[PHP] Urgent Help Needed
Hi, Its been a whiling i am searching for Sync Outlook with MySQL via PHP. I want to synchronize Calendar events and Contacts from Outlook with MySQL via PHP. Can you help me with it.. a very thanks in advance! ~Vinay Yahoo! oneSearch: Finally, mobile search that gives answers, not web links. http://mobile.yahoo.com/mobileweb/onesearch?refer=1ONXIC -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Urgent Help Needed
vingupta3 wrote: Its been a whiling i am searching for Sync Outlook with MySQL via PHP. I want to synchronize Calendar events and Contacts from Outlook with MySQL via PHP. Can you help me with it.. http://php.net/com Here endeth the PHP involvement. Look up the Outlook COM objects in the MSDN documentation for help with those. a very thanks in advance! A very you're welcome on time. -Stut -- http://stut.net/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Urgent Help Needed removing \n\r
Hi all, Slight problem, I have about 200 pages of HTML stored in a mysqlDB. I had to manually edit some information on EVERY single page. I used MySQL-Front and edited these pages. All of the tags were correct img and font I have run into a problem now, that every page I have edited has somehow corrupted most/all of the font and img tags. so they look like this font face=ARIAL, HELVETICA Now when the page is read it actually prints out font face=ARIAL, HELVETICA instead of changing the font or showing the IMAGE. The font I am not so much worried about it is the image that is not showing up and the there are too many pages to go through manually. Can someone help me with a PHP script to go through the db and replace the \n\r ? Thank you, --- -Daniel Negron // \\* / _ __ ___ \Lotus Notes Admin / Developer | | |/ /| _ ) | KB Electronic, Inc. | | ' | _ \ | 12095 NW 39th Street | |_|\ \|___/ | Coral Springs, FL. 33065 \ \_\ /954.346.4900 x 122 \\ // email: [EMAIL PROTECTED] - http://www.kbelectronics.com --- -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Urgent Help Needed removing \n\r
At 23:12 14.02.2003, Daniel Negron/KBE said: [snip] have run into a problem now, that every page I have edited has somehow corrupted most/all of the font and img tags. so they look like this font face=ARIAL, HELVETICA Now when the page is read it actually prints out font face=ARIAL, HELVETICA instead of changing the font or showing the IMAGE. The font I [snip] There must be a different problem since the tag is perfectly valid, even with the spaces/newlines you are showing. If it prints our the HTML tags have a look at the source if they didn't get encoded somehow (like lt;font face=quot;ARIALquot;gt;) -- O Ernest E. Vogelsinger (\)ICQ #13394035 ^ http://www.vogelsinger.at/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] URGENT HELP NEEDED - After Upgrade to MySQL 4.0.1.2
After upgrading to MySQL 4.0.1.2 I ma getting the message : Fatal error: Call to undefined function: mysql_connect() in /home/penpals/pub/mysql.php on line 3 Please help my production server is down! -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] URGENT HELP NEEDED - After Upgrade to MySQL 4.0.1.2
This one time, at band camp, Vernon [EMAIL PROTECTED] wrote: After upgrading to MySQL 4.0.1.2 I ma getting the message : Fatal error: Call to undefined function: mysql_connect() in /home/penpals/pub/mysql.php on line 3 hmm, did you install from RPM?? Kevin -- __ (_ \ _) ) | / / _ ) / _ | / ___) / _ ) | | ( (/ / ( ( | |( (___ ( (/ / |_| \) \_||_| \) \) Kevin Waterson Port Macquarie, Australia -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] URGENT HELP NEEDED - After Upgrade to MySQL 4.0.1.2
Yes -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] URGENT HELP NEEDED - After Upgrade to MySQL 4.0.1.2
On Thursday 13 February 2003 20:12, Vernon wrote: After upgrading to MySQL 4.0.1.2 I ma getting the message : Fatal error: Call to undefined function: mysql_connect() in Search archive on the above. Please help my production server is down! If it's critical shouldn't you test all upgrades on a development server before rolling out onto the production server? -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-general -- /* There never was a good war or a bad peace. -- B. Franklin */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] URGENT HELP NEEDED - After Upgrade to MySQL 4.0.1.2
This is not helping me. I know very well that I should not and I tried the upgrade on two other machines and all went wll. My problem still stands and a search does ntohiong but tell me to check that the path is correct. If it worked before then obviously the path is fine, unless something has changed. Please I need help getting this thing up now. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] URGENT HELP NEEDED - After Upgrade to MySQL 4.0.1.2
