Populating Select Field from Database - php

Question

I'm trying to populate a select field with PHP. The problem is I can't figure out how to display them because I'm getting the one that's value matches in the database showing up twice because I'm echoing it as selected and then looping it all the results. How can I just display the selected one that matched the fields value and then all the ones that don't match the selected one?
TABLE CATEGORIES
cat_id cat_name
1 soccer
2 baseball
3 basketball
TABLE ARTICLES
art_id art_cat_id
1 1
PHP / HTML
<select name="category">
<?php
$sql = "SELECT cat_id cat_name, art_id, art_cat_id
FROM categories LEFT JOIN articles
ON categories.cat_id = articles.art_cat_id
WHERE art_id = 1";
$result = query($sql);
if($result===false) {
echo("Query Fail");
}
else {
?>
<option value="<?php echo $data['art_cat_id'] ?>" selected="selected"><?php echo $data['cat_name'] ?></option>
<?php
while( $data = mysqli_fetch_array($result)) {
?>
<option value="<?php echo $data['cat_id'] ?>"><?php echo $data['cat_name'] ?></option>
<?php
}
}
?>
</select>
What it's returning
<select name="category">
<option value="1" selected="selected">soccer</option>
<option value="1">soccer</option>
<option value="2">baseball</option>
<option value="3">basketball</option>
</select>
What I'm looking for
<select name="category">
<option value="1" selected="selected">soccer</option>
<option value="2">baseball</option>
<option value="3">basketball</option>
</select>

Answer 1

Skip the row if the value matches the first one.
A snippet:
?>
<option value="<?php echo $data['art_cat_id'] ?>" selected="selected"><?php echo $data['cat_name'] ?></option>
<?php
while( $data = mysqli_fetch_array($result)) {
if ($data['art_cat_id'] == $data['cat_id']) continue;
?>
<option value="<?php echo $data['cat_id'] ?>"><?php echo $data['cat_name'] ?></option>
<?php
}