Load multiple pictures into MySQL?

[Michael martin]

I'm trying to create a picture database for a site that I'm working on and I
would like to do the following:
- (main goal) Have a php script load an entire directory of photos into my
mysql database in one swoop.
- Display thumbnails on a general viewing page.
- allow for viewing individual pictures by clicking on the thumbnails.

<?php
error_reporting(E_ALL);
$id = $_REQUEST["iid"];
$link = mysql_connect("localhost", "root", "") or die("Could not connect:
" . mysql_error());

mysql_select_db("mysql") or die(mysql_error());
$sql = "SELECT b1 FROM t1 where id in (1,2,3)";

$result = mysql_query("$sql") or die("Invalid query: " . mysql_error());
header("Content-type: image/jpeg");

while ($row = mysql_fetch_array($result))
{
$fileContent = $row['b1'];

$im = imagecreatefromstring($fileContent);
$width = imagesx($im);
$height = imagesy($im);
$imgw = 50;
$imgh = $height / $width * $imgw;
$thumb = ImageCreate($imgw,$imgh);

ImageCopyResized($thumb,$im,0,0,0,0,$imgw,$imgh,Im ageSX($im),ImageSY($im));
ImagejpeG($thumb);

imagedestroy ($im);
imagedestroy ($thumb);
mysql_close ($link);
}
?>

It is only displaying one image.

Thanks
M.

[Jerry Stuckle]

Michael martin wrote:

I'm trying to create a picture database for a site that I'm working on and I
would like to do the following:
- (main goal) Have a php script load an entire directory of photos into my
mysql database in one swoop.
- Display thumbnails on a general viewing page.
- allow for viewing individual pictures by clicking on the thumbnails.

<?php
error_reporting(E_ALL);
$id = $_REQUEST["iid"];
$link = mysql_connect("localhost", "root", "") or die("Could not connect:
" . mysql_error());

mysql_select_db("mysql") or die(mysql_error());
$sql = "SELECT b1 FROM t1 where id in (1,2,3)";

$result = mysql_query("$sql") or die("Invalid query: " . mysql_error());
header("Content-type: image/jpeg");

while ($row = mysql_fetch_array($result))
{
$fileContent = $row['b1'];

$im = imagecreatefromstring($fileContent);
$width = imagesx($im);
$height = imagesy($im);
$imgw = 50;
$imgh = $height / $width * $imgw;
$thumb = ImageCreate($imgw,$imgh);

ImageCopyResized($thumb,$im,0,0,0,0,$imgw,$imgh,Im ageSX($im),ImageSY($im));
ImagejpeG($thumb);

imagedestroy ($im);
imagedestroy ($thumb);
mysql_close ($link);
}
?>

It is only displaying one image.

Thanks
M.

You can't do it this way. The content-type header indicates a single
image will be generated.

You can split this into 2 files - the first one gets the rows and calls
the second once for each row to display the jpeg.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
js*******@attglobal.net
==================

[Michael Martin]

Not sure on how to do that as I am new to PHP.

m.

"Jerry Stuckle" <js*******@attglobal.netwrote in message
news:u8******************************@comcast.com. ..

Michael martin wrote:

>I'm trying to create a picture database for a site that I'm working on
and I would like to do the following:
- (main goal) Have a php script load an entire directory of photos into
my mysql database in one swoop.
- Display thumbnails on a general viewing page.
- allow for viewing individual pictures by clicking on the thumbnails.

<?php
error_reporting(E_ALL);
$id = $_REQUEST["iid"];
$link = mysql_connect("localhost", "root", "") or die("Could not
connect: " . mysql_error());

mysql_select_db("mysql") or die(mysql_error());
$sql = "SELECT b1 FROM t1 where id in (1,2,3)";

$result = mysql_query("$sql") or die("Invalid query: " .
mysql_error());
header("Content-type: image/jpeg");

while ($row = mysql_fetch_array($result))
{
$fileContent = $row['b1'];

$im = imagecreatefromstring($fileContent);
$width = imagesx($im);
$height = imagesy($im);
$imgw = 50;
$imgh = $height / $width * $imgw;
$thumb = ImageCreate($imgw,$imgh);
ImageCopyResized($thumb,$im,0,0,0,0,$imgw,$imgh,I mageSX($im),ImageSY($im));
ImagejpeG($thumb);

imagedestroy ($im);
imagedestroy ($thumb);
mysql_close ($link);
}
?>

It is only displaying one image.

Thanks
M.

You can't do it this way. The content-type header indicates a single
image will be generated.

You can split this into 2 files - the first one gets the rows and calls
the second once for each row to display the jpeg.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
js*******@attglobal.net
==================

[Jerry Stuckle]

Michael Martin wrote:

Not sure on how to do that as I am new to PHP.

m.

"Jerry Stuckle" <js*******@attglobal.netwrote in message
news:u8******************************@comcast.com. ..

>>Michael martin wrote:

>>>I'm trying to create a picture database for a site that I'm working on
and I would like to do the following:
- (main goal) Have a php script load an entire directory of photos into
my mysql database in one swoop.
- Display thumbnails on a general viewing page.
- allow for viewing individual pictures by clicking on the thumbnails.

<?php
error_reporting(E_ALL);
$id = $_REQUEST["iid"];
$link = mysql_connect("localhost", "root", "") or die("Could not
connect: " . mysql_error());

mysql_select_db("mysql") or die(mysql_error());
$sql = "SELECT b1 FROM t1 where id in (1,2,3)";

$result = mysql_query("$sql") or die("Invalid query: " .
mysql_error());
header("Content-type: image/jpeg");

while ($row = mysql_fetch_array($result))
{
$fileContent = $row['b1'];

$im = imagecreatefromstring($fileContent);
$width = imagesx($im);
$height = imagesy($im);
$imgw = 50;
$imgh = $height / $width * $imgw;
$thumb = ImageCreate($imgw,$imgh);
ImageCopyResized($thumb,$im,0,0,0,0,$imgw,$imgh ,ImageSX($im),ImageSY($im));
ImagejpeG($thumb);

imagedestroy ($im);
imagedestroy ($thumb);
mysql_close ($link);
}
?>

It is only displaying one image.

Thanks
M.

You can't do it this way. The content-type header indicates a single
image will be generated.

You can split this into 2 files - the first one gets the rows and calls
the second once for each row to display the jpeg.

The first file contains all the sql code. It gets the data from the
database and generates an <img...attribute in the HTML. This
attribute as the second file as its src and passes the filename of the
image being displayed to the second file.

The second file does all the work needed to create images. You probably
pulled the code you have out of a similar file. It gets I generates the
content header and all the image stuff itself. It knows which image to
use from the passed parameter.

Does this help with the process? Also, please don't top post.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
js*******@attglobal.net
==================