What are apache logs telling you? Vernon wrote: After upgrading to MySQL 4.0.1.2 I ma getting the message : Fatal error: Call to undefined function: mysql_connect() in /home/penpals/pub/mysql.php on line 3 Please help my production server is down! -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] URGENT HELP NEEDED - After Upgrade to MySQL 4.0.1.2
-Original Message- If it worked before then obviously the path is fine, unless something has changed. Obviously something changed if it worked on 2 other servers. Check all your logs. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] URGENT HELP NEEDED - After Upgrade to MySQL 4.0.1.2
For some reason I do not see any error logs for today. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] URGENT HELP NEEDED - After Upgrade to MySQL 4.0.1.2
On Thursday 13 February 2003 20:42, Vernon wrote: This is not helping me. I know very well that I should not and I tried the upgrade on two other machines and all went wll. My problem still stands and a search does ntohiong but tell me to check that the path is correct. If it worked before then obviously the path is fine, unless something has changed. Please I need help getting this thing up now. archives Call to undefined function mysql_connect() -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * -- Search the list archives before you post http://marc.theaimsgroup.com/?l=php-general -- /* With clothes the new are best, with friends the old are best. */ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP] URGENT HELP NEEDED - After Upgrade to MySQL 4.0.1.2
IMPERSONATE id='Ernest P. Worrel'Hey Vern/IMPERSONATE, PHP does not think you have the MySQL module installed. none of the mysql_* commands are going to work. check the output of phpinfo() to verify. What version of PHP are you running? Since you are using RPMs the only (helpful) advice I can give is try re-installing the PHP rpm. Possibly it needs to refresh itself. (I can offer some unhelpful advice like quit using RPMs and compile from source...switch to gentoo linux instead of RedHat for a better package manager...you know, stuff like that. But that's not helpful at the moment!) :) =C= * Cal Evans * Stay plugged into your audience. * http://www.christianperformer.com -Original Message- From: Vernon [mailto:[EMAIL PROTECTED]] Sent: Thursday, February 13, 2003 6:13 AM To: [EMAIL PROTECTED] Subject: [PHP] URGENT HELP NEEDED - After Upgrade to MySQL 4.0.1.2 After upgrading to MySQL 4.0.1.2 I ma getting the message : Fatal error: Call to undefined function: mysql_connect() in /home/penpals/pub/mysql.php on line 3 Please help my production server is down! -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Urgent help needed, sound scary when anyone did that ontitle :-)
On Thu, 25 Jan 2001, Jacky@lilst wrote:
$sql1 = "insert into firsttable (firstname, lastname) values('Jack','Chan')";
$resultsql1 = mysql_query($sql1);
$sqlLastID = "select LAST_INSERT_ID() from firsttable";
$resultlast = mysql_query($sqlLastID);
$FirstLast = mysql_result($resultlast,0,0);
snip
First of all, let's rewrite your code.
?PHP
// I'm assuming these two lines are in your code, just not
// in your post. Pay special note to the $CONN variable.
$CONN = mysql_connect("host", "user", "pass");
mysql_select_db("somedatabase", $CONN);
// Notice here that I don't care about the result mysql
// sends back. On inserts, it doesn't matter. Also
// notice that I'm telling mysql which connection to use.
$sql =
"INSERT INTO firsttable (firstname,lastname)\n".
"VALUES ('Jack', 'Chan')";
mysql_query($sql, $CONN);
// Now, we'll get your last insert. Notice that I changed
// your mysql_result into a mysql_fetch_array? The code is
// just a tad more scalable that way - in case you want the
// ID and something else if you redo your code.
$sql =
"SELECT LAST_INSERT_ID() AS nLast\n".
" FROM firsttable";
$result = mysql_query($sql, $CONN);
$row = mysql_fetch_array($result);
// ALWAYS clean up after a select. Especially if you do more than
// one in a page.
mysql_free_result($result);
// Here's your last insert id.
$nLastInsert = $row["nLast"];
?
It's *very important* that you specify which connection you want mysql to
use! I can't stress this enough. If more than one connection gets opened
(you use phplib for instance, it opens its own connection), and if you
don't specify which you're using, you'll get random results, and you'll
interfere with the *other* connection. Just because PHP says the
connection is optional doesn't mean it always is. Use it anyway.
