Re: [PHP-DB] No results from query

2002-04-09 Thread Marius Ursache



Lisi a écrit :

 That worked...why?


because in your table you wrote _SPACE_Hatzoloh_SPACE_
or _NEWLINE_Hatzoloh_NEWLINE_ ...any way you wrote Hatzoloh and some
caracter that you don't see it so easy (space, tab, new_line).

% is like * on regular expresion (wildcard)


 -Lisi

 At 09:23 AM 4/9/02 +0300, Marius Ursache wrote:

 Lisi a écrit :
 
   I have the following query:
  
   SELECT * FROM ayny WHERE organization LIKE 'Hatzoloh'
  
 
 try SELECT * FROM ayny WHERE organization LIKE '%Hatzoloh%' 
 
  
   which produces no results, when I set
  
   $numrows = mysql_num_rows($result);
  
   it equals 0.
  
   When I do the following:
  
   SELECT * FROM ayny
  
   and then loop through the results, echoing them to the screen with the
   following code:
  
   while ( list($key,$val) = each($result_array) ) {
   echo $key = $valBR;
   }
  
   I do get a line
  
   organization = Hatzoloh
  
   So why can't I pull just the record for Hatzoloh using my first query?
  
   Please let me know if you need more code to figure this out.
  
   Thanks,
  
   -Lisi
  
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[PHP-DB] How to avoid: Warning: Page has Expired

2002-04-09 Thread John Hughes

I have the first part of the kids' soccer photo database site up that I
mentioned in an earlier post.

The site displays nine image thumbnails. You click on the thumbnails and are
taken to a full size photo.  Everything works as expected for the first nine
images but when you go to the next page of thumbnails, you start getting:

Warning: Page has Expired The page you requested was created using
information
you submitted in a form. This page is no longer available. As a
security precaution,
Internet Explorer does not automatically resubmit your information
for you.

To resubmit your information and view this Web page, click the
Refresh button.

What causes this and how can I fix it so this doesn't happen?

The URL for this site is http://fuzzheads.org/barcelona

TIA

John Hughes



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[PHP-DB] PHP date calculations and db (slightly OT)

2002-04-09 Thread Markus Lervik


Hello list,

I have a database where people can reserve our computer class.
The columns are time_from, time_to, name_id, res_date, confirmed.

I have a table (HTML, not db *g*) that's supposed to print out a=20
month, starting from day 1 to num_days_in_month, and print the
numbers in the right spot as so:

May, 2002

+-+-+-+-+-+-+-+
| Mon | Tue | Wed | Thu | Fri | Sat | Sun |
+-+-+-+-+-+-+-+
|  29 |  30 |  1  |  2  |  3  =A6  4  |  5  |
+-+-+-+-+-+-+-+
|...etc...|
+-+-+-+-+-+-+-+

where 29 and 30 will be grayed out, and if there's a reservation
on, say, Friday, it'll show friday the 3rd bold (and as a link).

I've got a function that returns a $dates array where I've saved the
name of the first day of the month, number of days in month, name of the
month and number of the month. Now I'm confused as to how I would=20
parse the information when printing out the (HTML-)table. What I've
got so far is

for($w=3D1;$w=3D5;$w++) {  // max. 5 weeks in a month
 echo(tr);  // table row
  for($d=3D1;$d=3D7;$d++) { // days in week
for($res=3D0;$res=3D$num_rows;$res++) {   // counter for array from
db

now the problem is, that if I work further on this, I end up with
about four-five for-loops, which doesn't seem to be such a good idea.
The other problem is that I really don't have a clue as to where I'd
smack in a counter for the number of days in the month. However much
I ponder this I can't reach a sensible conclusion. It seems that putting
the loop just about anywhere would be a performance loss ;)

Anyone ever done anything like this?


Cheers,
Markus
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[EMAIL PROTECTED]
+358-6-325 3589 / +358-40-832 6709


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[PHP-DB] Finding highest number in auto increment

2002-04-09 Thread Lisi

Each record in my table has a unique ID number generated by auto increment 
each time a new record is added. This is stored in a field called ID. I 
have a page that displays records as they are selected, but I want the 
default to be the latest entry if none has been selected.

How do I write the query to find the highest number an display that record?

It should be something like this I think:

SELECT organization FROM ayny WHERE ID ...

how does this end?

