php-general Digest 27 Mar 2006 01:34:15 -0000 Issue 4036
php-general Digest 27 Mar 2006 01:34:15 - Issue 4036
Topics (messages 232574 through 232576):
Re: Oject passed via session error
232574 by: Chris
Re: PHP Links
232575 by: Kevin Waterson
Re: Creation and transfer of object thru session
232576 by: Chris
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--
---BeginMessage---
You need to declare the class before you call session_start(). Otherwise
it can't create the object properly.
Peter Lauri wrote:
Hi,
I do this:
?php
session_start();
require_once('classes/ns_cart.class.php');
if(!isset($_SESSION['ns_cart'])) $_SESSION['ns_cart'] = new NScart();
?
The output will be the following with print_r of the $_SESSION:
Array
(
[ns_cart] = nscart Object
(
[myProducts] = Array
(
[0] = ns_product Object
(
-
When I reloads the page I get the following:
Array
(
[ns_cart] = __PHP_Incomplete_Class Object
(
[__PHP_Incomplete_Class_Name] = nscart
[myProducts] = Array
(
[0] = __PHP_Incomplete_Class Object
-
What is this? I am confused...
/Peter
---End Message---
---BeginMessage---
This one time, at band camp, Thomas Bonham [EMAIL PROTECTED] wrote:
I'm trying to find out how make following happen
Example:
http://www.example.com/test.php?id=20
in test.php put this code
?php echo $_GET['id']; ?
Kevin
--
Democracy is two wolves and a lamb voting on what to have for lunch.
Liberty is a well-armed lamb contesting the vote.
---End Message---
---BeginMessage---
Peter Lauri wrote:
Hi,
I am creating an object from a Web Service. Throughout a session I would
like to keep the object that was created in the beginning of the session.
The reason for this is that the creation of the object takes some time
because of the communication with the Web Service. As I see it, there are
two possibilities:
1. Save the object in the $_SESSION variable, because the objects are not
big
2. Serialize the object and save it to a database, and use $_SESSION to keep
reference to the database
What is to prefer? What are the benefits of using 1 against using 2? Is
there any 3rd option?
I'd probably go with #1 mainly because it's always available through
$_SESSION - you don't have to hit the db to keep getting it.
--
Postgresql php tutorials
http://www.designmagick.com/
---End Message---
Re: [PHP] Creation and transfer of object thru session
Peter Lauri wrote: Hi, I am creating an object from a Web Service. Throughout a session I would like to keep the object that was created in the beginning of the session. The reason for this is that the creation of the object takes some time because of the communication with the Web Service. As I see it, there are two possibilities: 1. Save the object in the $_SESSION variable, because the objects are not big 2. Serialize the object and save it to a database, and use $_SESSION to keep reference to the database What is to prefer? What are the benefits of using 1 against using 2? Is there any 3rd option? I'd probably go with #1 mainly because it's always available through $_SESSION - you don't have to hit the db to keep getting it. -- Postgresql php tutorials http://www.designmagick.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Problem with app, condition not working.
Hey guys got a problem with an app i'm working on.
App: An app to track visitors and hosts visiting certain pages in a website.
Situation: the app is to populate a table (db) with IP numbers which are
to be concidered as filtered/excluded in the below snippet of code from
being tracked.
Problem: the below code is used to insert the actual data for the
tracker from the visitor on the page its requested.
However, before the actual information is submitted, there has to be a
condition to see if the current visitor's IP is listed in the excluded
list. My current code returns false to all, and will not input the data
whether the IP is in the list or not in the table.
Culprit: Seems to be the in_array causing it or an eye-flaw from my side
but cannot find it.
Any ideas?
Jimmie
//==
#-- Include db config
require_once(../includes/db.class.phtml);
#-- Connect to db
$sql = new conSQL();
//==
#-- Fetch current rules
$currentRulesQuery = $sql-query(SELECT host FROM filter);
if(!$currentRulesQuery){ print mysql_error(); }
#-- Result to array
#.
$currentHost = $_SERVER['REMOTE_ADDR'];
/* Obs: this loop runs all the rules *** do not edit */
while($currentRules = $sql-fetchArray($currentRulesQuery, MYSQL_ASSOC))
{
#-- If in array
if(!in_array($currentHost, $currentRules))
{
#---#
#-- Fetch data
$currentSettingsQuery = $sql-query(SELECT track_control FROM define);
if(!$currentSettingsQuery){ print mysql_error(); }
#-- Data to var
$currentSettings = $sql-fetchArray($currentSettingsQuery);
if($currentSettings['track_control']==1)
{
// Specify the tracking vars # Revision 1.0
//==
$page = $_SERVER['PHP_SELF']; // full path
$ip = $_SERVER['REMOTE_ADDR']; // interface protocol
$time = date(Y-m-d H:i:s); // current time and date
$request_method = $_SERVER['REQUEST_METHOD']; // refer agent method
$gateway_interface = $_SERVER['GATEWAY_INTERFACE']; // refer
agent gateway
$server_protocol = $_SERVER['SERVER_PROTOCOL']; // host server
protocol
$request_time = $_SERVER['REQUEST_TIME']; // time of request (
php 5.1.0 up )
$query_string = $_SERVER['QUERY_STRING']; // query string if
available
$accept_charset = $_SERVER['HTTP_ACCEPT_CHARSET']; // refer
agent charset
$accept_encoding = $_SERVER['HTTP_ACCEPT_ENCODING']; // refer
agent encoding
$accept_language = $_SERVER['HTTP_ACCEPT_LANGUAGE']; // refer
agent language
$http_connection = $_SERVER['HTTP_CONNECTION']; // service
connection
$user_agent = $_SERVER['HTTP_USER_AGENT']; // refer agent
#$remote_port = $_SERVER['REMOTE_PORT']; // refer agent port
//==
#-- Continue with the query
$insertQuery = $sql-query(INSERT INTO track
(page,ip,time,request_method,gateway_interface,server_protocol,request_time,query_string,accept_charset,
accept_encoding,accept_language,http_connection,user_agent)
VALUES ('$page',
'$ip',
'$time',
'$request_method',
'$gateway_interface',
'$server_protocol',
'$request_time',
'$query_string',
'$accept_charset',
'$accept_encoding',
'$accept_language',
'$http_connection',
'$user_agent'));
}
}
}
#---#
#-- Error check on query
if(!($insertQuery)){ print mysql_error(); }
else
{
#.
$sql-close();
}
//==
?
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Re: [PHP] Problem with app, condition not working.
PHP Mailer wrote:
Hey guys got a problem with an app i'm working on.
App: An app to track visitors and hosts visiting certain pages in a
website.
Situation: the app is to populate a table (db) with IP numbers which are
to be concidered as filtered/excluded in the below snippet of code from
being tracked.
Problem: the below code is used to insert the actual data for the
tracker from the visitor on the page its requested.
However, before the actual information is submitted, there has to be a
condition to see if the current visitor's IP is listed in the excluded
list. My current code returns false to all, and will not input the data
whether the IP is in the list or not in the table.
Culprit: Seems to be the in_array causing it or an eye-flaw from my side
but cannot find it.
//==
#-- Include db config
require_once(../includes/db.class.phtml);
#-- Connect to db
$sql = new conSQL();
//==
#-- Fetch current rules
$currentRulesQuery = $sql-query(SELECT host FROM filter);
if(!$currentRulesQuery){ print mysql_error(); }
#-- Result to array
#.
$currentHost = $_SERVER['REMOTE_ADDR'];
/* Obs: this loop runs all the rules *** do not edit */
while($currentRules = $sql-fetchArray($currentRulesQuery, MYSQL_ASSOC))
{
If you print our $currentRules:
var_dump($currentRules);
is it an array? What's in it?
--
Postgresql php tutorials
http://www.designmagick.com/
--
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Re: [PHP] Problem with app, condition not working.
Chris skrev:
PHP Mailer wrote:
Hey guys got a problem with an app i'm working on.
App: An app to track visitors and hosts visiting certain pages in a
website.
Situation: the app is to populate a table (db) with IP numbers which
are to be concidered as filtered/excluded in the below snippet of
code from being tracked.
Problem: the below code is used to insert the actual data for the
tracker from the visitor on the page its requested.
However, before the actual information is submitted, there has to be
a condition to see if the current visitor's IP is listed in the
excluded list. My current code returns false to all, and will not
input the data whether the IP is in the list or not in the table.
Culprit: Seems to be the in_array causing it or an eye-flaw from my
side but cannot find it.
//==
#-- Include db config
require_once(../includes/db.class.phtml);
#-- Connect to db
$sql = new conSQL();
//==
#-- Fetch current rules
$currentRulesQuery = $sql-query(SELECT host FROM filter);
if(!$currentRulesQuery){ print mysql_error(); }
#-- Result to array
#.
$currentHost = $_SERVER['REMOTE_ADDR'];
/* Obs: this loop runs all the rules *** do not edit */
while($currentRules = $sql-fetchArray($currentRulesQuery, MYSQL_ASSOC))
{
If you print our $currentRules:
var_dump($currentRules);
is it an array? What's in it?
Hello Chris,
Returned output from the array is:
Array ( [0] = 127.0.0.1 [host] = 127.0.0.1 )
Yes, it's fetched from the fields in the belonging table (filter)
- Jimmie
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Re: [PHP] opening .Z archive with gzopen
nicolas figaro wrote: Hi I'm using php 4.3.11 on linux RH9. I'd like to open a .Z archive with gzopen. I looked at the options of gzopen, but I'm not sure there is a way to specify the compression used by compress/uncompress. Has anyone ever tried and done this yet ? (without a system(uncompress $file) Since a compressed file isn't the same as a gzipp'ed file I'd guess this won't work. This might: http://pear.php.net/package/File_Archive but it doesn't mention .Z files specifically. -- Postgresql php tutorials http://www.designmagick.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] opening .Z archive with gzopen
Chris skrev: nicolas figaro wrote: Hi I'm using php 4.3.11 on linux RH9. I'd like to open a .Z archive with gzopen. I looked at the options of gzopen, but I'm not sure there is a way to specify the compression used by compress/uncompress. Has anyone ever tried and done this yet ? (without a system(uncompress $file) Since a compressed file isn't the same as a gzipp'ed file I'd guess this won't work. This might: http://pear.php.net/package/File_Archive but it doesn't mention .Z files specifically. Seems to me like it's the in_array causing this, I need a way to see if the current IP matches any in the Array - Jimmie -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Err wrong reply addr
Sorry guys -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Problem with app, condition not working.
PHP Mailer wrote:
Chris skrev:
PHP Mailer wrote:
Hey guys got a problem with an app i'm working on.
App: An app to track visitors and hosts visiting certain pages in a
website.
Situation: the app is to populate a table (db) with IP numbers which
are to be concidered as filtered/excluded in the below snippet of
code from being tracked.
Problem: the below code is used to insert the actual data for the
tracker from the visitor on the page its requested.
However, before the actual information is submitted, there has to be
a condition to see if the current visitor's IP is listed in the
excluded list. My current code returns false to all, and will not
input the data whether the IP is in the list or not in the table.
Culprit: Seems to be the in_array causing it or an eye-flaw from my
side but cannot find it.
//==
#-- Include db config
require_once(../includes/db.class.phtml);
#-- Connect to db
$sql = new conSQL();
//==
#-- Fetch current rules
$currentRulesQuery = $sql-query(SELECT host FROM filter);
if(!$currentRulesQuery){ print mysql_error(); }
#-- Result to array
#.
$currentHost = $_SERVER['REMOTE_ADDR'];
/* Obs: this loop runs all the rules *** do not edit */
while($currentRules = $sql-fetchArray($currentRulesQuery, MYSQL_ASSOC))
{
If you print our $currentRules:
var_dump($currentRules);
is it an array? What's in it?
Hello Chris,
Returned output from the array is:
Array ( [0] = 127.0.0.1 [host] = 127.0.0.1 )
Does it get into the part where it should insert the data?
a simple
echo __LINE__ . 'br/';
will tell you.
It could just be you have a bad insert query - you are not escaping any
data at all.
Print out the query:
$insert_qry = INSERT INTO
track(page,ip,time,request_method,gateway_interface,server_protocol,request_time,query_string,accept_charset,
accept_encoding,accept_language,http_connection,user_agent)
VALUES ('$page',
'$ip',
'$time',
'$request_method',
'$gateway_interface',
'$server_protocol',
'$request_time',
'$query_string',
'$accept_charset',
'$accept_encoding',
'$accept_language',
'$http_connection',
'$user_agent');
echo $insert_qry . 'br/';
$insertQuery = $sql-query($insert_qry);
and run it manually in mysql and see what happens.
--
Postgresql php tutorials
http://www.designmagick.com/
--
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Re: [PHP] Problem with app, condition not working.
Chris skrev:
PHP Mailer wrote:
Chris skrev:
PHP Mailer wrote:
Hey guys got a problem with an app i'm working on.
App: An app to track visitors and hosts visiting certain pages in a
website.
Situation: the app is to populate a table (db) with IP numbers
which are to be concidered as filtered/excluded in the below
snippet of code from being tracked.
Problem: the below code is used to insert the actual data for the
tracker from the visitor on the page its requested.
However, before the actual information is submitted, there has to
be a condition to see if the current visitor's IP is listed in the
excluded list. My current code returns false to all, and will not
input the data whether the IP is in the list or not in the table.
Culprit: Seems to be the in_array causing it or an eye-flaw from my
side but cannot find it.
//==
#-- Include db config
require_once(../includes/db.class.phtml);
#-- Connect to db
$sql = new conSQL();
//==
#-- Fetch current rules
$currentRulesQuery = $sql-query(SELECT host FROM filter);
if(!$currentRulesQuery){ print mysql_error(); }
#-- Result to array
#.
$currentHost = $_SERVER['REMOTE_ADDR'];
/* Obs: this loop runs all the rules *** do not edit */
while($currentRules = $sql-fetchArray($currentRulesQuery,
MYSQL_ASSOC))
{
If you print our $currentRules:
var_dump($currentRules);
is it an array? What's in it?
Hello Chris,
Returned output from the array is:
Array ( [0] = 127.0.0.1 [host] = 127.0.0.1 )
Does it get into the part where it should insert the data?
a simple
echo __LINE__ . 'br/';
will tell you.
It could just be you have a bad insert query - you are not escaping
any data at all.
Print out the query:
$insert_qry = INSERT INTO
track(page,ip,time,request_method,gateway_interface,server_protocol,request_time,query_string,accept_charset,
accept_encoding,accept_language,http_connection,user_agent)
VALUES ('$page',
'$ip',
'$time',
'$request_method',
'$gateway_interface',
'$server_protocol',
'$request_time',
'$query_string',
'$accept_charset',
'$accept_encoding',
'$accept_language',
'$http_connection',
'$user_agent');
echo $insert_qry . 'br/';
$insertQuery = $sql-query($insert_qry);
and run it manually in mysql and see what happens.
Hey Chris,
the query works fine if i remove the condition to check for the IP
- Jimmie
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Re: [PHP] Problem with app, condition not working.
PHP Mailer wrote:
Chris skrev:
PHP Mailer wrote:
Hey guys got a problem with an app i'm working on.
App: An app to track visitors and hosts visiting certain pages in a
website.
Situation: the app is to populate a table (db) with IP numbers which
are to be concidered as filtered/excluded in the below snippet of
code from being tracked.
Problem: the below code is used to insert the actual data for the
tracker from the visitor on the page its requested.
However, before the actual information is submitted, there has to be
a condition to see if the current visitor's IP is listed in the
excluded list. My current code returns false to all, and will not
input the data whether the IP is in the list or not in the table.
