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Start your free trialAmit Kumar
1,063 PointsMy genre dropdown is not showing .
<tr>
<th>
<label for="genre">Genre</label>
</th>
<td>
<select name="genre" id="genre">
<option value="">Select One</option>
<?php
$genre_arrayy = genre_array();
foreach ($genre_arrayy as $category=>$options) {
echo "<optgroup label=\"$category\">";
foreach ($options as $option) {
echo "<option value=\"$option\"";
if (isset($genre) && $genre==$option) {
echo " selected";
}
echo ">$option</option>";
}
echo "</optgroup>";
}
?>
</select>
</td>
</tr>
function genre_array($category = null) {
$category = strtolower($category);
include("connection.php");
try {
$sql = "SELECT genre, category"
. " FROM Genres "
. " JOIN Genre_Categories "
. " ON Genres.genre_id = Genre_Categories.genre_id ";
if (!empty($category)) {
$results = $db->prepare($sql
. " WHERE LOWER(category) = ?"
. " ORDER BY genre");
$results->bindParam(1,$category,PDO::PARAM_STR);
} else {
$results = $db->prepare($sql . " ORDER BY genre");
}
$results->execute();
} catch (Exception $e) {
echo "bad query";
}
$genres = array();
while ($row = $results->fetch(PDO::FETCH_ASSOC)) {
$genres[$row["category"]][] = $row["genre"];
}
return $genres;
}
Moderator edited: Markdown added so that the code will render properly in the forums.
jlampstack
23,932 PointsI only glimpsed at your code without reading the entire code, but right away I noticed you spelled 'array' where you have genre_array. There should be only 1 'y'.
There are multiple areas where you spelled array with 2 y's, but you have it spelled correct in your function.
Jennifer Nordell
Treehouse TeacherJennifer Nordell
Treehouse TeacherHi Amit! I added some markdown to your question so that the code will be easier to read for other students and hopefully assist you in getting help faster. Please see the Markdown Cheatsheet link at the bottom of the "Add an Answer" section for tips on how to format your text/code in your posts!