Question

*strong text*I don’t know if the title was clear or not,

but I want a PHP script that tells if a MySQL column has data, and if it does to print a line of text, but if it is NULL, to print something else…

Make sense?

Here is what I have, I know it is completely wrong…

$link = mysql_connect($DBHOST, $DBUSER, $DBPASSWORD);
mysql_select_db($DBNAME); 
if (!$link) {
    die('Could not connect: ' . mysql_error());
}

$result = mysql_query("SELECT * FROM members");

while($row = mysql_fetch_assoc($result))
{
     if($row['VIP'] = '1')
     {
        echo('VIP');
     }
     else
     {
        echo('Non-VIP');
     }
}

EDIT::

Here is the full code:

<?php

$DBTYPE = 'mysql';
$DBHOST = 'host';
$DBUSER = 'username';
$DBPASSWORD = 'password';
$DBNAME = 'database';

$link = mysql_connect($DBHOST, $DBUSER, $DBPASSWORD);
mysql_select_db($DBNAME); 
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
$member_id= '5'; // get value of the member, assumed that you have post the data into id. change if it is different into your code

$result = mysql_query("SELECT * FROM members where id=$member_id");

while($row = mysql_fetch_assoc($result))
{
     if($row['VIP'] == '1')
     {
        echo "<img src='../images/VIP.png' />";
     }
     else
     {
        echo 'Non-VIP';
     }
}
?>

Answer 1

use == to check equality, like this:

if($row['VIP'] == '1')
{
  echo('VIP');
}
else
{
 echo('Non-VIP');
}

EDIT

Use below code to get your desired result

$link = mysql_connect($DBHOST, $DBUSER, $DBPASSWORD);
mysql_select_db($DBNAME); 
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
$member_id=$_POST['id']; // get value of the member, assumed that you have post the data into id. change if it is different into your code

$result = mysql_query("SELECT * FROM members where id=$member_id");

while($row = mysql_fetch_assoc($result))
{
     if($row['VIP'] == '1')
     {
        echo "<img src='../images/VIP.png' />";
     }
     else
     {
        echo 'Non-VIP';
     }
}