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[SOLVED] When image is missing, how to display my own icon instead of 'red x' ?


JTapp

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ty this

<?PHP

$dbhandle = mysql_connect($hostname, $username, $password)
or die("Unable to connect to MySQL");

$selected = mysql_select_db("lodges",$dbhandle)
or die("Could not select lodges");

$id = $_GET['id'];

$query = mysql_query("SELECT a.strLodgeName, a.intLodgeNumber, a.intDistrictID, a.strLodgeWEB, a.strLodgeCounty, a.dtChartered, a.strLodgeMeetingPlace, a.strLodgeLocationCity, a.strLodgeLocationState, a.strLodgeLocationZip, a.strLodgeEmail, a.strLodgePhone, a.strLodgeFax, a.strDrivingDirectons, a.dtMeetingTime, a.strMeetingCodes, a.dtMealTime, a.strFloorSchool, a.strLodgeNews, a.strLodgePhotoPathAndName, b.strOfficerTitle, b.strFirstName, b.strLastName, b.BusinessPhone, b.PersEmail FROM tblLodges a LEFT JOIN tblOfficers b ON a.lngLodgeID = b.lngLodgeID WHERE a.intLodgeNumber=$id GROUP BY a.strLodgeName LIMIT 50")or die(mysql_error());
while ($row = @mysql_fetch_array($query))

{
$variable1=$row["strLodgeName"];
$variable2=$row["intLodgeNumber"];
$variable3=$row["intDistrictID"];
$variable4=$row["strLodgeWEB"];
$variable5=$row["strLodgeCounty"];
$variable6=$row["dtChartered"];
$variable7=$row["strLodgeMeetingPlace"];
$variable8=$row["strLodgeLocationCity"];
$variable9=$row["strLodgeLocationState"];
$variable10=$row["strLodgeLocationZip"];
$variable12=$row["strLodgeEmail"];
$variable13=$row["strLodgePhone"];
$variable14=$row["strLodgeFax"];
$variable15=$row["strDrivingDirectons"];
$variable16=$row["dtMeetingTime"];
$variable17=$row["dtMealTime"];
$variable18=$row["strFloorSchool"];
$variable19=$row["strLodgeNews"];
$variable20=$row["strLodgePhotoPathAndName"];
$variable21=$row["strMeetingCodes"];
$imagesFolder = '/lodgeimages/';

//table layout for results

echo "<center>\n";
echo "<p><H3>LODGE: $variable1</H4></p>";

//image to be displayed


$q = mysql_query($query) or die("Error running query:".mysql_error());
                                 if($q && mysql_num_rows($q) > 0)  {
                                $row = mysql_fetch_array($q);
                                if(!empty($row)) {
                                echo "<img src='" . $imagesFolder . $row['companyid'] . ".jpg' /></p>";
                                        }
                                        else {
                                    echo 'no image';
                                                }
																							                                            }
                                           

echo "<p><b>Lodge Number:</b> $variable2</p>";
echo "<p><b>District Name:</b> $variable3</p>";
echo "<a href=\"$variable4\">Click Here To Go To The Lodge Website</a>";
echo "<p><b>Lodge County:</b> $variable5</p>";
echo "<p><b>Lodge Chartered On:</b> $variable6</p>";
echo "<p><b>Lodge Address: </b> $variable7</p>";
echo "<p>$variable8, $variable9 $variable10</p>";
echo "<a href=\"mailto:$variable12\">Click Here To Email The Lodge</a>";
echo "<p><b>Lodge Phone Number: </b> $variable13</p>";
echo "<p><b>Lodge FAX Number: </b> $variable14</p>";
echo "<p><b>Lodge Driving Directions:</b><a href='http://maps.google.com/maps?saddr=&daddr=".$variable7.", ".$variable8.", ".$variable9."' target='_blank'>Link</a></p>";

echo "<p<b>Lodge Meetings:</b> $variable21</p>";
echo "<p><b>Lodge Meeting Time:</b> $variable16</p>";
echo "<p><b>Lodge Meal Time:</b> $variable17</p>";
echo "<p><b>Lodge Floor School:</b> $variable18</p>";
echo "<p><b>Lodge News:</b> $variable19</p>";

