JTapp

sasa - YOU ROCK! Thank you very much...hours and hours of frustration to an end! Have an awesome weekend!

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>"; }

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...

I got a blank page on that one...

runnerjp - no deal.. when there IS an image, the message takes over and the image isn't displayed. Thanks for your time!

I guess another work around could be I can make an image that says "Photo not available" and call it na.jpg and whenever the photo is missing it would replace it with my na.jpg... I just have NO IDEA AT ALL on how to do that....

I took out the extra echo and still have the problem. Just so you guys know the line of code I am trying to do this with goes as follows: Photos are stored in an 'Images' file on my server. Let's just say each photo is named 'companynameID.jpg' (really called 'intLodgeNumber') in that folder. In my database there is a 'companynameID' field. So my code looks for the ID in the database and matches it to the image in the folder. Hence: echo "<img src='" . $imagesFolder . $row['intLodgeNumber'] . ".jpg' /></p>"; So I'm trying to make the entire line completely disappear when there is no image in the Images folder. It may be this is impossible to do..?

Trying to troubleshoot all of the posted suggestions... poco, was the 'echo echo' in your last line a mistake?

This is what my path looks like - it the path for lodge number 147: /lodgeimages/147.jpg

It's not 'companyID' or 'companyid'- I just used that to explain myself.. so that's not the issue. Here is the path: $file = $imagesFolder . $row['intLodgeNumber'] . '.jpg'; if (file_exists($file)){ echo "<img src='" . $imagesFolder . $row['intLodgeNumber'] . ".jpg' /></p>"; } else { echo "<img src='/Images/footerbar.GIF' /></p>"; }

Does that make sense?

ahhh! Yes, in that particular case, it is displaying just the path but I've also tested echo "<img src='" . $imagesFolder . $row['companyID'] . ".jpg' /></p>"; and it works fine So the issue seems to be my personalized 'image not available' icon shows up when in fact there is an image available.. Hope this makes sense. thanks

did'nt work...

anybody?

can you explain this more?

Well, I'm able to see the icon, but the icon shows up all of the time - regardless of if an image is there or not. Does that make sense? Basically, I want the icon to show up only if the image is missing..

Ok - its hairy.. and I had changed my field name in my original post to make it easier for people to understand: <?php $username = "username"; $password = "pswd"; $hostname = "localhost"; $dbhandle = mysql_connect($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 echo "<img src='" . $imagesFolder . $row['intLodgeNumber'] . ".jpg' /></p>"; 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>"); } } ?>

I tried both recommendations and got a blank page... In the past it seems I have been advised to use the - if (file_exists($file)){ - But I could never get it working...

Here is my line.. if the image does not exist, I'm getting the standard missing icon on my webpage. Is there a way to customize/change the missing image icon when the image isn't there? echo "<img src='" . $imagesFolder . $row['companyid'] . ".jpg' /></p>";

added "or die" after your line and still got a blank page. i'm still learning this stuf..

I was hoping for a 'catch all'. But there are two addresses out there that are dead and they are .html I'm running Apache.. Thanks!

Thanks for the advice. I tried inserting that line and I got a blank page returned..? Any ideas around what I might have done wrong?

Below is my echo line.. I'm wanting the entire line to disappear if the field value is null. Does anybody know if there is a way I can do this? Thanks in advance for your time! echo "<p<b>Meeting Time:</b> $variable16</p>";

Does anybody know how I can make all old, dead webpages forward directly to my index.hml page? Instead of getting the "Page Not Found" error message? Thanks for your time.

Thanks for all of your insight. The script has an online demo if anybody wants to look at it. It is: Here is an online demo of what the end user would see: http://keskydee.com/demos/optin/optin.php Here is an online demo of the administrative piece: http://keskydee.com/demos/optin/admin/index.php