I say this because you had a later post complaining about how mysql said
you didn't select a database. Maybe you didn't.
This goes for everyone here. Not specifying which connection is in use is
*sloppy* and may lead to bugs you can't track down! At the last place I
worked, we started using phplib after everything was written, and it
*broke everything* because nobody was paying attention to the connections
they opened. We had to audit the code for days to track everything down.
Do yourself a favor *right now* and grep -r your codebase for all database
related functions. (grep -ir "mysql_" *) and FIX THEM NOW. That way,
you'll be safe knowing that it's *your* connection that's in use.
And always remember, mysql_error($CONN) is your friend.
Cheers!
--
+-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-+
| Shaun M. ThomasINN Database Programmer |
| Phone: (309) 743-0812 Fax : (309) 743-0830|
| Email: [EMAIL PROTECTED]AIM : trifthen |
| Web : www.townnews.com |
| |
| "Most of our lives are about proving something, either to |
| "ourselves or to someone else." |
| -- Anonymous |
+-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-+
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] Urgent help needed, sound scary when anyone did that on title :-)
Hi people,
I got here the syntax that is suppose to get the id from the "just inserted" record
and store it in value, did not work so far and I cannot see what is wrong in there,
can anyone give me a hint what is wrong here? ( And the reason I did not use
mysql_insert_id here is because the ID field at my tables are all BIGINT so
mysql_insert_id won't work, so I have to use LAST_INSERT_ID() instead). By the way,
the error after the page is executed keep saying that "Mysql warning : 0 is not Mysql
index" ( and point to the line "$FirstLast = mysql_result($resultlast,0,0);"). And I
did echo for the value of $FirstLast, it showed that there is no value in there.
Sniplet is like this: ( I tried to keep this down as much as try to give most detail
at the same time, so apologize for too long sniplet).
$sql1 = "insert into firsttable (firstname, lastname) values('Jack','Chan')";
$resultsql1 = mysql_query($sql1);
$sqlLastID = "select LAST_INSERT_ID() from firsttable";
$resultlast = mysql_query($sqlLastID);
$FirstLast = mysql_result($resultlast,0,0);
$sql2 = "insert into secondtable (FirsttableID,secfirstname, seclastname)
values('$FirstLast','Jacky','Chany')";
$resultsql2 = mysql_query($sql2);
$sqlLastIDsec = "select LAST_INSERT_ID() from secondtable";
$resultlast2 = mysql_query($sqlLastIDsec);
$secondLast = mysql_result($resultlast2,0,0);
$sql3 = "insert into Thirdtable (SecondTableID,FirsttableID,Thirdfirstname,
Thirdlastname) values('$secondLast','$FirstLast','Steve','Chan')";
$resultsql3 = mysql_query($sql3);
$sqlLastIDthird = "select LAST_INSERT_ID() from Thirdtable";
$resultlast3 = mysql_query($sqlLastIDthird);
$ThirdLast = mysql_result($resultlast3,0,0);
**
what have I done wrong? Please enlighten me here
Thanks
Jack
[EMAIL PROTECTED]
"There is nothing more rewarding than reaching the goal you set for yourself"
Re: [PHP] Urgent help needed, sound scary when anyone did that on title :-)
I don't think I can use mysql_insert_id() because the ID field in the tables
are all BIGINT Auto_Increment and I saw that in the manual, it said
mysql_insert_id will not work corerctly with this type of data.
I hope I am wrong though. But if it is as the manual said, what else could I
do?
Direct quote from the manul at the part about mysql_insert_id() is here:
***
mysql_insert_id() converts the return type of the native MySQL C API
function mysql_insert_id() to a type of long. If your AUTO_INCREMENT column
has a column type of BIGINT, the value returned by mysql_insert_id() will be
incorrect. Instead, use the internal MySQL SQL function LAST_INSERT_ID().
cheers
Jack
[EMAIL PROTECTED]
"There is nothing more rewarding than reaching the goal you set for
yourself"
- Original Message -
From: Joe Stump [EMAIL PROTECTED]
To: Jacky@lilst [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Wednesday, January 24, 2001 9:25 PM
Subject: Re: [PHP] Urgent help needed, sound scary when anyone did that on
title :-)
remove the result - so just type $id = mysql_insert_id() and it should
work
fine.
