Re: [PHP-DB] Re: php_mysql.dll (5.0.2) and libmysql.dll (4.0.20a) incompatibilities

2004-10-08 Thread Sadeq Naqashzade
Hi.
I do copy it to my Apache directory and it seems is ok now. but I can
not connect to database (php know the functions but raise error and
say my client is old)  I force use mysqli in replace of mysql
functions. and it work.

Is there any one know better solution?

Regards,
Sadeq


On Thu, 07 Oct 2004 11:47:55 -0400, Glenn Puckett [EMAIL PROTECTED] wrote:
 I just went through this myself.  PHP v5x comes with the correct
 libmysql.dll.  Copy that to System32 and it will work.  I haven't used
 it enough yet to know if it causes problems with MySQL, but PHP now works.
 
 
 
 Steve Olney wrote:
  Hi,
 
  I have an incompatibility problem when trying to use PHP 5.0.2 with an
  installation of MySQL version 4.0.20a.  The problem seems to be as follows:
 
  The php_mysql.dll file is compiled with support for two MySQL function calls
  (from the API) namely: mysql_drop_db and mysql_create_db (which are noted in
  the MySQL documentation as being deprecated for MySQL versions at least
  greater than 4.0.13 (the version I have installed at home)) but these
  functions are no longer compiled into the libmysql.dll (as of 4.x),
  obviously as these are deprecated MySQL have phased them out of the build of
  libmysql.dll.
 
  As of PHP 5.x I believe that you are required to obtain the libmysql.dll
  from the MySQL distribution as it's not in the php distribution, but this
  obviously poses as problem if you are working with a 4.x MySQL
  installation - the above incompatibility problem.
 
  Has anybody come across a solution to this problem apart from the obvious,
  recompiling the various lib's to remove this incompatibility (I'm thinking
  here that there might be compiled versions of the lib's out there that are
  'fixed') ?
 
  I'll check back soon to see if there are any solutions but in the meantime
  will attempt to 'fix' the problems with a recompile (which could take me
  some time as I need to install a compiler, the source, make the edits and
  test).  I'll post my results.
 
  Thanks,
 
  Steve


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RE: [PHP-DB] Importing Excel and Access data to MySQL

2004-10-08 Thread Bomgardner, Mark A
The best way I have found is convert the data to a comma separated value
file, use phpMyAdmin and insert the data into you database. A couple of
things that you have to watch out for is the number of records that you
are importing and that all your fields in your DB are of the correct
format and length.

Mark A. Bomgardner
Technology Specialist
KLETC


-Original Message-
From: Matthew Perry [mailto:[EMAIL PROTECTED] 
Sent: Thursday, October 07, 2004 9:46 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Importing Excel and Access data to MySQL

Simple question,
How does one import excel and access data to MySQL?
-- Matthew Perry

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[PHP-DB] Help: First Time Form with Sessions

2004-10-08 Thread Stuart Felenstein
Okay , so I'm working on the multi page form.  Insert
into DB will come at the last page.  
Couple of things I want to confirm:

1-Each page should have validiation for each field. 
Probably doesn't make any sense to wait for last page
? or maybe ?

Below is the first page of the form:
I only have the echo statements in there now to check
and make sure the variables were taking from the form
fields.  
Question though - when I go onto the next page, do I
need to echo the variables to retain the values or
just keep the variables listed ?  In other words as I
continue through the next form pages, do I need to
echo the previous ones ? 


?php
session_start();
$ListingName = $_REQUEST['ListingName'];
$AltP1  = $_REQUEST['AltP1'];
$AltP2  = $_REQUEST['AltP2'];
$Pgr = $_REQUEST['Pgr'];
$El2  = $_REQUEST['El2'];
$El3  = $_REQUEST['El3'];
?
?php
echo \$ListingName = .$ListingName.br/;
echo \$AltP1 = .$AltP1.br/;
echo \$AltP2 = .$AltP2.br/;
echo \$Pgr = .$Pgr.br/;
echo \$El2 = .$El2.br/;
echo \$El3 = .$El3.br/;
?


Thank you ,
Stuart

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RE: [PHP-DB] Help: First Time Form with Sessions

2004-10-08 Thread Bastien Koert
Hi Stu
1. why not check it on every page, then if it fails the user won't need to 
make it to the end and then have to go back to the beginning to fix 
something minor.

2. You could use hidden fields to pass the data back and for or just use a 
session, otherwise the variables expire on that page

$fieldname1 = $_REQUEST['fieldname1'];
//check field name for data correctness (regex or however you want to)
$_SESSION['user_input']['fieldname1']=$fieldname1;
...
$_SESSION['user_input']['fieldname3']=$fieldnameN;
The above is a 2-d array, that way if you are using other session vars, you 
won't get confused when deciding which one holds the data that you need.

