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Thread: show another image if image not found

  1. #1
    Join Date
    Dec 2007
    Posts
    57

    show another image if image not found

    Hi
    ive created a webpage which gets images from a database. I need to tell it to to display another image (which says no image available) if an image is not found. How would I do this????

    Your suggestions would be very much appreciated

  2. #2
    Join Date
    Jul 2007
    Posts
    357
    Can we see the code to see how you pull the images?

  3. #3
    Join Date
    Jun 2003
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    if your pulling images from the DB then it wont return any rows when looking, so use mysql_num_rows() to check the number of rows is 1 or more, otherwise redirect or load your default fall back image.

  4. #4
    Join Date
    Dec 2007
    Posts
    57

    Red face

    the code is shown below ellisgl
    PHP Code:
    <HTML>
    <HEAD>
    <TITLE>
    All Shoes
    </TITLE>
    </HEAD>

    <BODY>
    <H1>All Shoes</H1>

    <p>All shoes available for sale. You can search for certain sizes
    by using the search function.</p>

    <br>

    <TABLE CELLSPACING=1 CELLPADDING=4 bgcolor="#FFFFFF" WIDTH=666 BORDER=0>

    <?php

        
    #connect to MySQL
        
    $connpg_connect"host=pgdbs user=wwwuser dbname=dwt password=wwwuser")
                         or die( 
    "Err:Conn" );

             
        
    $query "SELECT * FROM shoes";
          
    $result pg_Exec ($conn$query);
          
          
    $rows pg_Numrows($result);

          if ((!
    $result) || ($rows 1)) {
                      echo 
    "<H1>ERROR - no rows returned</H1><P>";
                      exit;  
    //exit the script
                      
    }

         for (
    $i=0$i $rows$i++)
        {
        
    $code pg_result($result,$i,"code");        
        
    $colour pg_result($result,$i,"colour");
            
    $name pg_result($result,$i,"name");
        
    $sex pg_result($result,$i,"sex");
        
    $min_size pg_result($result,$i,"min_size");
        
    $max_size pg_result($result,$i,"max_size");
        
    $price pg_result($result,$i,"price");
        
    $image_url pg_result($result,$i,"image_url");
    ?>

        <TR>
        <TD colspan="7" align=center>
        <IMG src= "<?php echo $image_url ?>" width="100" height="100">
        </TD>
        <TD>
        No: <?php echo $code ?> , <?php echo $name ?>, Colour: <?php             echo $colour ?>, Min Size: <?php echo $min_size ?>,
            Max Size: <?php echo $max_size ?>, Price: <?php echo         $price ?>.00
        </TD>
        
    <?php
        
    }
    ?>

    </BODY>
    </HTML>
    thasnks for the advice scragar, im very new to php which is why I dont fully know how to do what you have suggested lol but I am grateful for the advice and am currently looking into how I would do what you have suggested.
    Thanks ellisgl and scragar

  5. #5
    Join Date
    Jul 2007
    Posts
    357
    if(empty($image_url)))
    {
    $image_url = 'http://www.whatever.com/defaultimage.jpg';
    }

  6. #6
    Join Date
    Jun 2003
    Location
    here
    Posts
    4,551
    try replacing:
    PHP Code:
    <IMG src= "<?php echo $image_url ?>" width="100" height="100">
    with something like:
    PHP Code:
    <IMG src= "<?php
    echo ($image_url=="")?"DEFAULT IMAGE URL GOES HERE":$image_url;
    ?>" width="100" height="100">
    you will need to fill in the default image URL and maybe change the empty quotes to something else if your DB using something other than a blank string is no URL is input.

  7. #7
    Join Date
    Dec 2007
    Posts
    57
    The code you gave works. Thanks for the help ellisgl and scragar

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