cdjsjoe Posted March 1, 2007 Share Posted March 1, 2007 Ok, I am new to both php and mysql so please bare with me. This is what I'm wanting to do.. I made a database that will return the results in a table. What I need help with is, in the one field name "type" it uses numbers to id the type of listing it is, 01, 02 and 03 What I want it to do is if the number is 01 I want it to show picture 01, if it's o2 I want it to show picture 02 if it's 03 I want it to show picture 3. I can't get it to change this for each row that is pulled. I'm sure it's something easy but I can't get it to work. Any help at all would be GREAT!!!! Thanks in advance. -Joe Quote Link to comment https://forums.phpfreaks.com/topic/40639-php-help-with-mysql-results/ Share on other sites More sharing options...
btherl Posted March 1, 2007 Share Posted March 1, 2007 Can you post your current code? Quote Link to comment https://forums.phpfreaks.com/topic/40639-php-help-with-mysql-results/#findComment-196572 Share on other sites More sharing options...
cdjsjoe Posted March 1, 2007 Author Share Posted March 1, 2007 Can you post your current code? here is what I have.. @mysql_select_db($database) or die( "Unable to select database"); $query="SELECT * FROM equipmentlist"; $result=mysql_query($query); \ $num=mysql_numrows($result); ?> <?php echo "<TABLE BORDER=\"0\" cellpadding\"0\" cellspacing=\"0\" width=\"100%\">\n"; echo "<tr width=\"100%\" bgcolor=\"#8383D5\"><td><font face=\"Arial\">Name</font></td><td><font face=\"Arial\">Type</font></td><td><font face=\"Arial\">Serial Number</font></td><td><font face=\"Arial\">In Service Date</font></td><td><font face=\"Arial\">Last Used By</font></td></tr>\n"; for($i = 0; $i < $num; $i++) { $row = mysql_fetch_array($result); if($i % 2) { echo "<TR bgcolor=\"ffffff\">\n"; } else { echo "<TR bgcolor=\"E6E6E6\">\n"; } echo "<td>".$row['name']."</td><td>".$row['type']."</td><td>".$row['serialnum']."</td><td>".$row['inservice']."</td><td>".$row['lastused']."</td>\n"; echo "</TR>\n"; } echo "</TABLE>\n"; ?> I am very very very lost from here! I'm sure the code is sloppy so please keep in mind I am brand new. Quote Link to comment https://forums.phpfreaks.com/topic/40639-php-help-with-mysql-results/#findComment-196577 Share on other sites More sharing options...
MadTechie Posted March 1, 2007 Share Posted March 1, 2007 this is untested but may help <?php @mysql_select_db($database) or die( "Unable to select database"); $query="SELECT * FROM equipmentlist"; $result=mysql_query($query); $num= mysql_numrows($result); ?> <?php echo "<TABLE BORDER=\"0\" cellpadding\"0\" cellspacing=\"0\" width=\"100%\">\n"; echo "<tr width=\"100%\" bgcolor=\"#8383D5\"><td><font face=\"Arial\">Name</font></td><td><font face=\"Arial\">Type</font></td><td><font face=\"Arial\">Serial Number</font></td><td><font face=\"Arial\">In Service Date</font></td><td><font face=\"Arial\">Last Used By</font></td></tr>\n"; $i=true; while ($row = mysql_fetch_array($result,MYSQL_NUM)) //loop all records { if($i) { echo "<TR bgcolor=\"ffffff\">\n"; }else{ echo "<TR bgcolor=\"E6E6E6\">\n"; } echo "<td>".$row['name']."</td><td>".$row['type']."</td><td>".$row['serialnum']."</td><td>".$row['inservice']."</td><td>".$row['lastused']."</td>\n"; echo "</TR>\n"; $i = !$i; //reverses the value } echo "</TABLE>\n"; ?> Quote Link to comment https://forums.phpfreaks.com/topic/40639-php-help-with-mysql-results/#findComment-196581 Share on other sites More sharing options...