Thanks,

-Lisi


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Re: [PHP-DB] Finding highest number in auto increment

2002-04-09 Thread Adam Voigt

SELECT * FROM whatever ORDER BY id DESC;

Ofcourse if you want Ascending order, it would be ASC instead of DESC

Adam Voigt
[EMAIL PROTECTED]

On Tue, 09 Apr 2002 15:48:45 +0200, Lisi [EMAIL PROTECTED] wrote:
 Each record in my table has a unique ID number generated by auto increment
 each time a new record is added. This is stored in a field called ID. I
 have a page that displays records as they are selected, but I want the
 default to be the latest entry if none has been selected.
 
 How do I write the query to find the highest number an display that record?
 
 It should be something like this I think:
 
 SELECT organization FROM ayny WHERE ID ...
 
 how does this end?
 
 Thanks,
 
 -Lisi
 
 
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Re: [PHP-DB] Finding highest number in auto increment

2002-04-09 Thread Adam Voigt

Oh, forget to include that if you only want that one record,
use the LIMIT 1 keyword as well.

Adam Voigt
[EMAIL PROTECTED]

P.S.: All of this is in the MySQL manual.

On Tue, 09 Apr 2002 15:48:45 +0200, Lisi [EMAIL PROTECTED] wrote:
 Each record in my table has a unique ID number generated by auto increment
 each time a new record is added. This is stored in a field called ID. I
 have a page that displays records as they are selected, but I want the
 default to be the latest entry if none has been selected.
 
 How do I write the query to find the highest number an display that record?
 
 It should be something like this I think:
 
 SELECT organization FROM ayny WHERE ID ...
 
 how does this end?
 
 Thanks,
 
 -Lisi
 
 
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[PHP-DB] Newbie attack...

2002-04-09 Thread Achilleas Maroulis

Hi. I'm Achilles from Greece and I have been into PHP for 3 or 4 weeks. So far I've 
read 2 books about it and a lot of the online manual. But now I've come onto a problem 
and I'm asking you for help.
I've designed a site which is completely based onto PHP and mySQL. What I mean is that 
all the main links are actually records into a table with columns such as: id, title, 
contents etc. Every time a link is clicked the main page rebuilts itself using the id 
parameter of the link and by an SQL command finds the title and the text to show. So 
far so good. Everything is working great except the fact that when I put some php code 
into the 'text' field, in order to run en extra sql command in the page, it hangs and 
just prints the code as if it was just text.
Propably because of some encoding that takes place by submitting the data from my form 
to the database.
I've tried some urlencode() and htmlspec... combination but didn't actually make it...

I hope that some of you are clever enough to understand what I'm saying (with my poor 
english) and kind enough to help me... 

Looking forward to hearing from you...



[PHP-DB] Re: Newbie attack...

2002-04-09 Thread Hugh Bothwell


  Everything is working great except the fact that when
 I put some php code into the 'text' field, in order to run
 en extra sql command in the page, it hangs and just
 prints the code as if it was just text.

(shrug) so far as the database is concerned, it *is* just
text.  If you want returned code evaluated, you will
have to explicitly evaluate it.


Possible approaches:

1.  Search for php code in your content: split the
string appropriately and evaluate as needed.
I would avoid this because it's a lot of extra work,
and because it's quite possible to write 'interleaved'
PHP code that requires several blocks working
together; evaluating a single block at a time will
just get you errors.

2.  Evaluate all your content.
You would have to be very careful about
letting people contribute content, and it might
be a bit slower; but it's very very simple.

3.  Have markers in your content, referring to
PHP code from another table.
Would be a bit slower (have to do multiple
queries to retrieve code as needed), but more
secure - people can safely contribute to the
HTML content table, you can let them
reference standardized PHP fragments.

4.  Half-way between #2 and #3: add a
flag field to your database table, if the flag
is set the content must be evaluated, otherwise
it is plain HTML and can just be echoed.
People can contribute content, but the flag
stays off until you double-check any PHP
code for safety.



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[PHP-DB] connect to filemaker via odbc

2002-04-09 Thread aanding

Hi,
has anybody experience with connecting to filemaker via odbc.
thanx

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Re: [PHP-DB] connect to filemaker via odbc

2002-04-09 Thread Joshua Hoover

I haven't tried it, but the best solution seems to be the FX.php classes for
connecting to FileMaker from PHP by Chris Hansen.  For more information and
to download, try:

http://www.iviking.org/projects/FXdownload.php

I believe you'll have to turn on Web Companion on FileMaker for this to
work.

