I'm lost!

<?php
$Ref=$_GET["Ref"];
$conn = mysql_connect("#######"); 
mysql_select_db("#######",$conn);
$result = mysql_query("Select FileName From ".$Ref."Where Enable = 1");
while($row = mysql_fetch_array($result));
{
[COLOR="Red"]	echo "<img src=".$row['FileName']." width='100'>\r";[/COLOR]
}
?>

Code (markup):

Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';' in /home/content/t/r/o/trollnest/html/bragflags/AJAX.php on line 8

?

 

Tri this wai,,,

<?php
$Ref=$_GET['Ref'];
$conn = mysql_connect("#######"); 
mysql_select_db("#######",$conn);
$result = mysql_query("Select FileName From $Ref  Where Enable = 1");
while($row = mysql_fetch_array($result));
{
	echo "<img src=\"$row['FileName']\" width='100'>";
}
?>

PHP:

 

try this....

<?php

$Ref=$_GET["Ref"];

$conn = mysql_connect("#######"); 

mysql_select_db("#######",$conn);

$result = mysql_query("Select FileName From ".$Ref."Where Enable = 1");

while($row = mysql_fetch_array($result)){

	echo "<img src=".$row['FileName']." width='100'>\r";

} ?>

Code (markup):

 

LOL... Learn php!

 

One other fix:

<?php
$Ref=$_GET["Ref"];

$conn = mysql_connect("#######"); 
mysql_select_db("#######",$conn);

$result = mysql_query("Select FileName From ".$Ref." Where Enable = 1");
while($row = mysql_fetch_array($result))
{
	echo "<img src=".$row['FileName']." width='100'>\r";

	// You can use this, too:
	// echo "<img src=\"{$row['FileName']}\" width='100'>\r";
}
?>

PHP:

Why are you taking the table name from the Query String? Some serious security problems going on there.

 

Tri this is working 100%

<?php
$Ref=$_GET["Ref"];
$conn = mysql_connect("localhost","user","password");
mysql_select_db("database",$conn);
$query = mysql_query("SELECT FileName from $Ref where Enable='1'");
$row = mysql_fetch_array($query);

    echo "<img src='$row[FileName]' width='100'>";

?>

PHP:

 

You should always enclose variables that are in quotation marks with { and }
example:

$rawr = "Hey!";
echo("Hello, the string is {$rawr}");

 

while($row = mysql_fetch_array($result));    //<-----you don't need that semicolon

PHP: