I have a script that calls up a picture stored in a database. I then have a script that displays said picture on another page. If there is no picture in the database I would like for their to be a no picture icon displayed.

New to php but I believe that some conditional statement is in order but I don't know where exactly to put it.

Here is the code to call up the picture. picscript1.php

<?php

$username = '';
$password = '';
$host = '';
$database = '';

mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

mysql_select_db($database) or die("Can not select the database: ".mysql_error());

$ID=$_GET['ID'];
$query = mysql_query("SELECT * FROM norestaurants WHERE ID='".$ID."'");
$row = mysql_fetch_array($query);
$content = $row['pic1'];


header('Content-type: image/jpg');
     echo $content;
   
?>

The code to then display the picture:

<?
$ID=$_GET['ID'];
echo "<img src='picscript1.php?ID=".$ID."' alt='$Name' width='224' height='168'>";
?>

Now I can do a conditional in picscript1.php but that would not stop the image icon from appearing. How can I do it in the second bit of code that displays the picture?

Any help will be greatly appreciated.

Thanks

you want to check to see if content is set else readfile another default image

$content = $row['pic1'];
if($content != "")
{
//show picture from database
echo $content;
}
else
{
//show generic picture
readfile(PATH_TO_GENERIC_IMAGE);
}

I changed my script to the following but I still get the broken image. The default picture I'm trying to use is stored in the same directory as the script but the image is stored in the database.

$content = $row['pic1'];
if ($content !="")
{

header('Content-type: image/jpg');
     echo $content;
}
else
{
readfile (CClogo.jpg);   
}

Is that the correct way to to the readfile?

you want to check to see if content is set else readfile another default image

$content = $row['pic1'];
if($content != "")
{
//show picture from database
echo $content;
}
else
{
//show generic picture
readfile(PATH_TO_GENERIC_IMAGE);
}

Put the image file name in quotes: "CClogo.jpg"
Make sure that the upper/lowercase spelling of the filename is correct.
Make sure that picscript1.php does not output any characters after the image data. It's good practice to delete the closing ?> php bracket from scripts which do not output HTML.

I've done a similar task for a project. here is a code which works fine. I hope it is help. You may need to tweak a bit

//The sql statement below may need to replaced with yours as it is combination of image name in db and actual file in a folder on server

$uploaddir = "images/clientsimages/thumbs/";

$get_data = "SELECT i.uploadedfile, i.clientid "
          . " FROM images AS i, ajax_client AS a "
          . " WHERE i.clientid = a.customerid "
          . " AND a.id = $_POST[sel_id]";

$get_data_res = mysql_query($get_data) or die(mysql_error() . '<br>sql : '.$get_data); 

if (mysql_num_rows($get_data_res) > 0) 
{ 
    while ($get_data_info = mysql_fetch_array($get_data_res)) 
    { 
        $uploadedfile = $uploaddir . $get_data_info["uploadedfile"]; 

        $clientID = $get_data_info["clientid"];
        
echo "<div style=\"position: absolute; width: 124px; height: 140px; z-index: 1; left: -200px; top: 8px; background-color:#000000\" id=\"layer3\">";$display_block .="<DIV style=\"position: absolute; left: 5px; top: 6px\" name=\"cutomerimage\" id=\"cutomerimage\"><image src=\"$uploadedfile\" border=\"1\"  width=\"112\" height=\"119\"/></DIV>";
echo "</div>";

    }
}
else//begin process for inserting the default image 
{
// in the line below you can create a default image to be displayed when there is none in the db
  echo"<div style=\"position: absolute; width: 124px; height: 140px; z-index: 1; left: -200px; top: 8px; background-color:#000000\" id=\"layer3\">";  //this a frame for the image
  echo "<DIV style=\"position: absolute; left: 5px; top: 6px\" name=\"cutomerimage\" id=\"cutomerimage\"><img border=\"0\" src=\"images/default_Image3.png\" width=\"112\" height=\"119\"></DIV>"; //place the default image here
echo"</div>";
}

Hope it is of some help!
Mossa

Thanks again. Works like a charm

Put the image file name in quotes: "CClogo.jpg"
Make sure that the upper/lowercase spelling of the filename is correct.
Make sure that picscript1.php does not output any characters after the image data. It's good practice to delete the closing ?> php bracket from scripts which do not output HTML.