--Joe
On Thu, Jan 25, 2001 at 10:21:17AM -0600, Jacky@lilst wrote:
Hi people,
I got here the syntax that is suppose to get the id from the "just
inserted" record and store it in value, did not work so far and I cannot see
what is wrong in there, can anyone give me a hint what is wrong here? ( And
the reason I did not use mysql_insert_id here is because the ID field at my
tables are all BIGINT so mysql_insert_id won't work, so I have to use
LAST_INSERT_ID() instead). By the way, the error after the page is executed
keep saying that "Mysql warning : 0 is not Mysql index" ( and point to the
line "$FirstLast = mysql_result($resultlast,0,0);"). And I did echo for the
value of $FirstLast, it showed that there is no value in there.
Sniplet is like this: ( I tried to keep this down as much as try to give
most detail at the same time, so apologize for too long sniplet).
$sql1 = "insert into firsttable (firstname, lastname)
values('Jack','Chan')";
$resultsql1 = mysql_query($sql1);
$sqlLastID = "select LAST_INSERT_ID() from firsttable";
$resultlast = mysql_query($sqlLastID);
$FirstLast = mysql_result($resultlast,0,0);
$sql2 = "insert into secondtable (FirsttableID,secfirstname,
seclastname) values('$FirstLast','Jacky','Chany')";
$resultsql2 = mysql_query($sql2);
$sqlLastIDsec = "select LAST_INSERT_ID() from secondtable";
$resultlast2 = mysql_query($sqlLastIDsec);
$secondLast = mysql_result($resultlast2,0,0);
$sql3 = "insert into Thirdtable
(SecondTableID,FirsttableID,Thirdfirstname, Thirdlastname)
values('$secondLast','$FirstLast','Steve','Chan')";
$resultsql3 = mysql_query($sql3);
$sqlLastIDthird = "select LAST_INSERT_ID() from Thirdtable";
$resultlast3 = mysql_query($sqlLastIDthird);
$ThirdLast = mysql_result($resultlast3,0,0);
**
what have I done wrong? Please enlighten me here
Thanks
Jack
[EMAIL PROTECTED]
"There is nothing more rewarding than reaching the goal you set for
yourself"
--
Joe Stump, PHP Hacker
[EMAIL PROTECTED]
http://www.miester.org/
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP] Urgent help needed, sound scary when anyone did that on title :-)
you know Jacky, there's another, less cool and less reliable way to get the
last inserted id:
SELECT id FROM table ORDER BY id DESC;
it will sort them all giving you the biggest id *number* (not what mysql
keeps) and you can keep it for as many milliseconds your script will run
more...
So if nothing else works for you - try this...
(depends on how secure stable you want your application to be, of
course...)
Cheers,
Maxim Maletsky
-Original Message-
From: Jacky@lilst [mailto:[EMAIL PROTECTED]]
Sent: Friday, January 26, 2001 1:21 AM
To: [EMAIL PROTECTED]
Subject: [PHP] Urgent help needed, sound scary when anyone did that on
title :-)
Hi people,
I got here the syntax that is suppose to get the id from the "just inserted"
record and store it in value, did not work so far and I cannot see what is
wrong in there, can anyone give me a hint what is wrong here? ( And the
reason I did not use mysql_insert_id here is because the ID field at my
tables are all BIGINT so mysql_insert_id won't work, so I have to use
LAST_INSERT_ID() instead). By the way, the error after the page is executed
keep saying that "Mysql warning : 0 is not Mysql index" ( and point to the
line "$FirstLast = mysql_result($resultlast,0,0);"). And I did echo for the
value of $FirstLast, it showed that there is no value in there.
Sniplet is like this: ( I tried to keep this down as much as try to give
most detail at the same time, so apologize for too long sniplet).