Hope that helps,

Joshua Hoover

 Hi,
 has anybody experience with connecting to filemaker via odbc.
 thanx


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[PHP-DB] undefined function

2002-04-09 Thread Nick

Hi:
in my code below, i get the error of undefined function. can anybody help
please?
Thanks

/* Connecting, selecting database */
$link = mysql_connect(localhost, $user_name, $password)
or die(Could not connect);
print Connected successfully;
mysql_select_db($my_Database) or die(Could not select database);

Fatal error: Call to undefined function: mysql_connect() in
/var/www/html/action.php on line 13



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Re: [PHP-DB] undefined function

2002-04-09 Thread Julie Meloni

N Fatal error: Call to undefined function: mysql_connect() in
N /var/www/html/action.php on line 13


This indicates you do not have MySQL support built into PHP.  Assuming
a non-Windows OS by the pathnames above, you must use --with-mysql when
issuing the configure command when building PHP.



- Julie

-- Julie Meloni
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http://www.amazon.com/exec/obidos/ASIN/0672323494/thickbookcom-20


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RE: [PHP-DB] How to avoid: Warning: Page has Expired

2002-04-09 Thread Peter Lovatt

If you change the form action  to 'get' instead of 'post' this avoids the
message.

form action=script.php3 method=get

'Get' data is displayed on the address line

eg http://domain.com/script.php3?photoID=123viewtype=full

so it should be avoided for sensistive data like passwords or credit card
numbers. This kind of data should always use 'post'

form action=script.php3 method=post

HTH

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

 -Original Message-
 From: John Hughes [mailto:[EMAIL PROTECTED]]
 Sent: 09 April 2002 08:30
 To: [EMAIL PROTECTED]
 Subject: [PHP-DB] How to avoid: Warning: Page has Expired


 I have the first part of the kids' soccer photo database site up that I
 mentioned in an earlier post.

 The site displays nine image thumbnails. You click on the
 thumbnails and are
 taken to a full size photo.  Everything works as expected for the
 first nine
 images but when you go to the next page of thumbnails, you start getting:

 Warning: Page has Expired The page you requested was created using
 information
 you submitted in a form. This page is no longer available. As a
 security precaution,
 Internet Explorer does not automatically resubmit your information
 for you.

 To resubmit your information and view this Web page, click the
 Refresh button.

 What causes this and how can I fix it so this doesn't happen?

 The URL for this site is http://fuzzheads.org/barcelona

 TIA

 John Hughes



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Re: [PHP-DB] undefined function

2002-04-09 Thread Nick

Hi:
thank you for you e mail
I am using redhat7.2
I am new to php and linux , do not know how to configure mysql into php
I wrote a simple function to ask the user for two numbers and add them
together
and print them out, and it worked. But when I used predefined function, I
get the error.
where do I go from here?
Thanks

- Original Message -
From: Julie Meloni [EMAIL PROTECTED]
To: Nick [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Sent: Tuesday, April 09, 2002 9:44 AM
Subject: Re: [PHP-DB] undefined function


 N Fatal error: Call to undefined function: mysql_connect() in
 N /var/www/html/action.php on line 13


 This indicates you do not have MySQL support built into PHP.  Assuming
 a non-Windows OS by the pathnames above, you must use --with-mysql when
 issuing the configure command when building PHP.



 - Julie

 -- Julie Meloni
 -- [EMAIL PROTECTED]
 -- www.thickbook.com

 Find Sams Teach Yourself MySQL in 24 Hours at
 http://www.amazon.com/exec/obidos/ASIN/0672323494/thickbookcom-20




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RE: [PHP-DB] How to avoid: Warning: Page has Expired

2002-04-09 Thread Jonathan Hilgeman

I've addressed this problem here:
http://php.sitecreative.com/faq.php

- Jonathan

-Original Message-
From: John Hughes [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, April 09, 2002 12:30 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] How to avoid: Warning: Page has Expired 


I have the first part of the kids' soccer photo database site up that I
mentioned in an earlier post.

The site displays nine image thumbnails. You click on the thumbnails and are
taken to a full size photo.  Everything works as expected for the first nine
images but when you go to the next page of thumbnails, you start getting:

Warning: Page has Expired The page you requested was created using
information
you submitted in a form. This page is no longer available. As a
security precaution,
Internet Explorer does not automatically resubmit your information
for you.

To resubmit your information and view this Web page, click the
Refresh button.

What causes this and how can I fix it so this doesn't happen?

