Question

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?

Answer 1

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.