JTapp Posted February 18, 2008 Share Posted February 18, 2008 Problem: I need to display an image (unique to my ID field) and display it on my webpage. At this point my stomach turns when I look at my existing code (I'm soooo new at this). But I have to somehow add to my code to get this done. Any help is greatly appreciated. There are three tables, the newest one is called 'tblLodgePics'- this is where I'm storing my image data. The code below is trying to find the 'link' and display it that way. (also known as $variable20 below). I have created a bin_data to upload the images to via another PHP page, but I haven't figured that out yet... I need a script for it. It currently has the following fields in my tblLodgePics: id varchar(15), description varchar(255), bin_data LONGBLOB, filename varchar(50), filesize varchar(50), filetype varchar(50), link varchar(255), shownno integer(5), clickcount integer(5), lngLodgeID (this is my unique field that can be found in the other 2 tables) Here is my current code to view it in (unformatted) action go to: http://www.la-mason.com/search15.html and search for 'Lodge Name' = "Abbeville". I basically need to embed Abbeville's photo into this page: //query details table begins $query = mysql_query("SELECT tblLodges.strLodgeName, tblLodges.intLodgeNumber, tblLodges.strDistrictName, tblLodges.strLodgeWEB, tblLodges.strLodgeCounty, tblLodges.dtChartered, tblLodges.strLodgeAddress, tblLodges.strLodgeAddress2, tblLodges.strLodgeLocationCity, tblLodges.strLodgeLocationState, tblLodges.strLodgeLocationZip, tblLodges.strLodgeEmail, tblLodges.strLodgePhone, tblLodges.strLodgeFax, tblLodges.strDrivingDirectons, tblLodges.dtMeetingTime, tblLodges.dtMealTime, tblLodges.strFloorSchool, tblLodges.strLodgeNews, tblOfficers.strFirstName, tblOfficers.strLastName, tblLodgePics.link FROM tblLodges LEFT JOIN tblOfficers ON tblLodges.lngLodgeID = tblOfficers.lngLodgeID LEFT JOIN tblLodgePics ON tblLodgePics.lngLodgeID = tblLodges.lngLodgeID WHERE $metode LIKE '%$search%' GROUP BY tblLodges.strLodgeName LIMIT 0, 50"); while ($row = @mysql_fetch_array($query)) { $variable1=$row["strLodgeName"]; $variable2=$row["intLodgeNumber"]; $variable3=$row["strDistrictName"]; $variable4=$row["strLodgeWEB"]; $variable5=$row["strLodgeCounty"]; $variable6=$row["dtChartered"]; $variable7=$row["strLodgeAddress"]; $variable8=$row["strLodgeAddress3"]; $variable9=$row["strLodgeLocationCity"]; $variable10=$row["strLodgeLocationState"]; $variable11=$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["link"]; //table layout for results print ("<tr>"); echo '<p><h3>' . $variable1 . '</h3></p><p>' . $variable2 . '</p>'; echo '<p>' . $variable3 . '</p>'; echo "<a href=\"$variable4\">Click Here To Go To The Lodge Website</a>"; echo '<p>' . $variable5 . '</p>' . $variable6 . '</p>'; echo '<p>' . $variable7 . '</p>' . $variable8 . '</p>' . $variable9 . '</p>' . $variable10 . '</p>' . $variable11 . '</p>'; echo "<a href=mailto:\"$variable12\">Click Here To Email The Lodge</a>"; echo '</p>' . $variable13 . '</p>' . $variable14 . '</p>'; echo '<p>' . $variable15 . '</p>'; echo '<p>' . $variable16 . '</p>' . $variable17 . '</p>' . $variable18 . '</p>'; echo '<p>' . $variable19 . '</p>'; echo '<p>' . $variable20 . '</p>'; print ("</tr>"); } ?> </tr> </table> <hr width=75% align=center size=4> <p><strong>Roster of Lodge Officers </strong></p> <p> <?php //query details table begins $query = mysql_query("SELECT tblLodges.strLodgeName, tblLodges.intLodgeNumber, tblLodges.strDistrictName, tblLodges.strLodgeLocationCity, tblLodges.strLodgeLocationZip, tblLodges.strLodgeCounty, tblOfficers.strFirstName, tblOfficers.strLastName, tblOfficers.BusinessPhone, tblOfficers.PersEmail FROM tblLodges LEFT JOIN tblOfficers ON tblLodges.lngLodgeID = tblOfficers.lngLodgeID WHERE $metode LIKE '%$search%' LIMIT 0, 50"); $results=mysql_query($query); echo "<table border='1'> <tr> <th>Officer First</th> <th>Officer last</th> <th>Officer Email</th> <th>Officer Phone</th> </tr>"; // while ($row = mysql_fetch_array($query)) { $variable1=$row["strFirstName"]; $variable2=$row["strLastName"]; $variable3=$row["PersEmail"]; $variable4=$row["BusinessPhone"]; //table layout for results print ("<tr>"); echo "<td class=\"td_id\"><h3>$variable1</h3></td>\n"; echo "<td class=\"td_id\"><h3>$variable2</h3></td>\n"; echo "<td class=\"td_id\"><h3>$variable3</h3></td>\n"; echo "<td class=\"td_id\"><h3>$variable4</h3></td>\n"; echo "<td class=\"td_id\"><h3>$variable5</h3></td>\n"; print ("</tr>"); } ?> Link to comment https://forums.phpfreaks.com/topic/91627-help-my-image-wont-display-can-you-check-my-code-for-me/ Share on other sites More sharing options...
53329 Posted February 18, 2008 Share Posted February 18, 2008 Well the generated source has no image tag. That could be a part of the problem. Is there one in the PHP code somewhere? If not thats the problem, if yes the problem is around that code. Link to comment https://forums.phpfreaks.com/topic/91627-help-my-image-wont-display-can-you-check-my-code-for-me/#findComment-469310 Share on other sites More sharing options...
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