php-windows Digest 15 Feb 2006 06:26:01 -0000 Issue 2889

Topics (messages 26705 through 26708):

Problem with 'imagecreatefromjpeg'
        26705 by: Ted Lorenz
        26707 by: Armando

PHP with https and ftps streams enabled and Apache on Win 2003 server
        26706 by: dpgirago.mdanderson.org

System integration Step 2
        26708 by: dale.kwikitsolutions.co.za

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----------------------------------------------------------------------
--- Begin Message ---
I'm trying to create a thumbnail from a jpeg file but the script stops when it 
encounters the function 'imagecreatefromjpeg'. I have 5.1.2 php loaded.

Below is the additional testing code and output which help me find the function 
causing the problem. Any input would be helpful!

Thanks
Ted

CODE-----
<?php
echo "start";
$type=false;
echo "<br>line_1";
$type=getimagesize("100.jpg");
echo "<br>line_2";
if($type)
{
    echo "<br>image type=" . $type;
    echo "<br> type[0]=" . $type[0];
    echo "<br> type[1]=" . $type[1];
    echo "<br> type[2]=" . $type[2];
    echo "<br> type[3]=" . $type[3];
}
else
{
    echo "<br>unknown type";
}
$src_img = imagecreatefromjpeg("100.jpg");
echo "<br>finish";
?>

OUTPUT
start
line_1
line_2
image type=Array
type[0]=1280
type[1]=922
type[2]=2
type[3]=width="1280" height="922"



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--- End Message ---
--- Begin Message --- Do you have the php_gd2.dll module loading? In your php.ini file under the modules, uncomment the line out then give it a try. Cheers.

Armando

Ted Lorenz wrote:
I'm trying to create a thumbnail from a jpeg file but the script stops when it 
encounters the function 'imagecreatefromjpeg'. I have 5.1.2 php loaded.

Below is the additional testing code and output which help me find the function 
causing the problem. Any input would be helpful!

Thanks
Ted

CODE-----
<?php
echo "start";
$type=false;
echo "<br>line_1";
$type=getimagesize("100.jpg");
echo "<br>line_2";
if($type)
{
    echo "<br>image type=" . $type;
    echo "<br> type[0]=" . $type[0];
    echo "<br> type[1]=" . $type[1];
    echo "<br> type[2]=" . $type[2];
    echo "<br> type[3]=" . $type[3];
}
else
{
    echo "<br>unknown type";
}
$src_img = imagecreatefromjpeg("100.jpg");
echo "<br>finish";
?>

OUTPUT
start
line_1
line_2
image type=Array
type[0]=1280
type[1]=922
type[2]=2
type[3]=width="1280" height="922"



__________________________________________________________________
Switch to Netscape Internet Service.
As low as $9.95 a month -- Sign up today at http://isp.netscape.com/register

Netscape. Just the Net You Need.

New! Netscape Toolbar for Internet Explorer
Search from anywhere on the Web and block those annoying pop-ups.
Download now at http://channels.netscape.com/ns/search/install.jsp


--- End Message ---
--- Begin Message ---
Howdy,

I have PHP 4.3.7 with https and ftps streams enabled via an alternate
php4ts.dll,  Apache 1.3.31 with mod_ssl, and Openssl 0.9.7d working on Win
2000 desktop. However the same exe's don't seem to work on a Win 2003
server as I get errors that don't make any sense given my successful
experience setting this up on Win 2K. I've been going in circles for a few
days here, and now understand that since Win 2003 is built on an XP
platform, perhaps I may need different exe versions to set this up.

If anyone has gotten this to work,  I would very much appreciate some info
regarding what versions of php, apache, and openssl I need to use for Win
2003, and any installation experiences.

Please cc me if responding as I only get the digest of this list.

Thanks very much.

David

--- End Message ---
--- Begin Message ---
Hi guys,

Got LDAP working and I am able to authenticate against Active Directory.

But now, does anyone now how to retrieve the username and password that
the user uses to logon to their PC so as to now authenticate for the
application against Active Directory?

Kind Regards,
Dale

--- End Message ---

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