The URL for this site is http://fuzzheads.org/barcelona

TIA

John Hughes



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[PHP-DB] If statement for sql statement

2002-04-09 Thread jas

Ok here is my question... I have a form that reads the contents of a
directory, places the resulting files into a select box within a form for
further processing, code is as follows:
?php
$dir_name = /virtual/path/to/images/directory/;
$dir = opendir($dir_name);
$file_list .= pFORM METHOD=\post\ ACTION=\index_done.php3\
SELECT NAME=\files\OPTION VALUE=\--\
NAME=\--\--/OPTION;
 while ($file_name = readdir($dir)) {
  if (($file_name != .)  ($file_name !=..)) {
  $file_list .= OPTION VALUE=\$file_name\
NAME=\$file_name\$file_name/OPTION;
  }
 }
 $file_list .= /SELECTbrbrINPUT TYPE=\submit\ NAME=\submit\
VALUE=\select\/FORM/p;
 closedir($dir);
?
This portion works great... now on index_done.php3 it takes the resulting
selection and places it into the db table, code is as follows:
?php
$file_var = http://localhost/images/;;
$db_name = database;
$table_name = image_path;
$connection = mysql_connect(localhost, user_name, password) or die
(Could not connect to database.  Please try again later.);
$db = mysql_select_db($db_name,$connection) or die (Could not select
database table. Please try again later.);
$sql = UPDATE $table_name SET image01_t=\$file_var$files\;
$result = mysql_query($sql, $connection) or die (Could not execute query.
Please try again later.);
?
What I need to know is if there is a way to create an if statement that
would either update or not update a selected field in the database.  Below
is a piece I would assume would work however I am not sure of the mysql
command to not update anything.  Any help would be great.
Jas

if( $file_name = '--' ) $sql = dont update table;
else $sql = UPDATE $table_name SET image01_t = \$files\;




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[PHP-DB] Mysql and memory issue

2002-04-09 Thread Trond Erling Hundal

Hi, maybe this list isn't the best place to ask for this but...
I keep getting memory exhausted errors from mysql, on a system running
as mail/mysql/apache+php server. (rh7.1)
Is it my queries that are inefficient? Is it my modelling-scheme that
sucks or could it be that mysql is leaking memory in some way?
Would it help to move mysql to a separate machine?



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Re: [PHP-DB] undefined function

2002-04-09 Thread Jason Wong

On Wednesday 10 April 2002 00:12, Nick wrote:
 Hi:
 thank you for you e mail
 I am using redhat7.2
 I am new to php and linux , do not know how to configure mysql into php

Install:

  php-mysql-4.0.6-7.i386.rpm

it's on the second disc.


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Re: [PHP-DB] Re: executing cgi

2002-04-09 Thread Jason Wong

On Tuesday 09 April 2002 10:49, David Robley wrote:
 In article 000d01c1df6c$962e8e20$0300a8c0@alex, [EMAIL PROTECTED]
 says...

  hey guys,
 
  kinda OT for the db list, but since most ppl know a lot about php can
  somehow tell me how to execute a cgi script within a php page?
 
  Thanks!

 I could, but I'll just point you to the manual and the Program Execution
 functions. For being OT.

 Mind turning word wrap on, please?


And if you want further info, this was discussed at some length on the 
php-general list just last week. Search the archives.


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[PHP-DB] AOL Problems

2002-04-09 Thread Bzdpltd

Hi group,

I wonder if anyone can help me with some problems I have just found out with a website 
we have just put together.

Our client unfortuntately uses the AOL browser on a MAC, so he is running version 5 of 
AOL for the mac.

Now we have a database driven website we have put together, and the problem occurs as 
the values do not seem to be getting sent to the database and so therefore he is not 
getting the correct results that he should be.

We have a bookstore, where you intially select your location in the world, and then 
also a category to select the books.

This is fine as it queries the database and returns the results to him. Then on the 
results page we have the category list so that they can go to another category, they 
do not have to select a new location as the values are already carried over each page. 
But what happens is that when they choose the new category and click submit, it does 
not change anything on the page, so they get the same results even though they have 
sent a new value to the server.

Has anybody had this problem, and is there a fix?

Help would be appreciated as this is the last problem we have to fix before this site 
can go live.

