    foreach($beenTo as $placeBeenTo) {
        echo "value='$place'";
        if ($placeBeenTo == $place) {
            $found == true;
            echo " selected=\"selected\" ";
            break;
        }
    }
    if (!$found) echo ">";
    echo $place . "</option>";
}
?> </select> <?php

There's probably a much more efficient way to do this.

1
votes

For clarification, you will want to have a name attribute for your select tag which allows for multiple selected options to function correctly.

<form method="post" action="">
<select name="places[]" multiple="multiple">
<?php
$_POST += array('places' => array());
$places = array('fr' => 'France', 'cn' => 'China', 'jp' => 'Japan');
$beenTo = array_flip($_POST['places']);
foreach ($places as $place => $place_label) {
  $selected = isset($beenTo[$place]) ? ' selected="selected"' : '';
  echo '<option value="' . $place . '"' . $selected . '>' . $place_label . '</option>';
}
?>
</select>
<input type="submit" value="Save Changes" />
</form>
0
votes
<option id = 'example' <? if ($_POST['example']) echo 'selected' ?>>

But you'll be using a $_SERVER['cookies'] or other cache to store the past visits, not the $_POST array.. edit with extreme predjudice