RE: [PHP-DB] Urent please

2002-07-07 Thread Beau Lebens

how about when u open the new window (using something like window.open(...))
you make it open a URL something like foobar.php?var1=foo and then on that
page (foobar.php) you can grab the value of "$var1" as "foo"

HTH

Beau

// -Original Message-
// From: its me [mailto:[EMAIL PROTECTED]]
// Sent: Monday, 8 July 2002 1:46 PM
// To: [EMAIL PROTECTED]
// Subject: [PHP-DB] Urent please
// 
// 
// maybe this is more a javescripy problem...but its related to 
// my php code:
// how can i open a new window with javacript and pass to it 
// php variables??
// thanks guys alot.
// 
// 
// Rehab M.Shouman
// 
// 
// 
// 
// 
// -
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[PHP-DB] Urent please

2002-07-07 Thread its me

maybe this is more a javescripy problem...but its related to my php code:
how can i open a new window with javacript and pass to it php variables??
thanks guys alot.


Rehab M.Shouman





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[PHP-DB] Oracle IDLE_TIME Profile Parameter & plogon

2002-07-07 Thread Steve Farmer

HI all,

I was wondering if anyone had experimented with the Oracle Profile 
parameters such as IDLE_TIME, CONNECT_TIME or SESSIONS_PER_USER in 
order to control the number of connections apache/php makes to the 
Oracle DB when using plogon?

I was thinking of experimenting with this and wondered if anyone had 
any pointers?

Regards
Steve Farmer
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RE: [PHP-DB] SQL prob (DATE_FORMAT)

2002-07-07 Thread Russ

George:

Thanks for that. That's pretty much what I realised anyway, while
composing the email. But I thought I'd send it off anyway!

Thanks a lot.
Russ :-)

-Original Message-
From: George Pitcher [mailto:[EMAIL PROTECTED]]
Sent: Friday, July 05, 2002 2:52 PM
To: PHP DB Mailing List (E-mail)
Subject: Re: [PHP-DB] SQL prob (DATE_FORMAT)


Russ,

I'm a bit of a novice, but I would have thought that the DATE_FORMAT
function would only work if the field had been defined as a DATE field
type.

If you are storing the date as an integer eg: 20020705 (today), why not
format your current date as a variable based on the same method? That
should
produce the same results. You only need to format the output but you can
do
that with PHP.

HTH

George
- Original Message -
From: "Russ" <[EMAIL PROTECTED]>
To: "PHP DB Mailing List (E-mail)" <[EMAIL PROTECTED]>
Sent: Friday, July 05, 2002 7:33 AM
Subject: [PHP-DB] SQL prob (DATE_FORMAT)


Hi all:

Thanks to Jason for his pointers - whatever the reason though, we need
to use INT(10) for our CMS to work! ;-) Anyways...

Using our MySQL INT(10) field the following query isn't behaving itself:

$event_sql = "SELECT *,DATE_FORMAT('EventDate','%Y-%m-%d') FROM events
WHERE EventDate>=CURDATE() ORDER BY EventDate DESC LIMIT 3";

This query should only return those event entries of today's date and
any future date right?

The MySQL manual points out that: CURDATE() returns dates thus:
mysql> SELECT CURDATE(); ->
 '1997-12-15'  (i.e.: with hyphens between the entities in the same
format as my DATE_FORMAT() arg's specify above)
Can anyone point out why the results of the queries continue to output
entries whose dates are prior to the value of CURDATE()? Do these
functions depend upon DATETIME, DATE, TITMESTAMP (etc) MySQL field types
or what??
Cheers all.
Russ


Mr Russ Michell
Web Applications Developer

Itomic.com
Email: [EMAIL PROTECTED]
Tel: +61 (0)8 9321 3844
Fax: +61 (0)8 6210 1364
Post: PO Box 228, Innaloo, WA 6918, Australia
Street: Suite 24, 158 William St, Perth, WA 6000, Australia

"No proof of existence is not proof of non-existence."
(Physicist: Stanton T. Friedman on Debunking Ufology)


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RE: [PHP-DB] disable backbutton

2002-07-07 Thread Beau Lebens

sander,
this isn't so much a problem with people using the back button or whatever,
but if you set all your pages not to cache

header("Expires: 0");
header("Cache-Control: no-cache");

will do that, then also require a cookie once they are logged in. during the
logout process (and have the cookie expire in 10 minutes or something) you
kill the cookie/session then if someone tries to access the secured pages,
they don't have a cookie and are denied access.

