CODEWITHSUNDEEP
phpmysql
Amir
Amir

Reputation: 181

Getting a error that my image cannot be displayed because of errors

I am getting an error message that says my image cannot be displayed because of errors. I looked at previous solutions to the problem but none has created any success. I made sure there are no blank lines, i have the html and the php seperate, and i have the right information before the header. Any help would be great as I think im missing something small. Code is below.

html:

            <?php
        $images = "SELECT product_images.img_name, product_images.img_tag, product_images.img_family, product_images.img_id FROM product_images
INNER JOIN product_list
ON product_images.img_tag = product_list.product_tag && product_images.img_family = product_list.name_family";
    $img_list = mysql_query($images);
    while($rows = mysql_fetch_array($img_list))  {
        $img_name = $rows['img_name'];
        $img_tag = $rows['img_tag'];
        $img_family = $rows['img_family'];
        $img_id = $rows['img_id'];
        //echo $img_name;
?>
                <div>
                  <img src="view.php?imgid=<?php echo $img_id; ?>" />
                </div>  
<?php 
    }
?>

php:

<?php
require_once('april25_connect.php');
$dir = "product_images/";
$img = $_GET['imgid'];
$query = "SELECT * FROM product_images WHERE img_id = '$img'";
//echo $query,'<br>';
$newid = mysql_query($query);
if (!$newid) {
    echo "Query '$newid' failed <br />\n";
    echo "Error: ".mysql_error()." <br />\n";
    exit;
}else{
    while($returned_id = mysql_fetch_array($newid)) {
    $name = $returned_id['img_name'];
    $tag =  $returned_id['img_tag'];
    $family = $returned_id['img_family'];
    header ("Content-type: image/jpeg; image/gif; image/png");
    $fullpath = $dir.$name;
    echo $fullpath;
    }
}
?>

Upvotes: 0

Views: 541

Answers (1)

Xhezairi
Xhezairi

Reputation: 554

I think you're approach is wrong. Shouldn't your echo the $fullpath at the IMG html tag instead of the $img_id?

<img src="view.php?imgid=<?php echo $fullpath; ?>" />

Maybe I'm misunderstanding it, but it's hard to tell by partly seeing the content of the script.

P.S. I don't see your script returning the $img_id var for use.

I checked your product_images/ directory, it's a level up, that makes it $dir = "../product_images/";

Upvotes: 1

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