hth
bastien
From: Stuart Felenstein [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Help: First Time Form with Sessions
Date: Fri, 8 Oct 2004 06:23:09 -0700 (PDT)
Okay , so I'm working on the multi page form.  Insert
into DB will come at the last page.
Couple of things I want to confirm:
1-Each page should have validiation for each field.
Probably doesn't make any sense to wait for last page
? or maybe ?
Below is the first page of the form:
I only have the echo statements in there now to check
and make sure the variables were taking from the form
fields.
Question though - when I go onto the next page, do I
need to echo the variables to retain the values or
just keep the variables listed ?  In other words as I
continue through the next form pages, do I need to
echo the previous ones ?
?php
session_start();
$ListingName = $_REQUEST['ListingName'];
$AltP1  = $_REQUEST['AltP1'];
$AltP2  = $_REQUEST['AltP2'];
$Pgr = $_REQUEST['Pgr'];
$El2  = $_REQUEST['El2'];
$El3  = $_REQUEST['El3'];
?
?php
echo \$ListingName = .$ListingName.br/;
echo \$AltP1 = .$AltP1.br/;
echo \$AltP2 = .$AltP2.br/;
echo \$Pgr = .$Pgr.br/;
echo \$El2 = .$El2.br/;
echo \$El3 = .$El3.br/;
?
Thank you ,
Stuart
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RE: [PHP-DB] Help: First Time Form with Sessions

2004-10-08 Thread Stuart Felenstein
See interspersed:
--- Bastien Koert [EMAIL PROTECTED] wrote:

 Hi Stu
 
 1. why not check it on every page, then if it fails
 the user won't need to 
 make it to the end and then have to go back to the
 beginning to fix 
 something minor.
 
 2. You could use hidden fields to pass the data back
 and for or just use a 
 session, otherwise the variables expire on that page

I think I'm stuck on sessions now :) .  So if there
were 3 pages and then a final 4th page where the
transactions to the database take place - without
passing the session variables around , they will still
be available on the 4th page ?

3 - Now I've run into an error: 
Warning: session_start(): Cannot send session cookie -
headers already sent by (output started at
/home/lurkkcom/public_html/Multi2Return.php:2) in
/home/lurkkcom/public_html/Multi2Return.php on line 3

Warning: session_start(): Cannot send session cache
limiter - headers already sent (output started at
/home/lurkkcom/public_html/Multi2Return.php:2) in
/home/lurkkcom/public_html/Multi2Return.php on line 3

On first page: 
I have session_start();

And on Multi2Return.php I have session_start();

I thought to continue session you need to put the
session_start function on each page.  If I remove
though , I get nothing.

Stuart

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Re: [PHP-DB] Help: First Time Form with Sessions

2004-10-08 Thread Matt M.
 Warning: session_start(): Cannot send session cookie -
 headers already sent by (output started at
 /home/lurkkcom/public_html/Multi2Return.php:2) in
 /home/lurkkcom/public_html/Multi2Return.php on line 3

read a little more about sessions.

session_start is trying to set a cookie but output has already
started, giving you this warining.

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RE: [PHP-DB] Help: First Time Form with Sessions

2004-10-08 Thread Murray @ PlanetThoughtful
Hi Stuart,

session_start() has to be the first actual codeline in your page.

So:

?
session_start()
/* rest of code */
?
html

Etc...

On the page  you're receiving the error on, move session_start() to the
first line executed in the script.

Much warmth,

Murray
http://www.planetthoughtful.org
Building a thoughtful planet,
One quirky comment at a time.


-Original Message-
From: Stuart Felenstein [mailto:[EMAIL PROTECTED] 
Sent: Saturday, 9 October 2004 1:57 AM
To: Bastien Koert; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Help: First Time Form with Sessions

See interspersed:
--- Bastien Koert [EMAIL PROTECTED] wrote:

 Hi Stu
 
 1. why not check it on every page, then if it fails
 the user won't need to 
 make it to the end and then have to go back to the
 beginning to fix 
 something minor.
 
 2. You could use hidden fields to pass the data back
 and for or just use a 
 session, otherwise the variables expire on that page

I think I'm stuck on sessions now :) .  So if there
were 3 pages and then a final 4th page where the
transactions to the database take place - without
passing the session variables around , they will still
be available on the 4th page ?

3 - Now I've run into an error: 
Warning: session_start(): Cannot send session cookie -
headers already sent by (output started at
/home/lurkkcom/public_html/Multi2Return.php:2) in
/home/lurkkcom/public_html/Multi2Return.php on line 3

Warning: session_start(): Cannot send session cache
limiter - headers already sent (output started at
/home/lurkkcom/public_html/Multi2Return.php:2) in
/home/lurkkcom/public_html/Multi2Return.php on line 3

On first page: 
I have session_start();

And on Multi2Return.php I have session_start();

I thought to continue session you need to put the
session_start function on each page.  If I remove
though , I get nothing.

Stuart

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RE: [PHP-DB] Help: First Time Form with Sessions

2004-10-08 Thread Stuart Felenstein
Yes, right before you answered this I found out about
the first line of code by googling. My apologies.

Stuart
--- Murray @ PlanetThoughtful
[EMAIL PROTECTED] wrote:

 Hi Stuart,
 
 session_start() has to be the first actual codeline
 in your page.
 
 So:
 
 ?
   session_start()
   /* rest of code */
 ?
 html
 
 Etc...
 
 On the page  you're receiving the error on, move
 session_start() to the
 first line executed in the script.
 
 Much warmth,
 
 Murray
 http://www.planetthoughtful.org
 Building a thoughtful planet,
 One quirky comment at a time.
 
 
 -Original Message-
 From: Stuart Felenstein [mailto:[EMAIL PROTECTED] 
 Sent: Saturday, 9 October 2004 1:57 AM
 To: Bastien Koert; [EMAIL PROTECTED]
 Subject: RE: [PHP-DB] Help: First Time Form with
 Sessions
 
 See interspersed:
 --- Bastien Koert [EMAIL PROTECTED] wrote:
 
  Hi Stu
  
  1. why not check it on every page, then if it
 fails
  the user won't need to 
  make it to the end and then have to go back to the
  beginning to fix 
  something minor.
  
