... sorry I thought it would be to much code.
Here is the complete code :
$inputImg = ImageCreateFromJPEG($picture_location);
# old size
$srcX = imagesx($inputImg);
$srcY = imagesy($inputImg);
# new size
$ratio = ($srcY / $dstY);
$dstX = ($srcX / $ratio);
$outputImg = ImageCreateTrueColor($maxX, $dstY);
imagefill($outputImg, 0, 0, ImageColorAllocate($outputImg, 0, 0,0));
imagecopyresampled($outputImg, $inputImg, (($maxX - $dstX) / 2),0,0,0,
$dstX, $dstY, $srcX, $srcY);
#
# save image to db into blob
// this does not work (outputImg is a the colorcorrected file)
$data = addslashes($outputImg);
// this one would work
# $data = addslashes(fread(fopen($picture_location, r),
filesize($picture_location)));
$stmt =
INSERT INTO test.picture_test
(file_name, file_type, picture)
VALUES
('$name', '$picture_location_type', '$data')
;
execute_stmt($stmt, $link);
#
Rasmus Lerdorf [EMAIL PROTECTED] schrieb im Newsbeitrag
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
There is no imagejpeg() call in the code snippet you provided. And you
don't say how it is failing. You haven't provided us with enough data to
answer this question.
-Rasmus
On Sat, 29 Jun 2002, andy wrote:
Hi there,
I would like to save a jpg into a blob field of mysql. The function
underneath works fine if I read the image from the temporary destination
where php did put it after uploading.
My problem is, that I would like to do some funky stuff to the image
like
changing colors or adding watermarks. So I have several functions ahead
before I used to store them successfully to the file system. Now I would
like to store it to a blob field, but this does not work. Like I said it
works to store the temp file, but not the other one. I guess it has to
do
with something regarding the imagejpeg function.
Here is what I tryed:
#
# save image to db into blob
// this does not work (outputImg is a the colorcorrected file)
$data = addslashes($outputImg);
// this one would work
# $data = addslashes(fread(fopen($picture_location, r),
filesize($picture_location)));
$stmt =
INSERT INTO test.picture_test
(file_name, file_type, picture)
VALUES
('$name', '$picture_location_type', '$data')
;
execute_stmt($stmt, $link);
#
Maybe some of you guy has a good idea on that.
Thanx for any help,
Andy
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