CODEWITHSUNDEEP
phpimage
Nathan
Nathan

Reputation: 319

Displaying dynamic images using php

I am trying to use php to display images based on the id of the item. This is for an ecommerce store, I am having difficulty figuring it out however. This is the php I have:

     <?php 
     $product_list = "";
     $sql = mysql_query("SELECT * FROM products ORDER BY date_added DESC");
     $productCount = mysql_num_rows($sql);
     if ($productCount > 0) {
         while($row = mysql_fetch_array($sql)){ 
             $pid = $row['id'];
             $product_name = $row["product_name"];
             $price = $row["price"];
             $date_added = strftime("%b %d, %Y", strtotime($row["date_added"]));
             $product_list .= "HERE IS WHERE THE IMAGE  &nbsp; <strong>$product_name</strong> - $$price - <em>Added $date_added</em> &nbsp; &nbsp; &nbsp; <a href='inventory_edit.php?pid=$id'>edit</a> &bull; <a href='inventory_list.php?deleteid=$id'>delete</a><br />";
         }
     } else {
         $product_list = "You have no products listed in your store yet";
     }
?>

In theory, I was trying to add, where the text HERE IS WHERE THE IMAGE is to use:

    <img src='../inventory_images/$pid.jpg' />

However this and other variations that I tried did not work. How the image is uploaded is that the variable "$pid" is the id of the product, and the .jpg ending is added, here is the code to explain that:

    $pid = mysql_insert_id();
    $newname = "$pid.jpg";
    move_uploaded_file( $_FILES['fileField']['tmp_name'], "../inventory_images/$newname");

So theoretically simply by having the product id I can get the image out using $pid... that just isn't working. Ideas? Thanks in advance!

Upvotes: 0

Views: 1442

Answers (2)

NaturalBornCamper
NaturalBornCamper

Reputation: 3886

It looks like you have an error in your output as well, could you try to replace that line:

$product_list .= "HERE IS WHERE THE IMAGE  &nbsp; <strong>$product_name</strong> - $$price - <em>Added $date_added</em> &nbsp; &nbsp; &nbsp; <a href='inventory_edit.php?pid=$id'>edit</a> &bull; <a href='inventory_list.php?deleteid=$id'>delete</a><br />";

With that line:

$product_list .= "HERE IS WHERE THE IMAGE  &nbsp; <strong>$product_name</strong> - \$$price - <em>Added $date_added</em> &nbsp; &nbsp; &nbsp; <a href='inventory_edit.php?pid=$pid'>edit</a> &bull; <a href='inventory_list.php?deleteid=$pid'>delete</a><br />";

Upvotes: 1

Luke Pittman
Luke Pittman

Reputation: 843

I see that you discovered that your image was not actually there so I thought I'd share this with you. You can check if this file exists quite easily:

if (file_exists ('../inventory_images/' . $pid . '.jpg')) {
  $product_image = '<img src="../inventory_images/' . $pid . '.jpg">';
} else {
  $product_image = '<img src="../inventory_images/image_not_found.jpg">';
}

Then simply use that $product_image variable in your $product_list where you need it. You could also modify the else statement and work in some error handling (update a field in a DB, send you an email, etc...).

Hope this helps.

Upvotes: 1

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