Elegant way to add "selected" into a dropdrown form using PHP - php

Question

I have a form that looks like this:
$guests = 2; // Just for testing
$form = 'Guests<br /><select name="guests">
<option value="0">0</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
<option value="6">6</option>
<option value="7">7</option>
<option value="8">8</option>
<option value="9">9</option>
</select>';
But the list of guests can be up to 100, so I probably shouldn't do a check for which value is selected manually.
What would be a good way for me to make the selected value become the selected value when the user sees this form?
Thanks!

Answer 1

What I normally do with dynamic select menus is use a foreach loop over an array, and just check each item.
Something like:
$dropdownName = "guests"; //Name of dropdown
$defaultValue = "0"; //value to select if there isn't one already set
$items = array('name'=>'value', 'anotherName'=>'anothervalue'); //items, name=>value
echo '<select name="'.$dropdownName.'">'; //Start Select
$selectedItem = (isset($_POST[$dropdownName])?$_POST[$dropdownName]:$defaultValue); //If a value is set, use it, otherwise use the default
foreach ($items as $name=>$value) //build select
{
echo '<option value="'.$value.'"'.(($selectedItem == $value)?' selected="selected"':'').'>'.$name.'</option>';
}
echo '</select>';

Answer 2

$guests = 2; // Just for testing
$form = 'Guests<br /><select name="guests">';
for($i = 0; $i <= 100; $i++)
$form .= "<option value='$i'{$guest == $i ? ' selected=\'selected\'' : ''}>$i</option>";
$form .= '</select>';
EDIT: $i < 101 was pretty lame...

Answer 3

You can store the value this value in a hidden field and on document.ready(), you can assign the selected value of drop down top the value of hidden value.
A solution that works but may not be efficient for some users...