I am struggling to produce a SQL result. I am trying to display my results in a table from one of the select options.
Any help?
HTML:
<form action="" method="post" name="parkname">
<select id="combobox" name="parkname" class="park-choice">
<option hidden value="C">C</option>
<option class="hidden" value="E">E</option>
<option class="hidden" value="W">W</option>
</select>
<span style="display:inline-block; width: 200px;"></span>
Capacity:
<select id="foo" style="display:inline-block; width: 80px;" id="combobox">
<input type="submit" value="submit">
</select>
</form>
PHP code:
<?php
$selectOption = $_POST['parkname'];
$query = "SELECT * FROM `ROOMS` WHERE `Park` = '$selectOption%';";
$result = mysql_query($query);
echo $result;
if ($result == FALSE) die ("could not execute statement $query<br />");
echo "<table>";
while($row = mysql_fetch_array($result)){
echo "<tr><td>" . $row['roomCode'] . "</td><td>" . $row['Park'] . "</td><td>" . $row['Capacity'] . "</td><td>" . $row['Style'] . "</td><td>" . $row['dataProjector'] . "</td><td>" . $row['Whiteboard'] . "</td><td>" . $row['OHP'] . "</td><td>" . $row['wheelchairAccess'] . "</td><td>" . $row['lectureCapture'] . "</td></tr>";
}
echo "</table>";
mysql_close();
?>
First, you need to clear your query from % sign.
$query = "SELECT * FROM `ROOMS` WHERE `Park` = '$selectOption';";
Or if you want to find all matches, you can use LIKE operator:
$query = "SELECT * FROM `ROOMS` WHERE `Park` LIKE '%$selectOption%';";
Second, for displaying html combobox with Park names (in your case), use code like:
<select name="park">
<option value="Park1">Display name</option>
<option value="Park2">Display name</option>
<option value="Park3">Display name</option>
</select>
Park1,Park2,Park3 - Values that would be sent to your php code inside $_REQUEST['park'] variable.
Display name - Just visually display name for options in select