Populate select drop down from a database table

Question

I have a table ("venues") that stores all the possible venues a volunteer can work, each volunteer is assigned to work one venue each.

I want to create a select drop down from the venues table.

Right now I can display the venue each volunteer is assigned, but I want it to display the drop down box, with the venue already selected in the list.

<form action="upd.php?id=7">
<select name="venue_id">
<?php //some sort of loop goes here
print '<option value="'.$row['venue_id'].'">'.$row['venue_name'].'</option>';
//end loop here ?>
</select>
<input type="submit" value="submit" name="submit">
</form>

For example, volunteer with the id of 7, is assigned to venue_id 4

<form action="upd.php?id=7">
<select name="venue_id">
    <option value="1">Bagpipe Competition</option>
    <option value="2">Band Assistance</option>
    <option value="3">Beer/Wine Pouring</option>
    <option value="4" selected>Brochure Distribution</option>
    <option value="5">Childrens Area</option>
    <option value="6">Cleanup</option>
    <option value="7">Cultural Center Display</option>
    <option value="8">Festival Merch</option>
</select>
<input type="submit" value="submit" name="submit">
</form>

Brochure Distribution option will already be selected when it displays the drop down list, because in the volunteers_2009 table, column venue_id is 4.

I know it will take a form of a for or while loop to pull the list of venues from the venues table

My query is:

$query = "SELECT volunteers_2009.id, volunteers_2009.comments, volunteers_2009.choice1, volunteers_2009.choice2, volunteers_2009.choice3, volunteers_2009.lname, volunteers_2009.fname, volunteers_2009.venue_id, venues.venue_name FROM volunteers_2009 AS volunteers_2009 LEFT OUTER JOIN venues ON (volunteers_2009.venue_id = venues.id) ORDER by $order $sort";

How do I populate the select drop down box with the venues (volunteers_2009.venue_id, venues.id) from the venues table and have it pre-select the venue in the list?

score 18 · Accepted Answer · edited Feb 01 '21 at 14:09

$query = "SELECT volunteers_2009.id, volunteers_2009.comments, volunteers_2009.choice1, volunteers_2009.choice2, volunteers_2009.choice3, volunteers_2009.lname, volunteers_2009.fname, volunteers_2009.venue_id, venues.venue_name FROM volunteers_2009 AS volunteers_2009 LEFT OUTER JOIN venues ON (volunteers_2009.venue_id = venues.id) ORDER by $order $sort";

$res = mysql_query($query);
echo "<select name = 'venue'>";
while (($row = mysql_fetch_row($res)) != null)
{
    echo "<option value = '{$row['venue_id']}'";
    if ($selected_venue_id == $row['venue_id'])
        echo "selected = 'selected'";
    echo ">{$row['venue_name']}</option>";
}
echo "</select>";

score 4 · Answer 2 · answered Oct 14 '08 at 01:45

assuming you have an array of venues...personally i don't like to mix the sql with other wizardry.

function displayDropDown($items, $name, $label, $default='') {
  if (count($items)) {
    echo '<select name="' . $name . '">';
    echo '<option value="">' . $label . '</option>';
    echo '<option value="">----------</option>';
    foreach($items as $item) {
      $selected = ($item['id'] == $default) ? ' selected="selected" : '';
      echo <option value="' . $item['id'] . '"' . $selected . '>' . $item['name'] . '</option>';
    }
    echo '</select>';
  } else {
    echo 'There are no venues';
  }
}

I have a table (venues) that store all the venues (id, venue_name) The venues.id is stored in the volunteers_2009 table (volunteers_2009.venue_id) — Brad, Oct 14 '08 at 02:02

score 2 · Answer 3 · answered Apr 10 '11 at 13:01

        <?php 
        $query = "SELECT * from blogcategory";
        //$res = mysql_query($query);
        $rows = $db->query($query);
        echo "<select name = 'venue'>";
        // while (($row = mysql_fetch_row($res)) != null)
        while ($record = $db->fetch_array($rows)) 
        {
            echo "<option value = '{$record['CategoryId']}'";
            if ($CategoryId == $record['CategoryId'])
                echo "selected = 'selected'";
            echo ">{$record['CategoryName']}</option>";
        }
        echo "</select>";
        ?>

score -1 · Answer 4 · answered Sep 21 '17 at 07:57

<!DOCTYPE html>
<html>
<head>
    <title>table binding</title>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>

</head>
<body>
    <div id="mydiv" style="width:100px;height:100px;background-color:yellow">

        <select id="myselect"></select>
    </div>

</body>
</html>


<?php
include('dbconnection.php');

$sql = "SHOW TABLES FROM $dbname";
$result = mysqli_query($conn,$sql);

if (!$result) {
    echo "DB Error, could not list tables\n";
    echo 'MySQL Error: ' . mysqli_error();
    exit;
}

while ($row = mysqli_fetch_row($result)) {
    echo "<script>
    var z = document.createElement('option');
    z.setAttribute('value', '".$row[0]."');
    var t = document.createTextNode('".$row[0]."');
    z.appendChild(t);
    document.getElementById('myselect').appendChild(z);</script>";

}



?>

Populate select drop down from a database table

4 Answers4

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