  2. You could use hidden fields to pass the data
 back
  and for or just use a 
  session, otherwise the variables expire on that
 page
 
 I think I'm stuck on sessions now :) .  So if there
 were 3 pages and then a final 4th page where the
 transactions to the database take place - without
 passing the session variables around , they will
 still
 be available on the 4th page ?
 
 3 - Now I've run into an error: 
 Warning: session_start(): Cannot send session cookie
 -
 headers already sent by (output started at
 /home/lurkkcom/public_html/Multi2Return.php:2) in
 /home/lurkkcom/public_html/Multi2Return.php on line
 3
 
 Warning: session_start(): Cannot send session cache
 limiter - headers already sent (output started at
 /home/lurkkcom/public_html/Multi2Return.php:2) in
 /home/lurkkcom/public_html/Multi2Return.php on line
 3
 
 On first page: 
 I have session_start();
 
 And on Multi2Return.php I have session_start();
 
 I thought to continue session you need to put the
 session_start function on each page.  If I remove
 though , I get nothing.
 
 Stuart
 
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RE: [PHP-DB] Help: First Time Form with Sessions

2004-10-08 Thread Bastien Koert
see interspered again ;)

From: Stuart Felenstein [EMAIL PROTECTED]
To: Bastien Koert [EMAIL PROTECTED], [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Help: First Time Form with Sessions
Date: Fri, 8 Oct 2004 08:56:48 -0700 (PDT)
See interspersed:
--- Bastien Koert [EMAIL PROTECTED] wrote:
 Hi Stu

 1. why not check it on every page, then if it fails
 the user won't need to
 make it to the end and then have to go back to the
 beginning to fix
 something minor.

 2. You could use hidden fields to pass the data back
 and for or just use a
 session, otherwise the variables expire on that page
I think I'm stuck on sessions now :) .  So if there
were 3 pages and then a final 4th page where the
transactions to the database take place - without
passing the session variables around , they will still
be available on the 4th page ?
yes, you will have them available...to add new elements to the session, you 
will need to put session-start(); on each page THAT REQUIRES sessions

3 - Now I've run into an error:
Warning: session_start(): Cannot send session cookie -
headers already sent by (output started at
/home/lurkkcom/public_html/Multi2Return.php:2) in
/home/lurkkcom/public_html/Multi2Return.php on line 3
Warning: session_start(): Cannot send session cache
limiter - headers already sent (output started at
/home/lurkkcom/public_html/Multi2Return.php:2) in
/home/lurkkcom/public_html/Multi2Return.php on line 3

You have some additional output here...could be a blank line, a cookie or 
html output to the browser...CAN COME from an INCLUDE file, check it out...


On first page:
I have session_start();
And on Multi2Return.php I have session_start();
I thought to continue session you need to put the
session_start function on each page.  If I remove
though , I get nothing.
Yes, you do need it on each page that uses session variables

Stuart
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Re: [PHP-DB] Dynamic pull down menus with PHP/Mysql

2004-10-08 Thread Antoine
On Fri, 8 Oct 2004 01:12:06 -0400, GH [EMAIL PROTECTED] wrote:
 For a project in college that I had once did using ColdFusion and
 MSAccess (please dont kill me :)) I had the simular situation
 
 What I did was when the First Drop Down was changed... it had an
 onChange action or something of that nature (do not have my exact code
 infront of me) that went to a JS function which sent the value to
 another frame (window) you can use a hidden 1x1 iframe to conduct
 this... it then returned the values to the other page via JS I
 have to Dig for the code which unfortinately is buried at the
 moment
 
 
 
 
 On Tue, 5 Oct 2004 08:22:22 -0700, Ed Lazor [EMAIL PROTECTED] wrote:
   -Original Message-
   This may be a more javascript related topic, but it's also php/mysql.
   Apologies in advance if this is too far off topic.
  
   I'm trying to pull data from MySQL using PHP to sort the results into a
   form with a pull down menu. That works fine; I can do that.
  
   But I have a second pull down menu whose items need to display based on
   the item chosen from the first pull down menu.
  
   I can do two lists seperately, but I need the second pull down menu to be
   a result of the first, and I can't figure a way to do it in PHP without
   reloading the page, which I don't really want to do.
  
   Is this a javascript-dependent function, in that a js will have to make
   the call to the database via some sort of scripted php/mysql request? I
   really like to avoid javascript if possible, but I'm unsure there's an
   alternative.
 
  You're describing what I think is called Dynamic Options.  Doing a Google
  for javascript dynamic option or javascript dynamic select will pull-up
  a few examples.
 
  Most of these solutions will expect you to load all data into javascript
  arrays.  In other words, you don't have to reload the page, because all of
  the data is already present.
 
  This approach doesn't work well when dealing with large amounts of data.  If
  you're running into this, use javascript's window.opener feature.  It allows
  you to spawn a second window that retrieves data and sends it to the first
  window.

Cripes! People are a bit slow today... what you are describing sounds
like a hierselect to me:
In pear,...

http://pear.php.net/package/HTML_QuickForm/docs/3.2.1/HTML_QuickForm/_HTML_QuickForm-3.2.1_QuickForm_hierselect_php.html

If you want to do it yourself just look at their code and get your
inspiration from there - and yes, their solution is javascript based.
Cheers
Antoine
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RE: [PHP-DB] Help: First Time Form with Sessions

2004-10-08 Thread Stuart Felenstein
Well I guess too much success at once is a bad thing. 

For some unapparent reason, the variables here are not
carrying over.  Anyone see anything wrong here ?

Page1 (Somewhat snipped)

?php
session_start();

if ( empty( $_SESSION['l_education'] ) ) {
$_SESSION['l_education']=array();
}

if ( is_array( $_REQUEST['LurkerEdu'] ) ) {
$_SESSION['l_education'] = array_unique(
array_merge( $_SESSION['l_education'],
 $_REQUEST['LurkerEdu'] )
);
}
?
td width=387select name=LurkerEdu[]
multiple=multiple size=3

Page 2:

?php
session_start();
?

!DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0
Transitional//EN
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;

html
head
titleUntitled Document/title
meta http-equiv=Content-Type content=text/html;
charset=iso-8859-1
/head
?php
if ( is_array( $_SESSION['l_education'] ) ) {
print bYour cart:/bol\n;
foreach ( $_SESSION['l_education'] as $p1 ) {
print li$p1/li;
}
print /ol;

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RE: [PHP-DB] Help: First Time Form with Sessions

2004-10-08 Thread Stuart Felenstein
Figured it out.  Page 2 had to have the field name in
the $SESSION parameters, not the holder from the Page1
array register.

Stuart


--- Stuart Felenstein [EMAIL PROTECTED] wrote:

 Well I guess too much success at once is a bad
 thing. 
 
 For some unapparent reason, the variables here are
 not
 carrying over.  Anyone see anything wrong here ?
 
 Page1 (Somewhat snipped)
 
 ?php
 session_start();
 
 if ( empty( $_SESSION['l_education'] ) ) {
 $_SESSION['l_education']=array();
 }
 
 if ( is_array( $_REQUEST['LurkerEdu'] ) ) {
 $_SESSION['l_education'] = array_unique(
 array_merge( $_SESSION['l_education'],
  $_REQUEST['LurkerEdu'] )
 );
 }
 ?
 td width=387select name=LurkerEdu[]
 multiple=multiple size=3
 
 Page 2:
 
 ?php
 session_start();
 ?
 
 !DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0
 Transitional//EN

http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;
 
 html
 head
 titleUntitled Document/title
 meta http-equiv=Content-Type content=text/html;
 charset=iso-8859-1
 /head
 ?php
 if ( is_array( $_SESSION['l_education'] ) ) {
 print bYour cart:/bol\n;
 foreach ( $_SESSION['l_education'] as $p1 ) {
 print li$p1/li;
 }
 print /ol;
 
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[PHP-DB] -14 Days Ago

2004-10-08 Thread Cole S. Ashcraft
I am trying to see whether a data in an array pulled from a MySQL DB 
(YEARMONTHDATE) is older than 14 days ago. I am trying to do this in 
PHP. My code looks like:

if($array['due'] = $today - 14)
{
echo h5Assignment In Void:/h5brh4Assignments in the void are 
read-only;
require('footer.php');
exit;
}.

I am having problems with the math. How do I do a date subtraction 
without ending up with something like 20040994 (not a valid date)?

Thanks,
Cole
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RE: [PHP-DB] -14 Days Ago

2004-10-08 Thread Wendell Frohwein
You can trying using the MySQL DATE_ADD() , DATE_SUB , and NOW()
functions.


-wendell frohwein

-Original Message-
From: Cole S. Ashcraft [mailto:[EMAIL PROTECTED] 
Sent: Friday, October 08, 2004 11:36 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] -14 Days Ago

I am trying to see whether a data in an array pulled from a MySQL DB 
(YEARMONTHDATE) is older than 14 days ago. I am trying to do this in 
PHP. My code looks like:

if($array['due'] = $today - 14)
{
echo h5Assignment In Void:/h5brh4Assignments in the void are 
read-only;
require('footer.php');
exit;
}.

I am having problems with the math. How do I do a date subtraction 
without ending up with something like 20040994 (not a valid date)?

Thanks,
Cole

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Re: [PHP-DB] -14 Days Ago

2004-10-08 Thread Cole S. Ashcraft
I needto do it in PHP, though.
Cole
Wendell Frohwein wrote:
You can trying using the MySQL DATE_ADD() , DATE_SUB , and NOW()
functions.
-wendell frohwein
-Original Message-
From: Cole S. Ashcraft [mailto:[EMAIL PROTECTED] 
Sent: Friday, October 08, 2004 11:36 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] -14 Days Ago

I am trying to see whether a data in an array pulled from a MySQL DB 
(YEARMONTHDATE) is older than 14 days ago. I am trying to do this in 
PHP. My code looks like:

if($array['due'] = $today - 14)
{
echo h5Assignment In Void:/h5brh4Assignments in the void are 
read-only;
require('footer.php');
exit;
}.

I am having problems with the math. How do I do a date subtraction 
without ending up with something like 20040994 (not a valid date)?

Thanks,
Cole
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[PHP-DB] Re: [PEAR] Pear installation error

2004-10-08 Thread Antoine
On 8 Oct 2004 17:03:15 -, Kevin Kraeer [EMAIL PROTECTED] wrote:
 I'm a one man band working for an ad agency...good times. At any rate, a
 client has requested that form values submitted through their website be
 written to an excel document.
 
