Pass PHP array through Select Option Fields - php

Question

I am writing a basic CMS system and have come across something which should be seemingly simple -but is beginning to frustrate me.!
I am trying to pass an array through a select option field to populate a list of categories in which I can save a post.
I have a 'posts' form which comprises of 3 fields. Title, content and Category ID (CatID).
When the user creates a post, they can select the category they wish to assign the post assigned to by using a drop down list - (this is populated by using a different form).
So the technical bit; -
MySQL DB:-
categories = catname (char60 PRIMARY), catid (INT10, AI)
posts = id (bigint20 PRIMARY), catid (int10 PRIMARY), title (text), content (varchar255)
Example of categories populates: catname = Home / catid = 1 ...etc
Output.php ;
<?php
function display_post_form($post = '') {
$edit = is_array($post);
?>
<form action="<?php echo $edit ? 'edit.php' : 'add.php' ; ?>" method="post">
<table border="0">
<tr>
<td> Title:</td>
<td> <input type="text" name="title" value="<?php echo $edit ? $post['title'] : '' ; ?>" size="60" /> </td>
</tr><tr>
<td> Content:</td>
<td> <textarea id="editor1" name="content" value="<?php echo $edit ? $post['content'] : '' ; ?>"> </textarea> </td>
</tr><tr>
<td> Category:</td>
<td><select name="catid">
<?php
$cat_array = get_categories($catid, $catname);
foreach($cat_array as $thiscat) {
echo "<option value=\"".$thiscat['catid']."\" ";
if (($edit) && ($thiscat['catid'] == $post['catid'])) {
echo " selected";
}
echo ">".$thiscat['catname']."</option>";
}
?>
</select>
</td>
</tr><tr>
<td> Button:</td>
<td <?php if (!$edit) { echo "colspan=2"; } ?> align="center">
<?php
if ($edit)
echo "<input type=\"hidden\" name=\"_id\" value=\"". $post['id'] ."\" />";
?>
<input type="submit" value="<?php echo $edit ? 'Update' : 'Add' ; ?> Post" />
</form></td>
</td>
</tr>
</table>
</form>
<?php
}
?>
Functions.php ;
function get_categories($catid, $catname) {
$conn = db_connect();
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL " .mysqli_connect_error();
}
$sql = "SELECT catname, catid FROM categories";
$result = mysqli_query($conn, $sql) or die(" Could not query database");
while($row = mysqli_fetch_assoc($result)) {
printf("\n %s %s |\n",$row["catname"],$row["catid"]);
}
mysqli_close($conn);
}
I am able to call in the 'get_cattegories()' function which generates a flat data of categories and their respective id's. I then combined this with the Select Option Field in the Output.php file and it doesn't generate anything.
Can anyone give some useful tips or advice? Many thanks :)

Answer 1

You are not returning the array but printing a string to the output. Change printf to return:
function get_categories($catid, $catname) {
$conn = db_connect();
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL " .mysqli_connect_error();
}
$sql = "SELECT catname, catid FROM categories";
$result = mysqli_query($conn, $sql) or die(" Could not query database");
$categories = array();
while($row = mysqli_fetch_assoc($result)) {
$categories[] = $row;
}
mysqli_close($conn);
return $categories;
}
Also I agree for the comments to your question. The arguments are useless.
You also may refactor the code, actually... alot. Move the mysql_connect() to the other place, probably at the beginning of your script.
I suggest to use some frameworks. I think KohanaPHP will be a good start. You will learn about architecture and some design patterns. Keep the good work and improve your skills ;-)