php-windows Digest 24 May 2008 13:56:12 -0000 Issue 3478
Topics (messages 28910 through 28912):
Re: Error Question!
28910 by: James Crow
28911 by: bitslayer.comcast.net
Re: WMI call to run a remote application
28912 by: Allain Lalonde
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----------------------------------------------------------------------
--- Begin Message ---
Matt,
You may want to watch out for is sql injection attacks. Just to be on
the safe side consider something like:
$projects = "select name, type, location, date_added from uploads where
user_id = '" . mysql_real_escape_string($_SESSION['user_id'], $conn) .
"'";
The reason for the error is $project_results != $projects_results.
Thanks,
James
On Fri, 2008-05-23 at 13:42 -0400, Matthew Gonzales wrote:
> 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 ---
--- Begin Message ---
Matthew,
It seems, from looking at your supplied code, that you have a variable-naming
typographic error.
The variables have to match:
$projects_results = mysql_query($projects, $conn) or die(mysql_error());
$project_data = mysql_fetch_array($project_results))
--
Jeff White is:
[EMAIL PROTECTED]
-------------- Original message --------------
From: Matthew Gonzales <[EMAIL PROTECTED]>
> 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
> $display_block = '
>
> Project Name
> Project Type
> View Project
> Project Created
>
';
>
> //Extract data to table
> while ($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 .='
>
> ' . $project_name .
> '
> ' . $project_type .
> '
> View
> class="heading">' . $project_date .
> '
>
';
> }
>
> //Close display block
> $display_block .='';
>
> 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]
> Phone: (706)542-9538
>
> --
> PHP Windows Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
--- End Message ---
--- Begin Message ---
To execute a shell statement on the server side use
http://ca3.php.net/manual/en/function.exec.php. That way you keep Web things
in PHP and Shell Script things in a more appriate language like cscript
(vbs) or batch files.
Even though PHP *can do it*, it doesn't mean it's the best language for
doing it.
My 2 cents,
Allain Lalonde
2008/5/22 John Arends <[EMAIL PROTECTED]>:
> 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
>
> --
> PHP Windows Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>
--- End Message ---
