Suchy Posted June 5, 2007 Share Posted June 5, 2007 I have a simple variable: $fulldir = 'photo_upload'; but when I moved the script that this variable is in, the adress should be $fulldir = '../photo_upload'; But this is giving me a error, what would be the correct way of specifing the directory ? Link to comment https://forums.phpfreaks.com/topic/54221-folders/ Share on other sites More sharing options...
zq29 Posted June 5, 2007 Share Posted June 5, 2007 Where is the directory in relation to the script? Link to comment https://forums.phpfreaks.com/topic/54221-folders/#findComment-268269 Share on other sites More sharing options...
saf Posted June 5, 2007 Share Posted June 5, 2007 Well this seems to work fine for me. Try using: $fulldir = './../photo_upload'; You might also want to specify your forwardslash (/) at the end of the variable, just so that you dont have to worry about it later in the code. Link to comment https://forums.phpfreaks.com/topic/54221-folders/#findComment-268360 Share on other sites More sharing options...
Suchy Posted June 5, 2007 Author Share Posted June 5, 2007 in my directory I have something like this: - index.php (webpage) - login.php (webpage) - photos.php (webpage) - photo_upload (folder) -1.jpg , 2.jpg , 3.jpg , 4.jpg... - functions (folder) - show.php , db_connect.php ... I have these variables in the script show.php in that folder "functions" and I want the $fulldir to point to the photo_upload folder Link to comment https://forums.phpfreaks.com/topic/54221-folders/#findComment-268366 Share on other sites More sharing options...
saf Posted June 5, 2007 Share Posted June 5, 2007 if you call show.php explicitly through the browser then your $fulldir has to be '../photo_upload', but if show.php is an include in index.php, login.php and/or photos.php, then your $full dir has to be 'photo_upload' Link to comment https://forums.phpfreaks.com/topic/54221-folders/#findComment-268397 Share on other sites More sharing options...
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