php/mysql query to retrieve dynamic value

While this is a mysql/database question I believe the answer to my question is using php, and that's why i'm posting it here instead of the database section.


I have a database with restaurants. They are located in different parts of the city, and have a 'area' row, which defines where in the city the place is.

I use this query to get the information and echo it on each page:

<?php
include 'include/opendb.php';

$query  = "SELECT * FROM restaurants WHERE restaurantid = '$id'";
$result = mysql_query($query);

while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
    echo "{$row['name']} in {$row['area']} ";
} 

include 'include/closedb.php';
?>

PHP:

Now, here's what I want to do (And my problem.)


I also want to show a list with, let's say, 5 other places in the same area (with the same value as the one on the page. To do this I'd have to run this query:

SELECT * FROM restaurants WHERE area = same-as-the-restaurant-above"

So, my question is: How do I get that value inside the SELECT-part of the mysql query? (As above.)

As you probably can tell I'm no database/php genious, so any answers would help a lot!

 

After the while:
On the while before of after the echo, set too:
$area=$row['area'];


After the while:

$sql="SELECT * FROM restaurants WHERE area = '".$area."' LIMIT 5;";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
echo "{$row['name']} in {$row['area']} ";
}

 

echo "{$row['name']} in {$row['area']} ";
$area = mysql_real_escape_string($row['area']);
$sql2 = "select name
   from restaurants
   where area = '{$area}'
   and id != {$id}
   limit 5";
$result2 = mysql_result($sql2) or die(mysql_error());
if (mysql_num_rows($result2))
{
   echo "<p>Some other fine restaurants in the area:</p><ul>";
   while ($row2 = mysql_fetch_assoc($result2))
      echo "<li>{$row2['name']}</li>";
   echo "</ul>";
}

Code (markup):

 

Thank you!

I'm using the code magiatar gave me, and it works out perfectly.

SmallPotatoes: Thanks aswell!

 

Magiatar's query will also show the same restaurant as being near itself sometimes. Better to add the:

and id != {$id}

to the query.