Trying to make dynamic option form - php

Question

OK I try this for days and I can't do it so I try to take it step by step first I need to get the info from the database correctly and show it in a form here is the database
CategoryID | CategoryName | ParentID
----------------------------------------
1 | FirstMenuCat1 | 0
2 | FirstMenuCat2 | 0
3 | SubMenuCat1 | 1
4 | SubMenuCat2 | 1
5 | SubMenuCat3 | 2
and here is my last attemp that worked so I know I get the correct data from database but I don't know how to make it a form:
<?php
require_once ('mysqli_connect.php');
$q = "SELECT CategoryName FROM menus where ParentID = '0' ORDER by CategoryID";
$r = #mysqli_query ($dbc, $q) or die("Error: ".mysqli_error($dbc));
/* associative array */
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)){
printf ("%s\n", $row["CategoryName"]);
}
?>
Here is my attempt to make it but its not working
/* associative array */
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)){
foreach($row){
echo "<option value=\"". urlencode( $_ ) ."\" $selected >$_</option>\n";
$selected = ""; // only the first element will be marked as selected
}
}
?>
Any ideas?

Answer 1

By looking at option tag i think you are trying to create a combobox list. If it is then First of all do not use \n. The options in a combobox list already shown line by line.
Secondly you have not printed the select tag. Without it the list will not be rendered properly. Try the following code:
echo "<select name='mylist'>";
$selected = "selected";
while (($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) !== FALSE){
echo "<option value='". $row['key']."' $selected >".$row['value']."</option>";
$selected = ""; // only the first element will be marked as selected
}
echo "</select>";
hopefully it may help.

Answer 2

Check this
echo '<select name="" id="">';
/* associative array */
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)){
// Check the $_ correctly
foreach($row){
echo "<option value='". urlencode( $_ ) ."'>$_</option>\n";
}
}
echo '</select>';
Note*: First element by default will be selected for the dd.

Answer 3

Not sure but I think this is what you need
$optionString = '';
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC))
{
$value = urlencode($row['CategoryName']);
$optionString .= "<option value='{$value}'>{$row['CategoryName']}</option>";
}
echo "<select>$optionString</select>";

Answer 4

$selected = 'selected';
$htmlSelect = '<select name="elementName" id="elementId">';
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)){
$htmlSelect .= '<option value="' . urlencode( $_ ) . '" ' . $selected . ' >' . $_ . '</option>';
$selected = ""; // only the first element will be marked as selected
}
$htmlSelect .= '</select>';
echo $htmlSelect;
Note 1: It's better to use double quote " instead of single quote ' on element attributes. Some browsers don't like single quotes.
Note 2: Couldn't understand why you're using foreach again in the while statement.
Note 3: If you will not remember which option was selected there is not any need to specify which element is selected, first one will be selected by default.