HTH

Beau

// -Original Message-
// From: Sander Peters [mailto:[EMAIL PROTECTED]]
// Sent: Monday, 8 July 2002 4:28 AM
// To: [EMAIL PROTECTED]
// Subject: [PHP-DB] disable backbutton
// 
// 
// Hello,
// 
// I would like to make a page expire or disable the backbutton 
// (clear the
// history).
// Why? I'm developping a database-related programm in 
// php/mysql and people can
// log in and out, but nobody should be able to use the 
// backbutton to be in
// the system again as the user that just logged of.
// 
// Can somebody help me?
// Thanks in advance!
// 
// Sander Peters
// 
// 
// 
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// PHP Database Mailing List (http://www.php.net/)
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RE: [PHP-DB] disable backbutton

2002-07-07 Thread Peter Lovatt

Hi

if you add




to your header that should stop the browser caching and if you do all your
links using a form with the method set to 'post' that should stop people
using the back button

HTH

Peter

---
Excellence in internet and open source software
---
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[EMAIL PROTECTED]
tel. 0121-242-1473
---

> -Original Message-
> From: Sander Peters [mailto:[EMAIL PROTECTED]]
> Sent: 07 July 2002 21:28
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] disable backbutton
>
>
> Hello,
>
> I would like to make a page expire or disable the backbutton (clear the
> history).
> Why? I'm developping a database-related programm in php/mysql and
> people can
> log in and out, but nobody should be able to use the backbutton to be in
> the system again as the user that just logged of.
>
> Can somebody help me?
> Thanks in advance!
>
> Sander Peters
>
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>


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[Fwd: Re: [PHP-DB] Having more problems]

2002-07-07 Thread John Coder




--- Begin Message ---

On Sun, 2002-07-07 at 01:07, Shiloh Madsen wrote:
> The newbie is still having troubles heh. Maybe some kind soul can tell 
> me what im doing wrong this time. This is the code for a page I am 
> working on. When I try to bring up the page in a browser, I just get a white page, 
>instead of having the HTML display. Anyone able to tell me why?
> 
> 
> 
> 
> 
> 
> 
> Login Page
> 
>  $dbhost = "127.0.0.1";
> $dbuser = "root";
> $db = "LoginInfo";
> 
> $LoginDB=@mysql_connect($dbhost, $dbuser, $dbpass);
your @ supresses error messages from php
> if (! $LoginDB) {
try 
if (! isset($LoginDB))
die(mysql_error());
instead
> print "Unable to connect to the database server at this time.";
> exit();
> } else {
> mysql_select_db("GameDB",$LoginDB);
> if (! @mysql_select_db("GameDB") )
same as above your @ supresses error messages from php
again try 
$ConnDB=mysql_select_db("GameDB")
if(! isset(($ConnDB))
die(mysql_error());
> print "Unable to locate the Game Database.";
> exit();
> }
> ?>
> 
> body { color: white; background: black; }
> 
>  alt="D&D Resource Page" border="0">"
> 
> 
> 
> Login: 
> Password: 
>  
> 
> 
> 
>  }
> ?>
> 
> 
> 
> 
> 
> -- 
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 



--- End Message ---

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[PHP-DB] disable backbutton

2002-07-07 Thread Sander Peters

Hello,

I would like to make a page expire or disable the backbutton (clear the
history).
Why? I'm developping a database-related programm in php/mysql and people can
log in and out, but nobody should be able to use the backbutton to be in
the system again as the user that just logged of.

Can somebody help me?
Thanks in advance!

Sander Peters



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RE: [PHP-DB] concatenating (was: RE: [PHP-DB] Retrieving/Displaying hyperlinked images with PHP)

2002-07-07 Thread César Aracena

Well, the string concatenation can be used to force PHP read and print a
variable within a echoed or printed string. As far as I know, it should
be used to make PHP handle a function inside a string. i.e.

echo "your total is:". total_function();

In your example, you might use as well:

echo " -Original Message-
> From: Jen Swofford [mailto:[EMAIL PROTECTED]]
> Sent: Sunday, July 07, 2002 2:31 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] concatenating (was: RE: [PHP-DB]
Retrieving/Displaying
> hyperlinked images with PHP)
> 
> Can somebody explain to me the rule on concatenating?  So far, I
haven't
> had
> to use those mysterious-looking dots.  ;)  My code goes like this:
> 
> echo " border=0>";
> 
> Does it work because I'm lucky?  Honestly I haven't been impressed
with
> the
> documentation I've found online.
> 
> Jen Swofford
> [EMAIL PROTECTED]
> 
> > [snip]
> > // Then show image and/or details
> > echo "";
> > [snip]
> 
> 
> 
> --
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[PHP-DB] Re: Re:dBASE connectivity

