Code on retrieving image (Help)

Php_Beginner2

Hi there,

I have written a code on retrieving image from the database. But I gt a error notice from my show_image.php file.

Here's the code:

<?php 
    //connect to database 
    $connection = mysql_connect("localhost", "root", "") or die('cant make connection : ' . mysql_error()); 
    $database = mysql_select_db ("mentormenteesystem", $connection) or die ("Could not select database"); 

//save the name of image in table 
$query = mysql_query("select * from tbl_img") or die(mysql_error()); 

//retrieve all image from database and store them in a variable 
while($row = mysql_fetch_array($query)) 
{ 
    $img_name = $row['img']; 
    $image = "<img src='site_images/$img_name' /><br />"; 

    //store all images in one variable 
    $all_img = $all_img . $image; 
} 
?> 

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> 
<html xmlns="http://www.w3.org/1999/xhtml"> 
<head> 
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> 
<title>Untitled Document</title> 
</head> 

<body> 
<?php echo $all_img;?> 

</body> 
</html>

The error notice is in this line

 $all_img = $all_img . $image;

How to correct it?

Derokorian

What is the notice? If its an undefined variables notice that's because you don't initialize the $all_img variable before the loop.

bradgrafelman

What is the error?

EDIT: Woops, forgot to refresh yet again.

Php_Beginner2

Yes, it is an 'undefined variable' notice. But for my case, I am not sure how I should define $all_img as I referring to tutorials and articles while I am writing this code. I seek no detail explanation for this.

Any advice will be appreciated.

Weedpacket

Since it's being used to build up a string (by appending each image tag to it), the reasonable thing to start with is an empty string.