RE: Table shows even when if () is false

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[snip]
 [snip]
$deferred_comments= "SELECT * FROM comments WHERE credit_card_id =
'$credit_card_id' AND request_type = 'D'";
$result_deferred_comments = mssql_query($deferred_comments) or
die(mssql_error());

if(!empty($result_deferred_comments)) {
[/snip]

$result_deferred_comments is not empty, a query resource was returned.
To see this;

echo $result_deferred_comments;

What you want to do is something like this;

$num_Deferred_Comments = mysql_num_rows($result_deferred_comments);

if(0 != $num_Deferred_Comments)) {....
[/snip]

My bad, just noticed you were using mssql, so use the proper functions
there, like mssql_num_rows 

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