 I investigated and decided PEAR was probably my best bet (there's an excel
 writer module if I understand correctly). So, I began what has become a
 quest to successfully install PEAR to our staging server.
 
 It's a Win2K Server running IIS and PHP 4.3.6. Whoever initially installed
 PHP 4 did not install PEAR along with it for some reason. So I decided to
 run the command line installer, with the following directories set up:
 
 Installation prefix - C:\PHP\pear
 Binaries Directory - $prefix\bin
 PHP code directory - $ prefix\docs
 Documentation base directory - $prefix\docs
 Data Base Directory - $prefix\data
 Tests Directory - $prefix\tests
 php.exe Path - C:\PHP\php.exe
 
 I then run the installer, and it does fine for a while. Then, after
 'Extracting Installer' appears I get this:
 
 Warning: main(PEAR.php): failed to open stream: No such file or directory
 in C:\PHP\pear\Archive\Tar.php on line 21
 
 Fatal error: main(): Failed opening required 'PEAR.php'
 (include_path='/C:\DOCUME~1\ADMIN~1.ORB\LOCALS~1\Temp\3\gop1c.tmp') in
 C:\PHP\pear\Archive\Tar.php on line 21
 
 So it seems pretty clear that it's an include path problem to me. In my
 PHP.ini, go-pear added
 
 include_path=.;C:\php\pear
 
 So maybe I need to change that to someplace else? But I'm also not so sure
 about that forward slash in front of 'C:\DOCUME~`.'
 
 Any ideas or advice would be greatly appreciated. As I said, I'm on a solo
 mission over here - no IT guys, no programming dept. Just ME. heh.

This is probably going to get me flamed terribly, or rather completely
ignored...
If it is just you, why on earth don't you upgrade to a real operating
system? *nix + apache[1-2] + php is go beautifully together..., won't
cost you anything in licencing (assuming you don't get a proprietary
unix), and is much less likely to be troubled by viruses and
whatnot... I say this because usually the installation of these three
packages is done with about 5 clicks, and gets configured pretty much
by itself (of course, if you want security and stuff, you have to
change the config, but I am sure that is not different to IIS...).
Just a suggestion.
Cheers
Antoine

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Re: [PHP-DB] -14 Days Ago

2004-10-08 Thread metin kale
http://dev.mysql.com/doc/mysql/en/Date_and_time_functions.html
That web page has tons of details.
metin
Cole S. Ashcraft wrote:
I am trying to see whether a data in an array pulled from a MySQL DB 
(YEARMONTHDATE) is older than 14 days ago. I am trying to do this in 
PHP. My code looks like:

if($array['due'] = $today - 14)
{
echo h5Assignment In Void:/h5brh4Assignments in the void are 
read-only;
require('footer.php');
exit;
}.

I am having problems with the math. How do I do a date subtraction 
without ending up with something like 20040994 (not a valid date)?

Thanks,
Cole
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RE: [PHP-DB] -14 Days Ago

2004-10-08 Thread Hutchins, Richard
Cole,

I don't know the full context of how you need to do your date math. However,
what if, instead of selecting a bunch of data and putting it in an array in
PHP then checking to see if its older than 14 days you wrote your query to
pull only those records that are older than 14 days?

Something like (assuming MySQL):

mysql select contentID, printDate from docmeta where TO_DAYS(NOW()) -
TO_DAYS(printDate)  14;

The above query subtracts a column's value from today's date. If the
difference is greater than 14, it get pulled into the query. If you only
wanted records newer than 14 days, you would just flip the comparison
operator like so:

mysql select contentID, printDate from docmeta where TO_DAYS(NOW()) -
TO_DAYS(printDate)  14;

(I hope I got that right)

Anyway, from there, you could do whatever you need to do with the result set
programmatically inside of your PHP script. My hunch is that it might be a
little faster to do the date math in MySQL (or your db of choice) and just
work with the exact result set you want inside your script.

Hope this helps.

Rich



 -Original Message-
 From: Cole S. Ashcraft [mailto:[EMAIL PROTECTED]
 Sent: Friday, October 08, 2004 2:36 PM
 To: [EMAIL PROTECTED]
 Subject: [PHP-DB] -14 Days Ago
 
 
 I am trying to see whether a data in an array pulled from a MySQL DB 
 (YEARMONTHDATE) is older than 14 days ago. I am trying to do this in 
 PHP. My code looks like:
 
 if($array['due'] = $today - 14)
 {
 echo h5Assignment In Void:/h5brh4Assignments in the void are 
 read-only;
 require('footer.php');
 exit;
 }.
 
 I am having problems with the math. How do I do a date subtraction 
 without ending up with something like 20040994 (not a valid date)?
 
 Thanks,
 Cole
 
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 To unsubscribe, visit: http://www.php.net/unsub.php
 
 

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RE: [PHP-DB] -14 Days Ago

2004-10-08 Thread Gryffyn, Trevor
That's probably the best way to go, but if you wanted another way:

$mySqlDate = 20041008;
$month = substr($mySqlDate,4,2);
$day = substr($mySqlDate,6,2);
$year = substr($mySqlDate,0,4);

$mySqlDateSerial = mktime(0,0,0,$month,$day,$year);
$twoWeeksAgoSerial = mktime(0,0,0,date(m),date(d)-14,date(Y));

If ($mySqlDateSerial  $twoWeeksAgoSerial) {
  echo h5Assignment In Void:/h5brh4Assignments in the void are
read-only;
  require('footer.php');
  exit;
}


But I always do things the hard way.  Haha.

