CODEWITHSUNDEEP
phpmysqlimage
Roger Williams
Roger Williams

Reputation: 159

PHP: How to display a default Image if the specified one doesn't exists?

I am working on a little project and I need to display the author's image for each author in my database. peep the code below:

--THE QUERY--

$getboth_sql = mysql_query(

"SELECT lyrics.lyrics_id, lyrics.lyrics_title, lyrics.lyrics_text,artists.artist_nane,artists.artist_id
FROM lyrics,artists
WHERE lyrics.artist_id = artists.artist_id
ORDER BY lyrics.lyrics_title");

while ($both = mysql_fetch_assoc($getboth_sql)) {

    $lyrics_id = $both[lyrics_id];
    $lyrics_title = $both[lyrics_title];
    $lyrics_text = $both[lyrics_text];
    $artist_name = $both[artist_nane];
    $artist_id = $both[artist_id];

    ?>
    <div class="artimg"><img src="images/artists/<?php echo $artist_name;?>.jpg"  height="50px" width="50px"/></div>        



    <?php
}//END While

The above code work but if I have not saved an image in the 'artists' directory no image show up for that artist.

<div class="artimg"><img src="images/artists/<?php echo $artist_name;?>.jpg"  height="50px" width="50px"/></div>

QUESTION: What is the BEST way to display a default image if the specified on is not found.

I have been playing around with the empty and file exists functions but i haven't gotten it right. Please Help.

PS I'm not a pro if this question seems stupid!

Upvotes: 1

Views: 18227

Answers (6)

Maizied Hasan Majumder
Maizied Hasan Majumder

Reputation: 1053

Try this way: For Laravel:

  <?php
                    $image = parse_url($user->image);
                    if(isset($image['host'])){
                         $image= $user->image;
                        }
                    else if($image==null){ 
                        $image= Request::root().'/uploads'.'/subsystems_icons/'.'democp.jpg';   
                    }
                    else {
                        $image= Request::root().'/uploads/'.$user->image;
                    }
                ?>
               
            <div class="">
                <img class="image" style="width: 100%; height: 300px; object-fit: contain ;"
                    src="{{$image}}">
            </div>

Upvotes: 0

Rishabh
Rishabh

Reputation: 586

<img id="currentPhoto" src="SomeImage.jpg" onerror="this.src='Default.jpg'" width="100" height="120">

Upvotes: 0

Zed
Zed

Reputation: 595

There are many ways to do this.

<?php

if(file_exists("images/artists/$artist_name.jpg"))
    $fileName = "$artist_name.jpg";
else
    $fileName = "default.jpg";
?>

<div class="artimg"><img src="images/artists/<?php echo $fileName;?>"  height="50px" width="50px"/></div>

Upvotes: 8

user895378
user895378

Reputation:

Disk Access = slow

You have a couple of options. One is to check that the file exists each and every time you need to display it. Unfortunately, this is less than ideal because disk reads can be a performance bottleneck. It's not a tenable solution if you're serving up hundreds of page loads a minute. In high-performance environments you want to avoid as many file stats as possible:

$img_dir = '/path/to/artist/images';
$expected_file = $img_dir . '/' . $artist_name . '.jpg';
$img = file_exists($expected_file) ? $artist_name : 'default';

Hard-code the names into your code

Another, if you have a small number of artists whose images you need to display is to hard-code in the names for the images you have:

$authors = array('Bill Shakespeare', 'Douglas Adams', 'Ralph Ellison');
$img = in_array($artist_name, $authors) ? $artist_name : 'default';

Of course, this isn't particularly helpful because then you'll need to edit your code each time your catalog of artist images changes.

The best idea ...

The preferred option in this instance would be this:

Since you're already hitting the database to get the artist records, store the relevant image filename in a database column of the artists table. That way you can avoid the superfluous disk access.

This method allows you to retrieve the filename from the database with your original query. If the field is empty you'll know to display the default image instead.

Upvotes: 1

Dr.Kameleon
Dr.Kameleon

Reputation: 22820

Code :

<?php
     if(file_exists("$your_image.jpg")) $filename = "$your_image.jpg";
     else $filename = "default.jpg";

     echo "<img src='$filename'/>";
?>

Upvotes: 0

vic
vic

Reputation: 1

try

if ( !file_exists( 'images/artists/' . $filename ) ) {
   $artist_name = 'holderthumbnail';
}

have holderthumbnail.jpg in your 'artists' directory

Upvotes: 0

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