Many thanks


Barry Zimmerman
www.bzdpltd.co.uk

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Re: [PHP-DB] If statement for sql statement

2002-04-09 Thread Jason Wong

On Wednesday 10 April 2002 12:32, jas wrote:
 Ok here is my question... I have a form that reads the contents of a
 directory, places the resulting files into a select box within a form for
 further processing, code is as follows:

[snip]

 What I need to know is if there is a way to create an if statement that
 would either update or not update a selected field in the database.  Below
 is a piece I would assume would work however I am not sure of the mysql
 command to not update anything.  Any help would be great.
 Jas

 if( $file_name = '--' ) $sql = dont update table;
 else $sql = UPDATE $table_name SET image01_t = \$files\;

Am I missing something? Psuedo-code:


 if ($file_name != '--') {
   $sql = UPDATE ;
   $result = $mysql_query($sql); 
   etc;
   etc; }
 else {
   don't even need an else!
 }


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[PHP-DB] Re: Join?

2002-04-09 Thread Lutz Brückner

Hi Volker,

after studying your code (not so easy to find  without formating and
lots of html), I think what you want to do is somthing like this:

$abfrage = 'SELECT A.id AS aid, A.autor AS aauthor, A.zeit,AS azeit'
.' A.betreff AS abetreff, A.inhalt AS ainhalt,'
.' U1.vorname AS autorv, U1.nachname AS autorn,'
.' K.autor AS kautor, K.zeit AS kzeit,
.' K.betreff AS kbetreff, K.inhalt AS kinhalt'
.' U2.vorname AS autorkv, U2.nachname AS autorkn,'
   .' FROM ts_artikel AS A'
  .' INNER JOIN ts_users AS U1 ON A.author=U1.id'
  .' INNER JOIN ts_kommentar AS K ON A.id=K.artikel_id'
  .' INNER JOIN ts_users AS U2 ON K.author=U2.id'
  .' ORDER BY A.zeit, K.zeit DESC ' . $limit;

$resultat = mysql_query($abfrage) or die (mysql_error());

while ($row = mysql_fetch_assoc($resultat)) {
// take care if the article id is changing 
echo 'whatever';
}

Ciao,
  Lutz


[EMAIL PROTECTED] (Volker Hartmann) writes:

 Hi,
 
 I've got a problem with database abstraction. There are two tables. One with
 articles (id,content,author,time) and one with comments on those articles
 (id,article_id,content,author,time). My script looks like this:
 
 function show_news($limit){
 /* get db parameters and connect to db */
 
 require (inc/db_parameter.inc.php);
 $connect = mysql_connect($dbhost,$dbuser,$dbpasswd) or die(Can't connect to
 database);
mysql_select_db($dbname,$connect) or die(mysql_error());
 
 $abfrage = SELECT id,autor,zeit,betreff,inhalt FROM ts_artikel ORDER BY
 zeit DESC . $limit;
 $resultat = mysql_query($abfrage) or die (mysql_error());
 
 /* get articles and display*/
 
 While ($row = mysql_fetch_array($resultat, MYSQL_ASSOC)) {
 
 /* get author */
 
 $abfrage2 = SELECT vorname,nachname FROM ts_users WHERE id=' .
 $row[autor] .';
 $resultat2 = mysql_query($abfrage2) or die (mysql_error());
 list($autorv, $autorn) = mysql_fetch_row($resultat2);
 
 /* get comments */
 
 $abfrage3 = SELECT autor,zeit,betreff,inhalt FROM ts_kommentar WHERE
 artikel_id=' . $row[id] . ' ORDER BY zeit ASC;
 $resultat3 = mysql_query($abfrage3) or die (mysql_error());
 
 /* generate display */
 
 echo table width='100%' cellspacing='0' cellpadding='0' class='formtable'
 style='width:100%;margin:0'trtd class='artikelhead'
 style='text-align:left;' . htmlentities($autorv).   .
 htmlentities($autorn);
 echo /tdtd class='artikelhead' style='text-align:right' . strftime(%A
 %d.%m.%Y,%H:%M:%S, $row[zeit]);
 echo /td/trtrtd colspan='2'b .  nl2br(chop($row[betreff])) .
 :/bbr / .  nl2br(chop($row[inhalt])) . /td/tr;
 /* get comments of there are some */
 if (!mysql_num_rows($resultat3)  1) {
 while ($row2 = mysql_fetch_array($resultat3)) {
 $abfrage4 = SELECT vorname,nachname FROM ts_users WHERE id=' .
 $row2[autor] . '; /*get author for comments */
 $resultat4 = mysql_query($abfrage4) or die (mysql_error());
 list($autorkv, $autorkn) = mysql_fetch_row($resultat4);
 echo trtd class='commenthead' style='text-align:left'img
 src='img/arrow.png' border='0' height='10' width='10'  .
 htmlentities($autorkv) .   . htmlentities($autorkn) . /tdtd
 class='commenthead' style='text-align:right' . strftime(%A
 %d.%m.%Y,%H:%M:%S, $row2[zeit]);
 echo /td/trtrtd colspan='2' class='commentcell'b .
 nl2br(chop($row2[betreff])) . :/bbr / .  nl2br(($row2[inhalt]));
 echo /td/tr;
 }
 }
 echo trtd colspan='2' class='artikelcell' style='border-top:1px solid
 #808080'a href='submit_comment.php?artikel= . $row[id] . 'Kommentar
 schreiben/a/td/tr/tablebr /;
 }
 }
 
 As you can see, there are some queries within while-loops.
 
