how can I display an image when php have no image to extract fromthe server?

I've recently started learning Javascript but still not a bit expert on the subject.

So I was wondering if anybody cloud help me out.

I'm building a portfolio site using MySql databases and PHP

My database table has 3 colons: name, small_image, description.

I've set up my PHP so that it extracts the name and the small_image.

$name       = htmlentities( $row['name'] );
$image_small    = "images/" . $row['image_small'];

and it echo like this:

$name
< img src='$image_small' with='$width' height='$height' />

my only problem is that when I go to the admin page and add a new work with no image on the site it appears an empty space.

what I would like to do is to have an image that can be replacing the missing image?

is there a way to make it work? or a better, easy way to do so?

I really appreciate it .

Is not working here is the full code.

< - - - - - - - - - - - - - - - - - - - - FULL CODE - - - - - - - - - - - - - - - - >

    // Loop through all of the records returned from the query
    while( $row = mysql_fetch_array( $results ) ) {
        // collect data from each field
        $id         = $row['id开发者_如何学编程'];
        $name       = htmlentities( $row['name'] );
        $desc_short     = htmlentities( $row['desc_short'] );           

        if(isset( $row['image_small'])){
            $image_small = "images/jewelry/small/" . $row['image_small'];
        }

        else{
            $image_small = "images/blank.jpg";
        }


echo "

        <li class='column_$x $last_row_class'>
        <a href='details.php?id=$id'>
            <img src='$image_small' with='$width' height='$height' />
        </a>
        <p class='product_name'><a href='details.php?id=$id'>$name</a></p>
        <p class='product_desc'>$desc_short</p>
        $rating_div
        </li>";

---

you should display blank image(default image) if there is not exists image from database. e.g.

$name = htmlentities( $row['name'] ); 
if(isSet($row['image_small'])){
         $image_small = "images/" . $row['image_small'];
}
else{
    $image_small = "images/blank.jpg";
}

Note: You should put blank.jpg in your image folder.

I suggest you to save image name in database when image uploaded in system folder. Or you can check your image file in folder as follows :

<?php
$filename = "images/" . $row['image_small'];

if (!file_exists($filename)) {
     $image_small = "images/blank.jpg";
} 
?>

Enjoy!!!!!!!

---

Try changing the code in your first code block above to something like this:

$name         = htmlentities( $row['name'] );
$image_small  = ($row['image_small'] != '') ? "images/" . $row['image_small'] : '/path/to/default.png';

What it does is check if $row['image_small'] has a value and if so use it for $image_small, otherwise set $image_small to the path of a default small image.

---

Alternatively to the other suggestions, you can use the ISNULL() function in MySQL to return the default image name. For example, create a default.png image in your images directory, and use this in your query:

SELECT
  name,
  IF(ISNULL(small_image),'default.png',small_image) AS "small_image",
  description
FROM
  images;

Not saying it's necessarily a better answer, just an alternative.

---

A client side solution is to use "onerror" attribute of img tag. This goes like :

 <img src='$image_small' with='$width' height='$height' onerror='this.src = "/images/default.jpg" ;' />