cdjsjoe Posted March 1, 2007 Author Share Posted March 1, 2007 this is untested but may help Sorry, I really didn't understand that.. what I want to do is replace the value of one recordset (I guess thats the correct name) with 1 of 3 graphics. The first graphic is of a speaker to show that row is audio related. The next graphic is of a TV to show that row is video related and the last graphic is of a light to show that row is related to Lighting. So while row 1 and row 2 come back with the type as "01" I want it to display graphic1 in place of "01". Row 3 and 5 may have the type value of "02" which would then display the TV graphic and row 4 may show the type as "03" which would then replace "03" with graphic 03. I'm sure I'm not explaning this right... Here is an example.. Name Type Serial Number In Service Date Last Used By EAW Vdocs Array 2 01 12-3456789 03152006 T.S.O. Midias 50 Ch Mixer 01 123456789 11162006 NA Sony 42" Plasma 02 123456789 06212005 Matell Toys Martin Mac 500 03 123456789 05292003 M.H.S. Sanyo PLC9000 Proj. 02 123456789 02132003 P and G Dev. What I want to do is replace all the numbers in the "type" cat. with graphics.. so all of the 01's will use one graphic while 02 and 03 use a different.. I'm sorry I'm not explaning this right.. Thanks again for all your help. Quote Link to comment https://forums.phpfreaks.com/topic/40639-php-help-with-mysql-results/#findComment-196591 Share on other sites More sharing options...
MadTechie Posted March 1, 2007 Share Posted March 1, 2007 ok this is how i would do it.. note the line $images = array("01" => "path_to_image\image1.jpg", "02" => "path_to_image\image2.jpg", "03" => "path_to_image\image3.jpg", ) #<---------NEW "01" => "path_to_image\image1.jpg" 01 = the type "path_to_image\image1.jpg" = a path to an image <?php @mysql_select_db($database) or die( "Unable to select database"); $query="SELECT * FROM equipmentlist"; $result=mysql_query($query); $num= mysql_numrows($result); ?> <?php echo "<TABLE BORDER=\"0\" cellpadding\"0\" cellspacing=\"0\" width=\"100%\">\n"; echo "<tr width=\"100%\" bgcolor=\"#8383D5\"><td><font face=\"Arial\">Name</font></td><td><font face=\"Arial\">Type</font></td><td><font face=\"Arial\">Serial Number</font></td><td><font face=\"Arial\">In Service Date</font></td><td><font face=\"Arial\">Last Used By</font></td></tr>\n"; $images = array("01" => "path_to_image\image1.jpg", "02" => "path_to_image\image2.jpg", "03" => "path_to_image\image3.jpg", ) #<---------NEW $i=true; while ($row = mysql_fetch_array($result,MYSQL_NUM)) //loop all records { if($i) { echo "<TR bgcolor=\"ffffff\">\n"; }else{ echo "<TR bgcolor=\"E6E6E6\">\n"; } echo "<td>".$row['name']."</td>"; //echo "<td>".$row['type']."</td>"; #<-----------------OLD echo "<td><img src='".$images[$row['type']]."' /></td>"; #<-----------------NEW echo "<td>".$row['serialnum']."</td>"; echo "<td>".$row['inservice']."</td>"; echo "<td>".$row['lastused']."</td>\n"; echo "</TR>\n"; $i = !$i; //reverses the value } echo "</TABLE>\n"; ?> Quote Link to comment https://forums.phpfreaks.com/topic/40639-php-help-with-mysql-results/#findComment-196593 Share on other sites More sharing options...
cdjsjoe Posted March 1, 2007 Author Share Posted March 1, 2007 ok this is how i would do it.. note the line $images = array("01" => "path_to_image\image1.jpg", "02" => "path_to_image\image2.jpg", "03" => "path_to_image\image3.jpg", ) #<---------NEW "01" => "path_to_image\image1.jpg" 01 = the type "path_to_image\image1.jpg" = a path to an image I bet that will work.. thanks so much!!!! Quote Link to comment https://forums.phpfreaks.com/topic/40639-php-help-with-mysql-results/#findComment-196595 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.