-TG


 -Original Message-
 From: Wendell Frohwein [mailto:[EMAIL PROTECTED] 
 Sent: Friday, October 08, 2004 2:57 PM
 To: 'Cole S. Ashcraft'; [EMAIL PROTECTED]
 Subject: RE: [PHP-DB] -14 Days Ago
 
 
 You can trying using the MySQL DATE_ADD() , DATE_SUB , and NOW()
 functions.
 
 
 -wendell frohwein
 
 -Original Message-
 From: Cole S. Ashcraft [mailto:[EMAIL PROTECTED] 
 Sent: Friday, October 08, 2004 11:36 AM
 To: [EMAIL PROTECTED]
 Subject: [PHP-DB] -14 Days Ago
 
 I am trying to see whether a data in an array pulled from a MySQL DB 
 (YEARMONTHDATE) is older than 14 days ago. I am trying to do this in 
 PHP. My code looks like:
 
 if($array['due'] = $today - 14)
 {
 echo h5Assignment In Void:/h5brh4Assignments in the void are 
 read-only;
 require('footer.php');
 exit;
 }.
 
 I am having problems with the math. How do I do a date subtraction 
 without ending up with something like 20040994 (not a valid date)?
 
 Thanks,
 Cole

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Re: [PHP-DB] -14 Days Ago

2004-10-08 Thread Cole S. Ashcraft
That would probably work, and its not too hard to do a rows returned check.
Richard Hutchins wrote:
Cole,
I don't know the full context of how you need to do your date math. However,
what if, instead of selecting a bunch of data and putting it in an array in
PHP then checking to see if its older than 14 days you wrote your query to
pull only those records that are older than 14 days?
Something like (assuming MySQL):
mysql select contentID, printDate from docmeta where TO_DAYS(NOW()) -
TO_DAYS(printDate)  14;
The above query subtracts a column's value from today's date. If the
difference is greater than 14, it get pulled into the query. If you only
wanted records newer than 14 days, you would just flip the comparison
operator like so:
mysql select contentID, printDate from docmeta where TO_DAYS(NOW()) -
TO_DAYS(printDate)  14;
(I hope I got that right)
Anyway, from there, you could do whatever you need to do with the result set
programmatically inside of your PHP script. My hunch is that it might be a
little faster to do the date math in MySQL (or your db of choice) and just
work with the exact result set you want inside your script.
Hope this helps.
Rich


-Original Message-
From: Cole S. Ashcraft [mailto:[EMAIL PROTECTED]
Sent: Friday, October 08, 2004 2:36 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] -14 Days Ago
I am trying to see whether a data in an array pulled from a MySQL DB 
(YEARMONTHDATE) is older than 14 days ago. I am trying to do this in 
PHP. My code looks like:

if($array['due'] = $today - 14)
{
echo h5Assignment In Void:/h5brh4Assignments in the void are 
read-only;
require('footer.php');
exit;
}.

I am having problems with the math. How do I do a date subtraction 
without ending up with something like 20040994 (not a valid date)?

Thanks,
Cole
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Re: [PHP-DB] -14 Days Ago

2004-10-08 Thread Janet Valade
Cole S. Ashcraft wrote:
I am trying to see whether a data in an array pulled from a MySQL DB 
(YEARMONTHDATE) is older than 14 days ago. I am trying to do this in 
PHP. My code looks like:

if($array['due'] = $today - 14)
{
echo h5Assignment In Void:/h5brh4Assignments in the void are 
read-only;
require('footer.php');
exit;
}.

I am having problems with the math. How do I do a date subtraction 
without ending up with something like 20040994 (not a valid date)?

Thanks,
Cole
You need dates in timestamp format to subtract them. You can use the 
function strtotime() to convert to timestamp. For instance, you can use:

$old_date = strtotime(2 weeks ago);
You then just need to convert the date from the database to a timestamp. 
strtotime will do something like:

$db_date = strtotime(October 10 2004);
or
$db_date = strtotime(10 October 2004);
But, I don't think it will do it your way. You may have to change the 
order of your string, as well as add spaces. Check the manual at 
http://us4.php.net/strtotime. IF the database value is in some kind of 
MySQL date format, you can perhaps retrieve it as a timestamp. MySQL has 
several DATE formats and date/time functions.

Janet
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[PHP-DB] Help: Arrays with Session Variables not happening

2004-10-08 Thread Stuart Felenstein
Back again , sorry :)

I had my form set up with 7 multi selects , on the
return page, all 7  arrays were returned correctly. 
Now that I've changed back 5 to regular menu selects
(1 selection), the 2 arrays left won't return the
values.   The menus work fine.

