Question

I’ve database table with the following pageid’s 1,2,3,4,5,6,7,8 –> will added more.
and i’m using

$sql ="select * from pages order by pageid";
$result= mysql_query($sql);
while($page= mysql_fetch_array($result)){
echo "$page[pagetitle]";
}

Now i want to select all except pageid=4 ! so that it won’t show the page title that have pageid = 4 and show the others 1,2,3,5,6 —> and any other added.

How to say select all except !

Answer 1

Simply add a WHERE clause WHERE pageid <> 4:

$sql ="select * from pages WHERE pageid <> 4 order by pageid";
$result= mysql_query($sql);
while($page= mysql_fetch_array($result)){
  echo $page['pagetitle'];
}

Or replace with a PHP variable to make it dynamic:

$exclpage = 4;
$sql ="SELECT * FROM pages WHERE pageid <> $exclpage ORDER BY pageid";

If you have multiple pageids to exclude, use a NOT IN() clause:

// Exclude pages 4,5,9
$sql ="SELECT * FROM pages WHERE pageid NOT IN (4,5,9) ORDER BY pageid";

Update:

Since this answer has more recently had some attention, it is advisable to use prepared statements and parameterize the variables instead of variables directly in the query, particularly if the variables are the result of any user input.

$sql ="SELECT * FROM pages WHERE pageid <> ? ORDER BY pageid";
// Bind ? according to your preferred API's method, such as MySQLi::bind_param() or PDO::bindParam()