crazy_jayne Posted September 18, 2006 Share Posted September 18, 2006 Hi I am not a PHP programmer but am tring to use PHP through the Dreamweaver interface. WHich I why I may seem rather stupid.I would like to display images and this little bit of code works fine when the images are in the same directory as the php file.[color=blue]<img src="<?php echo $row_Recordset6['Name']; ?>">[/color]However, I don't seem to be able to get my head around the code required to be able view the images if they are in another diectory.Can someone please help.Crazy Jayne Quote Link to comment Share on other sites More sharing options...
wildteen88 Posted September 18, 2006 Share Posted September 18, 2006 If you want to get image/file from another directory you'll either want to use a relative path or an absolute path[b]Relative path:[/b][code=php:0]<img src="../other_dir_name/<?php echo $row_Recordset6['Name']; ?>">[/code]The ../ goes up one level from the current working direcotry (where the script is running) and then goes into the other_dir_name directory[b]Absolute path:[/b][code=php:0]<?php$fullPath = 'http://' . $_SERVER['HOST_NAME' . '/other_dir_name/' . $row_Recordset6['Name'];echo '<img src="http://' . $fullPath . '">';?>[/code]Absolute path is where you provide the full path to the file/folder. Quote Link to comment Share on other sites More sharing options...
crazy_jayne Posted September 18, 2006 Author Share Posted September 18, 2006 Thanks for such a quick response.Up and working now - It's easy when you know how!CheersJayne Quote Link to comment Share on other sites More sharing options...
weknowtheworld Posted January 7, 2007 Share Posted January 7, 2007 Recordset6 means? Quote Link to comment Share on other sites More sharing options...
coffear Posted January 8, 2007 Share Posted January 8, 2007 it is just the name they gave to the array which can be anything containing a-z A-Z 0-9 _ as long as it starts with a letter.The content of the var seems to be an associative array of a row taken from the database. Quote Link to comment Share on other sites More sharing options...
cyrixware Posted April 17, 2007 Share Posted April 17, 2007 hehehe.. you cannot display images in this source code: <img src="<?php echo $row_Recordset6['Name']; ?>">wildteen88 is correct. And Recordset6 use for displaying the values or the data in a certain field of your dbase.. :) Quote Link to comment Share on other sites More sharing options...
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