php-windows Digest 23 May 2008 17:45:04 -0000 Issue 3477
Topics (messages 28907 through 28909):
Re: I use Content-Disposition to open up a save as window, can I use an
absolute path?
28907 by: Stut
WMI call to run a remote application
28908 by: John Arends
Error Question!
28909 by: Matthew Gonzales
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--- Begin Message ---
On 22 May 2008, at 18:13, googling1000 wrote:
Now, I have header("Content-Disposition: filename=file1.txt");
I tried to do header("Content-Disposition: filename=C:\Program
Files\file1.txt");
But, it's not working. The save as window pops up, but it doesn't
give the
file the right filename (which is file1), and it doesn't know the type
(which is .txt)
So, we can't use an absolute path with Content-Disposition?
Shockingly not, no. I can't actually think of a legitimate reason why
you'd want to.
-Stut
--
http://stut.net/
--- End Message ---
--- Begin Message ---
I'd like to have a PHP script running on my web server (running IIS 6)
that can make a WMI call and start a specific remote process on a remote
computer.
For what its worth, the remote process I'd like to trigger is a batch file.
I've googled this extensively and can't find anything that points me in
the right direction. All the sample scripts are in vbscript, but I
assume I can probably do something like this with PHP as well since it
has COM libraries.
Any suggestions?
-John
--- End Message ---
--- Begin Message ---
Hello Everyone,
I keep getting this error: *Warning*: mysql_fetch_array(): supplied
argument is not a valid MySQL result resource in
*C:\wamp\www\login\testroom\upload.php* on line *107, *for this
particular area of code.
Code:
//Check to see what files they have uploaded in the past
$projects = "select name, type, location, date_added from uploads
where user_id = '" . $_SESSION['user_id'] . "' ";
$projects_results = mysql_query($projects, $conn) or die(mysql_error());
//Create table for projects
$display_block = '<table>
<tr>
<th>Project Name</th>
<th>Project Type</th>
<th>View Project</th>
<th>Project Created</th>
</tr>';
//Extract data to table
while ($project_data = mysql_fetch_array($project_results)) * Error
occurs here!*
{
$project_name = $project_data['name'];
$project_type = $project_data['type'];
$project_link = $project_data['loction'];
$project_date = $project_data['fmt_date_added'];
//insert project data
$display_block .='
<tr>
<td class="heading">' . $project_name .
'</td>
<td class="heading">' . $project_type .
'</td>
<td class="heading"><a href="' .
$project_link . '">View</a></td>
<td> class="heading">' . $project_date .
'</td>
</tr>';
}
//Close display block
$display_block .='</table>';
Does anyone have any idea what is going on. I have used a similar code
liek this before and it worked. I can output the mysql string to the
browser and then copy and paste it to PHPmyadmin and do an sql query and
it comes back with the correct results. I don't know what is up.
Thanks!
Matt G
**
--
Matthew Gonzales
IT Professional Specialist
Enterprise Information Technology Services
University of Georgia
Email: [EMAIL PROTECTED] <mailto:[EMAIL PROTECTED]>
Phone: (706)542-9538
--- End Message ---
