My dropdown list would not drop down - php

Question

I tried to use php to create a dropdown list populated by a SELECT query but it would not dropdown. Here is my php script
<?php
require_once ('mysqli_connect.php');
#mysqli_select_db($dbc,"test");
$query = "SELECT category_id, category_name FROM forum_category ORDER BY category_name";
$result = #mysqli_query($dbc, $query);
if(!$result)
{
echo "query error: " . mysqli_error($dbc);
}
// while($row = #mysqli_fetch_array($result))
// {
// echo "<p>$row[0] $row[1]</p>\n"; //this works
// }
echo "<td bgcolor=\"#E6E6E6\"";
echo "<strong>Category:</strong>";
echo "<select name=\"category\">";
while($row = #mysqli_fetch_array($result))
{
// echo "<p><option value='".$row[0]."'>".$row[1]."</option></p>\n";
echo "<option value='".$row[0]."'>".$row[1]."</option>";
}
mysqli_close($dbc);
echo "</select>";
echo "</td>";
?>
The query worked (the commented while loop outside the dropdown list displayed 15 records). But the dropdown list only showed one category_name and would not dropdown. Can someone help me figure out the problem? Thanks.

Answer 1

There are a few problems with your code.
You can either remove both echo "<td bgcolor=\"#E6E6E6\""; and echo "</td>";
which is the reason why your data is not showing.
Or, add the appropriate <table></table> tags.
Plus this echo "<td bgcolor=\"#E6E6E6\""; should be echo "<td bgcolor=\"#E6E6E6\">"; there was a missing > so that alone is breaking your <td>.
So in order to have the proper table syntax do:
Sidenote: I added . "\n" in order to get clean and well-aligned HTML source.
<?php
require_once ('mysqli_connect.php');
#mysqli_select_db($dbc,"test");
$query = "SELECT category_id, category_name FROM forum_category ORDER BY category_name";
$result = #mysqli_query($dbc, $query);
if(!$result)
{
echo "query error: " . mysqli_error($dbc);
}
echo "<table>" . "\n";
echo "<tr>" . "\n";
echo "<td bgcolor=\"#E6E6E6\">" . "\n";
echo "<strong>Category:</strong>" . "\n";
echo "<select name=\"category\">" . "\n";
while($row= mysqli_fetch_array($result)){
echo "<option value='".$row[0]."'>".$row[1]."</option>" . "\n";
}
echo "</select>" . "\n";
echo "</td>" . "\n";
echo "</tr>" . "\n";
echo "</table>" . "\n";
mysqli_close($dbc);
?>
Edit: (Try this) since I don't know what your full code looks like.
<form id="form1" name="form1" method="post" action="form1_handler.php">
<?php
require_once ('mysqli_connect.php');
#mysqli_select_db($dbc,"test");
if(isset($_POST['submit'])){
$query = "SELECT category_id, category_name FROM forum_category ORDER BY category_name";
$result = #mysqli_query($dbc, $query);
if(!$result)
{
echo "query error: " . mysqli_error($dbc);
}
echo "<table>" . "\n";
echo "<tr>" . "\n";
echo "<td bgcolor=\"#E6E6E6\">" . "\n";
echo "<strong>Category:</strong>" . "\n";
echo "<select name=\"category\">" . "\n";
while($row= mysqli_fetch_array($result)){
echo "<option value='".$row[0]."'>".$row[1]."</option>" . "\n";
}
echo "</select>" . "\n";
echo "</td>" . "\n";
echo "</tr>" . "\n";
echo "</table>" . "\n";
mysqli_close($dbc);
} // brace for if(isset($_POST['submit']))
?>
<input type="submit" name="submit" value="Submit">
</form>