2002-07-07 Thread labhras sammin

Thanks again Miles

I tried all the numerous permutations for assigning the value to the record
but I couldnt get it to work

However I took your advice on ODBC and got it to work!!!
The only
problem I have is entering a date through a SQL insert command.
I am using the following sql string (for may 21 2002) which works fine for
strings, booleans
and double data types

insert into M_EVENTS (FROM) values ('20020521')


I have tried the following formats with and without quotes to no avail:

21/05/2002
05/21/2002
21052002
20022105
20020521

This is the error I am getting:
SQL error: [Microsoft][ODBC dBase Driver] Syntax error in INSERT INTO
statement., SQL state 37000 in SQLExecDirect

nearly there!
I would appreciate any suggestions
thanks
Labhras



"Miles Thompson" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Labrhas,
>
> If you echo $new_record, and $newrecord["name"], what displays? I'm really
> trying to confirm that $new_record is an array, and that
> $new_record["name"] contains what you expect. Is it possible that you have
> to assign like this:
>  $new_record['name'] = "'xx'";
> which would put a quoted string in the array. I know this seems weird, but
> I started with dBase II and I've been working with FoxPro since 1.01 and
> every so often you get into weird scenarios where the dBase/FP engine
> thinks you're passing a var not a literal. (This is also "grasping at
> staws" territory.)
>
> I want to reiterate that the dbase functions are not recommended for
> production work. See the note under dbase. AS an alternative, what about
> using odbc? You may be able to run an insert query.
>
> Miles
>
> At 03:17 PM 7/4/2002 +0100, labhras sammin wrote:
> >thanks for your help
> >I have tested that everything works, valid handles, returned method
values,
> >etc.
> >Line 23 is the dbase_add_record line. Do you know of anywhere there may
be
> >an example
> >thanks Labhras
> >
> >
> >"Miles Thompson" <[EMAIL PROTECTED]> wrote in message
> >[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > > 1. Please don't cross post.
> > > 2. Which line is #23?
> > > 3. Have you tested to ensure that you have a valid handle? ($db_num)
> > > 4. Have you tried adding error code to see the result of
dbase_create(),
> > > dbase_open()
> > > 5. I seem to remember reading in the docs, a long time ago (2 yr) that
the
> > > dbase functions were not really reliable and were intended mostly to
> > > extract info for export to another database. I could not be 100% wrong
on
> > > that, and I've not worked with the dbase functions in PHP.
> > >
> > > After this reply, please keep this focused on the php-db list.
> > >
> > > Regards - Miles Thompson
> > >
> > > At 12:14 PM 7/4/2002 +0100, labhras sammin wrote:
> > > >Hi I need to use PHP (ver 4 windows) to add a record to a dBase iv
> >database.
> > > >
> > > >I can  create, open, read and close a database but when I try to add
a
> > > >record I always get  the following error:
> > > >Warning: unexpected error in C:\php\phpstuff\addTest1.php on line 23
> > > >
> > > >Here is the code [shortened] that I use to create the database table,
> >open
> > > >the table, create the record and add the record- the database table
has
> >one
> > > >field a 10 char string:
> > > >
> > > >$dbname = "test1.dbf";
> > > >$dbflag = 2;
> > > >$def = array( array("name", "C",  10));
> > > >dbase_create($dbname, $def));
> > > >$db_num = dbase_open($dbname, $dbflag);
> > > >$new_record['name'] = 'xx';
> > > >dbase_add_record ($db_num, $new_record);
> > > >
> > > >any help, advice or code samples would be very appreciated
> > > >thanks
> > > >Labhras
> > > >
> > > >
> > > >
> > > >
> > > >--
> > > >PHP Database Mailing List (http://www.php.net/)
> > > >To unsubscribe, visit: http://www.php.net/unsub.php
> > >
> > >
> >
> >
> >
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>
>





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[PHP-DB] concatenating (was: RE: [PHP-DB] Retrieving/Displaying hyperlinked images with PHP)

2002-07-07 Thread Jen Swofford

Can somebody explain to me the rule on concatenating?  So far, I haven't had
to use those mysterious-looking dots.  ;)  My code goes like this:

echo "";

Does it work because I'm lucky?  Honestly I haven't been impressed with the
documentation I've found online.

