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Unexpected T VARIABLE?


Unholy Prayer

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Ok, I am trying to make an affiliates management program, but I keep getting this error on the index display page... here it is:
[code]Parse error: syntax error, unexpected T_VARIABLE in /home/mutantde/public_html/programs/index.php on line 4[/code]

This is line 4:
[code]$result = mysql_query("SELECT * from affiliates");[/code]

This is probably a noob question, but this is my first PHP program so I guess I am a PHP Noob.

Thanks in advance for the help.
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This is the whole page:
[code]<?php
require_once('database.php')

$result = mysql_query("SELECT * from affiliates");

while($r=mysql_fetch_array($result))
{  
  $site_name=$r["site_name"];
  $site_url=$r["site_url"];
  $banner_url=$r["banner_url"];
  echo "<a href='$site_url' target='new'><img src='$banner_url' width='88' height='31'></a>"
}
?>[/code]

I know... not that much coding, right?  But hey, with a PHP noob like me, this kind of stuff is expected.
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[quote author=thorpe link=topic=104587.msg417245#msg417245 date=1155795777]
Jeremysr, a simple yes / no. Do you understand the meaning of your (our) avitar?
[/quote]

Yes. I read the "How to be a Hacker" article and the emblem faqs.

And Unholy Prayer, semi-colons signify the end of a line.
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Jeremysr, cool. I have the book.

[quote]that's wierd because semicolons have never been a problem in my other scripts[/quote]

You should have... there are a required part of php synatx.

Yes, you'll need them at the end.
[code=php:0]
<?php
require_once('database.php');

$result = mysql_query("SELECT * from affiliates");

while($r=mysql_fetch_array($result))
{  
  $site_name=$r["site_name"];
  $site_url=$r["site_url"];
  $banner_url=$r["banner_url"];
  echo "<a href='$site_url' target='new'><img src='$banner_url' width='88' height='31'></a>";
}
?>
[/code]
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Ok, there's no error now, but it's not displaying anything...  Here it is:
http://www.mutant-designs.com/programs/index.php

Note:  That is the correct URL, I just forgot to add the "affiliates" directory and put all the files in it.  I'll get to it when I'm done with the program.
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You never check to see if your query actually works before using it. Get in the habbit.

[code=php:0]
<?php
require_once('database.php');

if ($result = mysql_query("SELECT * from affiliates")) {
  while($r=mysql_fetch_array($result))
    $site_name=$r["site_name"];
    $site_url=$r["site_url"];
    $banner_url=$r["banner_url"];
    echo "<a href='$site_url' target='new'><img src='$banner_url' width='88' height='31'></a>";
  }
} else {
  echo "query failed ".mysql_error();
}
?>
[/code]
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