echo "</center>\n";
}
?>          
          <hr>
          <p>
            <?php
//query details table begins
$query = mysql_query("SELECT tblLodges.strLodgeName, tblLodges.intLodgeNumber, tblLodges.intDistrictID, tblLodges.strLodgeLocationCity, tblLodges.strLodgeLocationZip, tblLodges.strLodgeCounty, tblOfficers.lngLodgeID, tblOfficers.strOfficerTitle, tblOfficers.strFirstName, tblOfficers.strLastName, tblOfficers.BusinessPhone, tblOfficers.PersEmail FROM tblLodges LEFT JOIN tblOfficers ON tblLodges.lngLodgeID = tblOfficers.lngLodgeID WHERE tblLodges.intLodgeNumber=$id GROUP BY tblOfficers.lngOfficerTitleID LIMIT 0, 50")or die(mysql_error());


echo "<center>\n";
echo "<H2>Roster of Lodge Officers</H2>\n";
echo "<table border='1'>
   <tr>
<th>Lodge Number</th>
<th>Officer Title</th>
<th>Officer First</th>
<th>Officer Last</th>
<th>Officer Email</th>
<th>Officer Phone</th>

</tr>";

if (mysql_num_rows($query)) {
    while ($row = mysql_fetch_array($query)) {
        $variable1=$row["intLodgeNumber"];
        $variable2=$row["strOfficerTitle"];
        $variable3=$row["strFirstName"];
        $variable4=$row["strLastName"];
        $variable5=$row["PersEmail"];
        $variable6=$row["BusinessPhone"];
        
        //table layout for results
        
        print("<tr>");
        echo "<tr align=\"center\" bgcolor=\"#EFEFEF\">\n";
        echo "<td class=\"td_id\">$variable1</td>\n";
        echo "<td class=\"td_id\">$variable2</td>\n";
        echo "<td class=\"td_id\">$variable3</td>\n";
        echo "<td class=\"td_id\">$variable4</td>\n";
        echo "<td class=\"td_id\">$variable5</td>\n";
        echo "<td class=\"td_id\">$variable6</td>\n";
        print("</tr>");
    }
}

?>

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Photos are stored in an 'Images' folder on my server.

Let's just say each photo is named 'companyID.jpg' (really called 'intLodgeNumber') in that folder.

In my database there is a 'companyID' field - this is a primary field and can't be changed.

So my code looks for the ID in the database and matches it to the image in the folder.

Here is my code:

 

echo "<img src='" . $imagesFolder . $row['companyID'] . ".jpg' /></p>";

 

So I can't change the 'companyID' field.. there will always be a value.

if I create a 0.jpg - I've got to figure out a way to tell it to search for the 'companyID.jpg' in the Images folder and if there isn't one, to use 0.jpg.    Its starting to seem impossible...

 

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If all your images are in a central folder (and assuming your using Apache) you can add a

 

.htaccess file

 

Just have

 

ErrorDocument 404 yourimage.php

You can have the PHP file detect if the user was trying to get a page or an image, say if they where after a page redirect somewhere and if it is a image display the error image using GD if you need help with that just let me know

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This ought to be a trivial exercise. I suspect the problem lies in the confusion of exactly where the images are, relevant to the script looking for them.

 

You tell us that /lodgeimages/147.jpg is an example of one image. So is the value of the variable $imagesFolder actually /lodgeimages/ ?  If it is then pocobueno's code for file_exists is going to work.

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Yes, you are correct - this is the line that refers to it:

 

$imagesFolder = '/lodgeimages/';

 

Still, P's code isn't working..?

 

 

Here is a suggestion from another forum - but it doesn't make sense to me.. and it isn't working.. any thoughts on it?

 

if (file_exists($imagesFolder.$row['intLodgeNumber'].".jpg")) {
  echo "<img src='" . $imagesFolder . $row['intLodgeNumber'] . ".jpg' /></p>";
} else 
{ 
//just because it was on the end of your line... echo the </p> if no image
  echo "</p>";
}

 

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sasa - YOU ROCK!

Thank you very much...hours and hours of frustration to an end!

 

Have an awesome weekend!

 

change line

if (file_exists($imagesFolder.$row['intLodgeNumber'].".jpg")) {

to

if (file_exists($_SERVER['DOCUMENT_ROOT'].$imagesFolder.$row['intLodgeNumber'].".jpg")) {

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