Jen Swofford
[EMAIL PROTECTED]

> [snip]
> // Then show image and/or details
> echo "";
> [snip]



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RE: [PHP-DB] Newbie Reminder on HTML Variable Passing - quick fix

2002-07-07 Thread Peter Lovatt

hi

The enclosed will read all incoming variables and convert them, effectively
turning on 'register_globals'

It is a quick fix for broken scripts but does remove the security provided
by turning register_globals off !

use it as an include at the beginning of scripts.

Peter

http://www.sunmaia.net for full licence terms  #
#supporthttp://www.sunmaia.net/support/ #
#   #
#   You can do whatever you like with this so long  #
#   as this header remains in place #
#   #
#   It worked last time I used it, but it is not#
#   guaranteed in any way!  #
#   #
#   #
/

while(list($key, $val) = each($HTTP_POST_VARS))
{
$$key = stripslashes($val);
}


while(list($key, $val) = each($HTTP_GET_VARS))
{
$$key = stripslashes($val);
}


while(list($key, $val) = each($HTTP_COOKIE_VARS))
{
$$key = stripslashes($val);
}



?>



---
Excellence in internet and open source software
---
Sunmaia
www.sunmaia.net
[EMAIL PROTECTED]
tel. 0121-242-1473
---

> -Original Message-
> From: Allens [mailto:[EMAIL PROTECTED]]
> Sent: 07 July 2002 16:53
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Newbie Reminder on HTML Variable Passing
>
>
> I just recently got back to work on my db's and found that html variable
> passing isn't working anymore. I did some research and fould that the
> core setting in PHP for register_globals is off by default now since
> 4.1.x. As a newbie, this stumped me. There are some new variables
> created by the PHP team to address this, but we need to change our code
> some. The url below helps explain some of this. Hope this helps some of
> the other newbies entering this environment that have been confused by
> this as I had been. :)
>
> http://www.php.net/release_4_1_0.php
> :)
>Gale L. Allen Jr
>Macintosh Support Specialist
> 865/947-5740
>
>"Remember, Love wins over all"
> (:
>
>
> --
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> To unsubscribe, visit: http://www.php.net/unsub.php
>


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[PHP-DB] Newbie Reminder on HTML Variable Passing

2002-07-07 Thread Allens

I just recently got back to work on my db's and found that html variable 
passing isn't working anymore. I did some research and fould that the 
core setting in PHP for register_globals is off by default now since 
4.1.x. As a newbie, this stumped me. There are some new variables 
created by the PHP team to address this, but we need to change our code 
some. The url below helps explain some of this. Hope this helps some of 
the other newbies entering this environment that have been confused by 
this as I had been. :)

http://www.php.net/release_4_1_0.php
:)
   Gale L. Allen Jr
   Macintosh Support Specialist
865/947-5740

   "Remember, Love wins over all"
(:


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RE: [PHP-DB] Retrieving/Displaying hyperlinked images with PHP

2002-07-07 Thread Rich Hutchins

Just in case you were still looking for info on your question, I've put all
of the code from a page I have working that does pretty much what you're
looking for. I'm no guru, but the code is commented pretty well. The page
does a little more than what you have mentioned you're looking for, but
maybe some of the "extras" will be of use to you.


##

");

//set the directory. needs to be pulled in as a variable from a menu
choice or previous page
$dir = $incomingDirName;

//Handles the error in the case that the incoming directory can't be 
found
for some reason.
if (! @openDir($dir))
{
echo ("I'm sorry, but the image directory
".$incomingDirName." can not be found on this server.");
exit();
}

/*
output the current directory information. testing purposes only.
$thisDir=getcwd();
echo "Results of getcwd(): ".$thisDir."\n";
echo "Files in ".$dir.":\n";
*/

//open up a table on the page
echo("


");

//scan the directory and load its contents into an array
$handle=opendir($dir);
while (false !== ($file = readdir($handle)))
{
$retVal[] = $file;
}

//close the directory read operation then sort the array
closedir($handle);
sort($retVal);
//return $retVal; //required if a Function

//initialize a counter for use in formatting the table
$counter = 0;

//iterate over array
for ($i = 0; $i < count($retVal); $i++)
{
//strip out the . and .. level directory information
if ($retVal[$i] != "." && $retVal[$i] != "..")
{
//look for directory contents that start with "tn_"
if (strstr($retVal[$i], "tn_"))
{
//if the file starts with tn_ set it for use 
as the img src""
$thumImg = $retVal[$i];

//then strip off the tn_ and use the result as 
the image to link to
$linkTo = substr_replace($retVal[$i], "", 0, 
3);

//combine directory and file information into 
one variable for ease of
coding
$fullThumPath = $dir.$retVal[$i];
$fullLinkToPath = $dir.$linkTo;

//counter starts at 0. for iterations 0-3, put 
images into cells in
current row
if ($counter <= 3)
{
//drop the two variables into a line 
to create table information
echo ("");

$counter++;
}

//for iterations greater than 3 (four images) 
close the current row,
start another and drop current image in first cell
else
{
echo ("");

$counter = 1;
}
}
}

//This is the basis for checking for a directory without image 
files.
It's a bit broken as it reports an error whenever a file that soesn't start
with "tn_" is encountered. Which is exactly what the if() statement tells it
to do. Needs a little work.

