snowdog Posted February 13, 2013 Share Posted February 13, 2013 I have to display 4 images in a table. The images are held in a directory structure that goes like this: franchises\franchise_id_1\lotImages\dealer_id_1\lot_id_1\wheelFL.jpeg Each 1 is actually a variable that comes from the database and depends on what dealer and franchise is logged in. Here is how I would write it: I can only assume it is not the proper way to do it. Any suggestions on the proper way to write something like this? <?php echo("<tr><td> $run['vinNumber'] </td><td> $run['carYear'] $nbsp; $run['carModel'] </td><td><img src='franchises/franchise_id_."$franchise_id"./lotImages/dealer_id_."$dealer_id"./lot_id."$run['id']".wheelFL.jpeg'></td></tr>"); ?> Link to comment https://forums.phpfreaks.com/topic/274427-help-with-concatination-and-proper-code-layout/ Share on other sites More sharing options...
scootstah Posted February 13, 2013 Share Posted February 13, 2013 Have you tried the manual? Link to comment https://forums.phpfreaks.com/topic/274427-help-with-concatination-and-proper-code-layout/#findComment-1412145 Share on other sites More sharing options...
snowdog Posted February 13, 2013 Author Share Posted February 13, 2013 yes and not understanding it Link to comment https://forums.phpfreaks.com/topic/274427-help-with-concatination-and-proper-code-layout/#findComment-1412147 Share on other sites More sharing options...
scootstah Posted February 13, 2013 Share Posted February 13, 2013 You have the concatenation backwards. It should be " . $var . " (not . "$var" . ). Link to comment https://forums.phpfreaks.com/topic/274427-help-with-concatination-and-proper-code-layout/#findComment-1412150 Share on other sites More sharing options...
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