On 5/5/06, James Nunnerley <[EMAIL PROTECTED]> wrote:
Doing a "readfile ($file_location)" outputs the binary... can someone point
me in the direction of being able to translate that binary into a viewable
image file?
Drop the appropriate content-type header:
header('Content-type: image/png');
James Nunnerley wrote:
I'm creating a file manager application, from which I want the user to be
able to edit/view files.
On the text side of things, it's pretty easy, however from the image side of
things, I'm not sure how to allow the user to view files outside of the web
folder.
Doing a "rea
I'm creating a file manager application, from which I want the user to be
able to edit/view files.
On the text side of things, it's pretty easy, however from the image side of
things, I'm not sure how to allow the user to view files outside of the web
folder.
Doing a "readfile ($file_location)" o
> 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 t
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 Li
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 . "" . $comment;
or, instead of using Header, crea
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 $d
7 matches
Mail list logo