/*
if (! strstr($retVal[$i], "tn_"))
{
echo ("I'm sorry, but this directory 
does not contain
any thumbnail images. Please contact Rich and let him know he screwed
up!");
}
*/
}

//ending image table tags
echo("


");

//ending page div tag
echo("

");
?>


##

Rich

-Original Message-
From: markbm [mailto:[EMAIL PROTECTED]]
Sent: Saturday, July 06, 2002 7:3

RE: [PHP-DB] Having more problems

2002-07-07 Thread Howard Picken

Several things I've noticed.

One, you have not body tags in your code anywhere.
Just applying a style to the body doesn't (aas far as
I know doesn't declare it in your code.

Secondly, and I don't konw if you left it out
to protect the innocent, but you haven't supplied
a password for the database (assuming your using
passwords, which is advisable).

You are not doing anything with database either.
Are you trying to select a record from it or just
checking if exists?

Howard


-Original Message-
From: Peter Lovatt [mailto:[EMAIL PROTECTED]]
Sent: Sunday, 7 July 2002 5:26 PM
To: Shiloh Madsen; [EMAIL PROTECTED]
Subject: RE: [PHP-DB] Having more problems


You are missing a  tag to mark the start of the body, so all your
content will be part of the header, I think!

HTH

Peter

---
Excellence in internet and open source software
---
Sunmaia
www.sunmaia.net
[EMAIL PROTECTED]
tel. 0121-242-1473
---

> -Original Message-
> From: Shiloh Madsen [mailto:[EMAIL PROTECTED]]
> Sent: 07 July 2002 06:07
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Having more problems
>
>
> The newbie is still having troubles heh. Maybe some kind soul can tell
> me what im doing wrong this time. This is the code for a page I am
> working on. When I try to bring up the page in a browser, I just
> get a white page, instead of having the HTML display. Anyone able
> to tell me why?
>
>
>
> 
>
> 
> 
> Login Page
> 
>  $dbhost = "127.0.0.1";
> $dbuser = "root";
> $db = "LoginInfo";
>
> $LoginDB=@mysql_connect($dbhost, $dbuser, $dbpass);
> if (! $LoginDB) {
> print "Unable to connect to the database server at this time.";
> exit();
> } else {
> mysql_select_db("GameDB",$LoginDB);
> if (! @mysql_select_db("GameDB") )
> print "Unable to locate the Game Database.";
> exit();
> }
> ?>
> 
> body { color: white; background: black; }
> 
>  alt="D&D Resource Page" border="0">"
> 
> 
> 
> Login: 
> Password: 
>  
>
> 
>
>  }
> ?>
>
> 
>
>
>
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RE: [PHP-DB] Having more problems

2002-07-07 Thread Peter Lovatt

You are missing a  tag to mark the start of the body, so all your
content will be part of the header, I think!

HTH

Peter

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> -Original Message-
> From: Shiloh Madsen [mailto:[EMAIL PROTECTED]]
> Sent: 07 July 2002 06:07
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Having more problems
>
>
> The newbie is still having troubles heh. Maybe some kind soul can tell
> me what im doing wrong this time. This is the code for a page I am
> working on. When I try to bring up the page in a browser, I just
> get a white page, instead of having the HTML display. Anyone able
> to tell me why?
>
>
>
> 
>
> 
> 
> Login Page
> 
>  $dbhost = "127.0.0.1";
> $dbuser = "root";
> $db = "LoginInfo";
>
> $LoginDB=@mysql_connect($dbhost, $dbuser, $dbpass);
> if (! $LoginDB) {
> print "Unable to connect to the database server at this time.";
> exit();
> } else {
> mysql_select_db("GameDB",$LoginDB);
> if (! @mysql_select_db("GameDB") )
> print "Unable to locate the Game Database.";
> exit();
> }
> ?>
> 
> body { color: white; background: black; }
> 
>  alt="D&D Resource Page" border="0">"
> 
> 
> 
> Login: 
> Password: 
>  
>
> 
>
>  }
> ?>
>
> 
>
>
>
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