Here is my code:

Page 1:

?php
session_start();

if ( empty( $_SESSION['l_industry'] ) ) {
$_SESSION['l_industry']=array();
}

if ( is_array( $_REQUEST['LurkerIndustry'] ) ) {
$_SESSION['l_industry'] = array_unique(
array_merge( $_SESSION['l_industry'],
 $_REQUEST['LurkerIndustry'] )
);
}
if ( empty( $_SESSION['l_tterm'] ) ) {
$_SESSION['l_tterm']=array();
}

if ( is_array( $_REQUEST['LurkerTaxTerm'] ) ) {
$_SESSION['l_tterm'] = array_unique(
array_merge( $_SESSION['l_tterm'],
 $_REQUEST['LurkerTaxTerm'] )
);
}
?
?php
$LurkerEdu = $_REQUEST['LurkerEdu'];
$LurkerAuth  = $_REQUEST['LurkerAuth'];
$LurkerExperience  = $_REQUEST['LurkerExperience'];
$LurkerLevel  = $_REQUEST['LurkerLevel'];
$LurkerSecurity = $_REQUEST['LurkerSecurity'];
?


Page 2:

?php
session_start();
?



!DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0
Transitional//EN
http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;

body
html
head
titleUntitled Document/title
meta http-equiv=Content-Type content=text/html;
charset=iso-8859-1
/head
?php

echo \$LurkerEdu = .$LurkerEdu.br/;
echo \$LurkerAuth = .$LurkerAuth.br/;
echo \$LurkerExperience =
.$LurkerExperience.br/;
echo \$LurkerLevel = .$LurkerLevel.br/;
echo \$LurkerSecurity = .$LurkerSecurity.br/;



if ( is_array( $_SESSION['LurkerIndustry'] ) ) {
print bYour industry:/bol\n;
foreach ( $_SESSION['LurkerIndustry'] as $p2 ) {
print li$p2/li;
}
print /ol;
}

if ( is_array( $_SESSION['LurkerTaxTerm'] ) ) {
print bYour tax type:/bol\n;
foreach ( $_SESSION['LurkerTaxTerm'] as $p6 ) {
print li$p6/li;
}
print /ol;
}
?

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Re: [PHP-DB] Help: Arrays with Session Variables not happening

2004-10-08 Thread Stuart Felenstein
K ..my problem seems to be combining echo on the menu
values with the print array.  If I echo $array[0];
that seems to work fine !?


Stuart
--- Stuart Felenstein [EMAIL PROTECTED] wrote:

 Back again , sorry :)
 
 I had my form set up with 7 multi selects , on the
 return page, all 7  arrays were returned correctly. 
 Now that I've changed back 5 to regular menu selects
 (1 selection), the 2 arrays left won't return the
 values.   The menus work fine.
 
 Here is my code:
 
 Page 1:
 
 ?php
 session_start();
 
 if ( empty( $_SESSION['l_industry'] ) ) {
 $_SESSION['l_industry']=array();
 }
 
 if ( is_array( $_REQUEST['LurkerIndustry'] ) ) {
 $_SESSION['l_industry'] = array_unique(
 array_merge( $_SESSION['l_industry'],
  $_REQUEST['LurkerIndustry'] )
 );
 }
 if ( empty( $_SESSION['l_tterm'] ) ) {
 $_SESSION['l_tterm']=array();
 }
 
 if ( is_array( $_REQUEST['LurkerTaxTerm'] ) ) {
 $_SESSION['l_tterm'] = array_unique(
 array_merge( $_SESSION['l_tterm'],
  $_REQUEST['LurkerTaxTerm'] )
 );
 }
 ?
 ?php
 $LurkerEdu = $_REQUEST['LurkerEdu'];
 $LurkerAuth  = $_REQUEST['LurkerAuth'];
 $LurkerExperience  = $_REQUEST['LurkerExperience'];
 $LurkerLevel  = $_REQUEST['LurkerLevel'];
 $LurkerSecurity = $_REQUEST['LurkerSecurity'];
 ?
 
 
 Page 2:
 
 ?php
 session_start();
 ?
 
 
 
 !DOCTYPE html PUBLIC -//W3C//DTD XHTML 1.0
 Transitional//EN

http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd;
 
 body
 html
 head
 titleUntitled Document/title
 meta http-equiv=Content-Type content=text/html;
 charset=iso-8859-1
 /head
 ?php
 
 echo \$LurkerEdu = .$LurkerEdu.br/;
 echo \$LurkerAuth = .$LurkerAuth.br/;
 echo \$LurkerExperience =
 .$LurkerExperience.br/;
 echo \$LurkerLevel = .$LurkerLevel.br/;
 echo \$LurkerSecurity = .$LurkerSecurity.br/;
 
 
 
 if ( is_array( $_SESSION['LurkerIndustry'] ) ) {
 print bYour industry:/bol\n;
 foreach ( $_SESSION['LurkerIndustry'] as $p2 ) {
 print li$p2/li;
 }
 print /ol;
 }
 
 if ( is_array( $_SESSION['LurkerTaxTerm'] ) ) {
 print bYour tax type:/bol\n;
 foreach ( $_SESSION['LurkerTaxTerm'] as $p6 ) {
 print li$p6/li;
 }
 print /ol;
 }
 ?
 
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[PHP-DB] Why does this conditional run, even if not true?