 Is there a more elegant and performant way to do those queries? I thought of
 JOIN, but I cannot see how to do it.
 
 Any help would be appreciated
 
 TIA Volker

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Re: [PHP-DB] PHP tags in a MySQL echo

2002-04-09 Thread Joshua Stein

 Is there any way to work around this and get it to treat all the echoed rows
 as PHP not just only text?

Wrap the print around an eval():

   print(eval($string));


   -j.

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[PHP-DB] Re: Mysql and memory issue

2002-04-09 Thread Yasuo Ohgaki

Trond Erling Hundal wrote:
 Hi, maybe this list isn't the best place to ask for this but...

This is the right place.

 I keep getting memory exhausted errors from mysql, on a system running
 as mail/mysql/apache+php server. (rh7.1)
 Is it my queries that are inefficient? Is it my modelling-scheme that
 sucks or could it be that mysql is leaking memory in some way?
 Would it help to move mysql to a separate machine?

If you do normal query, mysql store whole result in memory!
You can avoid that. It's in the manual.

--
Yasuo Ohgaki




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[PHP-DB] LOBS with CURSOR_SHARING=FORCE gives core dump

2002-04-09 Thread Prince

I have the following php code. If I add the parameter CURSOR_SHARING=FORCE

in my init.ora and while executing the program I get the Segmentation
fault.

$ /usr/local/php4/bin/php -q curshare.php

Segmentation fault (core dumped)

Is there any work-around for this? Is this a bug? This seems to be happening
only when LOB with returning clause is present.

I tested a similar program from C program and is fine.

Note: currently I have set the parameter  (CURSOR_SHARING=FORCE ) only in
the program/session level.

Thanks,

Prince.





$ cat curshare.php

#!/usr/local/php4/bin/php -q

?

//program name curshare.php

$conn = OCILogon($dbuser, $dbpass, $dbname);

//This is added to affect only the current session.

$sql = alter session set cursor_sharing=force;

$stmt = ociparse($conn,$sql);

OCIExecute($stmt,OCI_DEFAULT);

$lob = OCINewDescriptor($conn, OCI_D_LOB);

$sqlstmt = insert into blob_test ( packageid, packagebody )
values('12344321123efd', EMPTY_BLOB()) returning packagebody into
:packagebody;

$stmt = OCIParse($conn, $sqlstmt) ;

OCIBindByName($stmt, ':packagebody', $lob, -1, OCI_B_BLOB);

OCIExecute($stmt, OCI_DEFAULT);

OCICommit($conn);

OCIFreeDesc($lob);

OCIFreeStatement($stmt);

OCILogoff($conn);

?




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Re: [PHP-DB] How to avoid: Warning: Page has Expired

2002-04-09 Thread olinux

Using the GET method is a great solution, but
sometimes you just gotta POST! I have found the best
solution to be using a separate file to do the
processing.

For example login.php submits 'user' and 'password' to
auth.php. auth.php queries the db for 'user' and
'password' and redirects [via header()] to
whatever.php or back to login.php if login was
unsuccessful.

olinux


--- John Hughes [EMAIL PROTECTED] wrote:
 I have the first part of the kids' soccer photo
 database site up that I
 mentioned in an earlier post.
 
 The site displays nine image thumbnails. You click
 on the thumbnails and are
 taken to a full size photo.  Everything works as
 expected for the first nine
 images but when you go to the next page of
 thumbnails, you start getting:
 
 Warning: Page has Expired The page you
 requested was created using
 information
 you submitted in a form. This page is no
 longer available. As a
 security precaution,
 Internet Explorer does not automatically
 resubmit your information
 for you.
 
 To resubmit your information and view this
 Web page, click the
 Refresh button.
 
 What causes this and how can I fix it so this
 doesn't happen?
 
 The URL for this site is
 http://fuzzheads.org/barcelona
 
 TIA
 
 John Hughes
 
 
 
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