$sql1 = "insert into firsttable (firstname, lastname)
values('Jack','Chan')";
$resultsql1 = mysql_query($sql1);
$sqlLastID = "select LAST_INSERT_ID() from firsttable";
$resultlast = mysql_query($sqlLastID);
$FirstLast = mysql_result($resultlast,0,0);
$sql2 = "insert into secondtable (FirsttableID,secfirstname, seclastname)
values('$FirstLast','Jacky','Chany')";
$resultsql2 = mysql_query($sql2);
$sqlLastIDsec = "select LAST_INSERT_ID() from secondtable";
$resultlast2 = mysql_query($sqlLastIDsec);
$secondLast = mysql_result($resultlast2,0,0);
$sql3 = "insert into Thirdtable (SecondTableID,FirsttableID,Thirdfirstname,
Thirdlastname) values('$secondLast','$FirstLast','Steve','Chan')";
$resultsql3 = mysql_query($sql3);
$sqlLastIDthird = "select LAST_INSERT_ID() from Thirdtable";
$resultlast3 = mysql_query($sqlLastIDthird);
$ThirdLast = mysql_result($resultlast3,0,0);
**
what have I done wrong? Please enlighten me here
Thanks
Jack
[EMAIL PROTECTED]
"There is nothing more rewarding than reaching the goal you set for
yourself"
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] Urgent help needed, sound scary when anyone did that on title :-)
I did try mysql_error() to see waht exactly is my problem and here hwat I
really need to know. My sniplet was all correct about using last_insert_id()
and the message from mysql_error(0 tells me that "no selected database".
Although I did use mysql_select_db already and I am sure there is nothing
wrong with that part of code. The real problem is when I don't think I added
user to this database properly previously so I simply went to grab another
username that provide me the access to the database, but unfortunately
different one, stupid me.
So what I cannot do here is that now I have database on mysql server call
"FreeSale" and I need to add user in it, how do I do that properly?
cheers
Jack
[EMAIL PROTECTED]
"There is nothing more rewarding than reaching the goal you set for
yourself"
- Original Message -
From: Maxim Maletsky [EMAIL PROTECTED]
To: 'Jacky@lilst' [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Wednesday, January 24, 2001 10:22 PM
Subject: RE: [PHP] Urgent help needed, sound scary when anyone did that on
title :-)
you know Jacky, there's another, less cool and less reliable way to get
the
last inserted id:
SELECT id FROM table ORDER BY id DESC;
it will sort them all giving you the biggest id *number* (not what mysql
keeps) and you can keep it for as many milliseconds your script will run
more...
So if nothing else works for you - try this...
(depends on how secure stable you want your application to be, of
course...)
Cheers,
Maxim Maletsky
-Original Message-
From: Jacky@lilst [mailto:[EMAIL PROTECTED]]
Sent: Friday, January 26, 2001 1:21 AM
To: [EMAIL PROTECTED]
Subject: [PHP] Urgent help needed, sound scary when anyone did that on
title :-)
Hi people,
I got here the syntax that is suppose to get the id from the "just
inserted"
record and store it in value, did not work so far and I cannot see what is
wrong in there, can anyone give me a hint what is wrong here? ( And the
reason I did not use mysql_insert_id here is because the ID field at my
tables are all BIGINT so mysql_insert_id won't work, so I have to use
LAST_INSERT_ID() instead). By the way, the error after the page is
executed
keep saying that "Mysql warning : 0 is not Mysql index" ( and point to the
line "$FirstLast = mysql_result($resultlast,0,0);"). And I did echo for
the
value of $FirstLast, it showed that there is no value in there.
Sniplet is like this: ( I tried to keep this down as much as try to give
most detail at the same time, so apologize for too long sniplet).
$sql1 = "insert into firsttable (firstname, lastname)
values('Jack','Chan')";
$resultsql1 = mysql_query($sql1);
$sqlLastID = "select LAST_INSERT_ID() from firsttable";
$resultlast = mysql_query($sqlLastID);
$FirstLast = mysql_result($resultlast,0,0);
$sql2 = "insert into secondtable (FirsttableID,secfirstname, seclastname)
values('$FirstLast','Jacky','Chany')";
$resultsql2 = mysql_query($sql2);
$sqlLastIDsec = "select LAST_INSERT_ID() from secondtable";
$resultlast2 = mysql_query($sqlLastIDsec);
$secondLast = mysql_result($resultlast2,0,0);
$sql3 = "insert into Thirdtable
(SecondTableID,FirsttableID,Thirdfirstname,
Thirdlastname) values('$secondLast','$FirstLast','Steve','Chan')";
$resultsql3 = mysql_query($sql3);
$sqlLastIDthird = "select LAST_INSERT_ID() from Thirdtable";
$resultlast3 = mysql_query($sqlLastIDthird);
$ThirdLast = mysql_result($resultlast3,0,0);
**
what have I done wrong? Please enlighten me here
Thanks
Jack
[EMAIL PROTECTED]
"There is nothing more rewarding than reaching the goal you set for
yourself"
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]