2004-10-08 Thread Karen Resplendo
$c is a field in an array I have loaded in from a text file. For some reason, none of 
the conditionals branch off and I end up printing out error messages for every row in 
the text file. My brackets are balanced, I just didn't include the bottom part:

If ($c==9)
{
If($fieldarray[3]==P) 
{

 $reject=validate($fieldarray[0],$c,$connectionSDWIS, $fieldarray[$c],$c);
 
This If statement runs even if $reject = YES. Can't figure out why:
 
 
 if ($reject==NO)
  {
   
//
   //loop through the rows in the this text file checking for Original ID of Repeat
   
//
   $handle2 = fopen ($uploadfileandpath,r);
   while ($field2array = fgetcsv ($handle2, $userfile_size, ,)) 
   { 
   If($field2array[2]!=$fieldarray[$c])
{
 echo brField 2 of array 2 text row: .$field2array[2].br;
 echo brField 9 of array 1 text row: .$fieldarray[$c].br;
 $acceptOrReject = R;  
 $displayrows.=font color=redbSee below: Original Sample ID for 
Repeat does not exist : .$fieldarray[$c]./b/fontbr;
   
}
  
  } //end of while loop
  
fclose($handle2);
  
  }//end of If($reject ==No)


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Yahoo! Mail Address AutoComplete - You start. We finish.

RE: [PHP-DB] Why does this conditional run, even if not true?

2004-10-08 Thread Bastien Koert
hard to say without the code for the function...what is returned by the 
fuction determines if the code runs, right? Are you sure the code is 
returning YES? Maybe post the code for the function

bastien

From: Karen Resplendo [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Why does this conditional run, even if not true?
Date: Fri, 8 Oct 2004 18:33:02 -0700 (PDT)
$c is a field in an array I have loaded in from a text file. For some 
reason, none of the conditionals branch off and I end up printing out error 
messages for every row in the text file. My brackets are balanced, I just 
didn't include the bottom part:

If ($c==9)
{
If($fieldarray[3]==P)
{
 $reject=validate($fieldarray[0],$c,$connectionSDWIS, 
$fieldarray[$c],$c);

This If statement runs even if $reject = YES. Can't figure out why:
 if ($reject==NO)
  {
   
//
   //loop through the rows in the this text file checking for Original 
ID of Repeat
   
//
   $handle2 = fopen ($uploadfileandpath,r);
   while ($field2array = fgetcsv ($handle2, $userfile_size, ,))
   {
   If($field2array[2]!=$fieldarray[$c])
{
 echo brField 2 of array 2 text row: 
.$field2array[2].br;
 echo brField 9 of array 1 text row: 
.$fieldarray[$c].br;
 $acceptOrReject = R;
 $displayrows.=font color=redbSee below: Original Sample 
ID for Repeat does not exist : .$fieldarray[$c]./b/fontbr;

}
  } //end of while loop
fclose($handle2);
  }//end of If($reject ==No)
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RE: [PHP-DB] Re: [PEAR] Pear installation error

2004-10-08 Thread Bastien Koert
You could make life really simple and write csv document (excel loves those 
as long as you handle the data right)force the file to download with the 
additional add-type header set to vnd/ms-excel and excel opens it up 
automagically...much simpler, no?

we do this for (gasp) asp data to excel all the time
bastien

From: Antoine [EMAIL PROTECTED]
Reply-To: Antoine [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Re: [PEAR] Pear installation error
Date: Fri, 8 Oct 2004 21:01:29 +0200
On 8 Oct 2004 17:03:15 -, Kevin Kraeer [EMAIL PROTECTED] 
wrote:
 I'm a one man band working for an ad agency...good times. At any rate, a
 client has requested that form values submitted through their website be
 written to an excel document.

 I investigated and decided PEAR was probably my best bet (there's an 
excel
 writer module if I understand correctly). So, I began what has become a
 quest to successfully install PEAR to our staging server.

 It's a Win2K Server running IIS and PHP 4.3.6. Whoever initially 
installed
 PHP 4 did not install PEAR along with it for some reason. So I decided 
to
 run the command line installer, with the following directories set up:

 Installation prefix - C:\PHP\pear
 Binaries Directory - $prefix\bin
 PHP code directory - $ prefix\docs
 Documentation base directory - $prefix\docs
 Data Base Directory - $prefix\data
 Tests Directory - $prefix\tests
 php.exe Path - C:\PHP\php.exe

 I then run the installer, and it does fine for a while. Then, after
 'Extracting Installer' appears I get this:

 Warning: main(PEAR.php): failed to open stream: No such file or 
directory
 in C:\PHP\pear\Archive\Tar.php on line 21

 Fatal error: main(): Failed opening required 'PEAR.php'
 (include_path='/C:\DOCUME~1\ADMIN~1.ORB\LOCALS~1\Temp\3\gop1c.tmp') in
 C:\PHP\pear\Archive\Tar.php on line 21

 So it seems pretty clear that it's an include path problem to me. In my
 PHP.ini, go-pear added

 include_path=.;C:\php\pear

 So maybe I need to change that to someplace else? But I'm also not so 
sure
 about that forward slash in front of 'C:\DOCUME~`.'

 Any ideas or advice would be greatly appreciated. As I said, I'm on a 
solo
 mission over here - no IT guys, no programming dept. Just ME. heh.

This is probably going to get me flamed terribly, or rather completely
ignored...
If it is just you, why on earth don't you upgrade to a real operating
system? *nix + apache[1-2] + php is go beautifully together..., won't
cost you anything in licencing (assuming you don't get a proprietary
unix), and is much less likely to be troubled by viruses and
whatnot... I say this because usually the installation of these three
packages is done with about 5 clicks, and gets configured pretty much
by itself (of course, if you want security and stuff, you have to
change the config, but I am sure that is not different to IIS...).
Just a suggestion.
Cheers
Antoine
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