PHP populate drop down list with SQL data

Question

I want to populate a drop down list with values from database.

<?php
require 'conn.php';

$filter=mysql_query("select distinct fuel_type from car");
while($row = mysql_fetch_array($filter)) {
$options ="<option>" . $row['fuel_type'] . "</option>";

$menu="<form id='filter' name='filter' method='post' action=''>
  <p><label>Filter</label></p>
    <select name='filter' id='filter'>
      " . $options . "
    </select>
</form>";

echo $menu;
}
?>

The issue is that I get two lists instead of one list with the values inside. Please advise

drop down list

John Conde · Accepted Answer · 2012-04-16 02:11:55Z

2

You need to remove the $menu from your loop:

<?php
require 'conn.php';

$options = '';
$filter=mysql_query("select distinct fuel_type from car");
while($row = mysql_fetch_array($filter)) {
    $options .="<option>" . $row['fuel_type'] . "</option>";
}

$menu="<form id='filter' name='filter' method='post' action=''>
  <p><label>Filter</label></p>
    <select name='filter' id='filter'>
      " . $options . "
    </select>
</form>";

echo $menu;

?>

answered Apr 16, 2012 at 2:11

John Conde

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1 Comment

John Conde Over a year ago

You're probably missing the . in the while loop. Copy and paste my code and try again.

Soufiane Hassou · Accepted Answer · 2012-04-16 02:09:55Z

2

echo $menu; should be outside your loop.

answered Apr 16, 2012 at 2:09

Soufiane Hassou

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Comments

Menztrual · Accepted Answer · 2012-04-16 02:16:34Z

1

<?php
require 'conn.php';

$filter=mysql_query("select distinct fuel_type from car");
$menu="
<form id='filter' name='filter' method='post' action=''>
  <p><label>Filter</label></p>
    <select name='filter' id='filter'>";

// Add options to the drop down
while($row = mysql_fetch_array($filter))
{
  $menu .="<option>" . $row['fuel_type'] . "</option>";
}

// Close menu form
$menu = "</select></form>";

// Output it
echo $menu;
?>

answered Apr 16, 2012 at 2:16

Menztrual

41.7k12 gold badges59 silver badges70 bronze badges

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