[PHP] Showing an image outside of the web folder

2006-05-05 Thread James Nunnerley
I'm creating a file manager application, from which I want the user to be able to edit/view files. On the text side of things, it's pretty easy, however from the image side of things, I'm not sure how to allow the user to view files outside of the web folder. Doing a readfile ($file_location)

Re: [PHP] Showing an image outside of the web folder

2006-05-05 Thread John Nichel
James Nunnerley wrote: I'm creating a file manager application, from which I want the user to be able to edit/view files. On the text side of things, it's pretty easy, however from the image side of things, I'm not sure how to allow the user to view files outside of the web folder. Doing a

Re: [PHP] Showing an image outside of the web folder

2006-05-05 Thread Rabin Vincent
On 5/5/06, James Nunnerley [EMAIL PROTECTED] wrote: Doing a readfile ($file_location) outputs the binary... can someone point me in the direction of being able to translate that binary into a viewable image file? Drop the appropriate content-type header: header('Content-type: image/png');

[PHP] showing an image with php

2005-03-11 Thread Ross Hulford
I want to asssign an image to a variable and show it if certain conditions are met. E.g If condition is met { $myimage= retrieve image (images/myimage.jpg) echo $myimage } Have not used image creation before with php just dipping my toe in the water. Thanks R. -- PHP General Mailing

Re: [PHP] showing an image with php

2005-03-11 Thread Richard Lynch
I want to asssign an image to a variable and show it if certain conditions are met. E.g If condition is met { $myimage= retrieve image (images/myimage.jpg) echo $myimage } Have not used image creation before with php just dipping my toe in the water. Not sure you need to use image

[PHP] Showing an Image

2001-07-02 Thread kachaloo
HI, I want to show an image from a database but to do that I am using : $result = @mysql_query($sql); $data = @mysql_result($result, 0, PICTURE); $type = @mysql_result($result, 0, FILETYPE); Header(Content-type: $type); echo $data;

Re: [PHP] Showing an Image

2001-07-02 Thread mike cullerton
how about something like, $result = @mysql_query($sql); $data = @mysql_result($result, 0, PICTURE); $type = @mysql_result($result, 0, FILETYPE); $comment = @mysql_result($result, 0, COMMENT); Header(Content-type: $type); echo $data . br . $comment; or, instead of using Header, create a