Echo row when $variable is duplicate but primary key is unique

I'm creating a list of stores within cities. So where there's only one store in a city, I want it to display just the one result. When there are two or more stores within the city, I want it to display the information on the multiple stores. Here's what I have:

$result = mysql_query("SELECT * FROM stores where StoreID=$id order by city");
if (!$result) { 
die('Invalid query: ' . mysql_error());
$id=mysql_result($result,"StoreID");
mysql_close();
}

<?php echo "$city"; ?>
<?php echo "$street"; ?> etc.

This gives me just one result and I'm not sure how to modify it to look for multiple records. I'm a PHP/MySql beginner so I real开发者_开发知识库ly appreciate your help and apologize if this is a stupid question or ridiculous question. Thank you in advance for your help!

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Use this basic loop:

<?while($row = mysql_fetch_assoc($result):?>
  city: <?=$row['city']?><BR>
  State: <?=$row['state']?><BR>
<?endwhile;?>

Get rid of the mysql_result() part - use mysql_fetch_assoc() it's easier.

Also, you should be using the http://www.php.net/mysqli library instead of MySQL in PHP.

PS. I like short tags, but you can change to long tags if you prefer.

Also instead of:

$result = mysql_query("SELECT * FROM stores where StoreID=$id order by city");
if (!$result) { 
die('Invalid query: ' . mysql_error());

I like:

$result = mysql_query("SELECT * FROM stores where StoreID=$id order by city") or die('Invalid query: ' . mysql_error());

It's easier to read.

---

$result = mysql_query("SELECT * FROM stores where StoreID=$id order by city");
if (!$result) { 
    die('Invalid query: ' . mysql_error());
}else{
    while($row = mysql_fetch_assoc($result)){
         echo $row['city']." : ".$row['street']."<br />";
    }

}

while() is fetching results step by step, so it will give it your result