图像现在显示从数据库调用时

SOLVED!

so ive created an image upload page which saves the image to a folder and sends the file name to the DB, ive checked to see if they are actually being added to the DB and folder which they are, but when i call the data to another page to display the images i get broken images.

below is my code to call the images, its some code ive scraped together from various tutorials as they gave me the same problem as im having now.

UPDATE: ive managed to get the images showing but now im faced with being shown the same image for each row of data called, the id and img_name and right for each row but the image is always the same as the first listed.

UPDATED CODE:

<?php
//connect to database
include ('connect.php');

//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 ($row = mysql_fetch_assoc($query))
{
    $img_name = $row['img'];
}
?>



<?php

include ('connect.php');

$img_id = mysql_query("SELECT * FROM tbl_img");

while ($row = mysql_fetch_assoc($img_id))
{

$id = $row['img_id'];

echo "
$id<br>
$img_name<br>
<img src='http://localhost/testarea/include/site_images/$img_name' />
";

echo "<br><br><br></p>
";

}

?>

If you start the path of image with / you mean an absolute path where / is the DocumentRoot folder (or the directory of virtualhost)

With src ="includes/xxx/image.png" you mean that includes is in the same folder with the php script. If it is not you can use relative path like

src="../includes/xxx/image.png" for example

<?php

include ('connect.php');

$img_id = mysql_query("SELECT * FROM tbl_img");

while ($row = mysql_fetch_assoc($img_id))
{
$img_name = $row['img'];
$id = $row['img_id'];

echo "
$id<br>
$img_name<br>
<img src='http://localhost/testarea/include/site_images/$img_name' />
";

echo "<br><br><br></p>
";

}