Culprit: Seems to be the in_array causing it or an eye-flaw from my
side but cannot find it.
//==
#-- Include db config
require_once(../includes/db.class.phtml);
#-- Connect to db
$sql = new conSQL();
//==
#-- Fetch current rules
$currentRulesQuery = $sql-query(SELECT host FROM filter);
if(!$currentRulesQuery){ print mysql_error(); }
#-- Result to array
#.
$currentHost = $_SERVER['REMOTE_ADDR'];
/* Obs: this loop runs all the rules *** do not edit */
while($currentRules = $sql-fetchArray($currentRulesQuery, MYSQL_ASSOC))
{
If you print our $currentRules:
var_dump($currentRules);
is it an array? What's in it?
Hello Chris,
Returned output from the array is:
Array ( [0] = 127.0.0.1 [host] = 127.0.0.1 )
If you think it's the in_array - check it:
if (in_array($currentHost, $currentRules)) {
echo Host: $currentHost is in rulesbr/;
var_dump($currentRules);
}
--
Postgresql php tutorials
http://www.designmagick.com/
--
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To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Problem with app, condition not working.
Chris skrev:
PHP Mailer wrote:
Chris skrev:
PHP Mailer wrote:
Hey guys got a problem with an app i'm working on.
App: An app to track visitors and hosts visiting certain pages in a
website.
Situation: the app is to populate a table (db) with IP numbers
which are to be concidered as filtered/excluded in the below
snippet of code from being tracked.
Problem: the below code is used to insert the actual data for the
tracker from the visitor on the page its requested.
However, before the actual information is submitted, there has to
be a condition to see if the current visitor's IP is listed in the
excluded list. My current code returns false to all, and will not
input the data whether the IP is in the list or not in the table.
Culprit: Seems to be the in_array causing it or an eye-flaw from my
side but cannot find it.
//==
#-- Include db config
require_once(../includes/db.class.phtml);
#-- Connect to db
$sql = new conSQL();
//==
#-- Fetch current rules
$currentRulesQuery = $sql-query(SELECT host FROM filter);
if(!$currentRulesQuery){ print mysql_error(); }
#-- Result to array
#.
$currentHost = $_SERVER['REMOTE_ADDR'];
/* Obs: this loop runs all the rules *** do not edit */
while($currentRules = $sql-fetchArray($currentRulesQuery,
MYSQL_ASSOC))
{
If you print our $currentRules:
var_dump($currentRules);
is it an array? What's in it?
Hello Chris,
Returned output from the array is:
Array ( [0] = 127.0.0.1 [host] = 127.0.0.1 )
If you think it's the in_array - check it:
if (in_array($currentHost, $currentRules)) {
echo Host: $currentHost is in rulesbr/;
var_dump($currentRules);
}
Strangest thing is:
if i add the excluded IP to the field in the DB it will insert the
visitor info to the db, however if i remove the excluded ip from the
filter, it wont
It's upsidedown!
- Jimmie
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Re: [PHP] Problem with app, condition not working.
PHP Mailer wrote:
Chris skrev:
PHP Mailer wrote:
Chris skrev:
PHP Mailer wrote:
Hey guys got a problem with an app i'm working on.
App: An app to track visitors and hosts visiting certain pages in a
website.
Situation: the app is to populate a table (db) with IP numbers
which are to be concidered as filtered/excluded in the below
snippet of code from being tracked.
Problem: the below code is used to insert the actual data for the
tracker from the visitor on the page its requested.
However, before the actual information is submitted, there has to
be a condition to see if the current visitor's IP is listed in the
excluded list. My current code returns false to all, and will not
input the data whether the IP is in the list or not in the table.
Culprit: Seems to be the in_array causing it or an eye-flaw from my
side but cannot find it.
//==
#-- Include db config
require_once(../includes/db.class.phtml);
#-- Connect to db
$sql = new conSQL();
//==
#-- Fetch current rules
$currentRulesQuery = $sql-query(SELECT host FROM filter);
if(!$currentRulesQuery){ print mysql_error(); }
#-- Result to array
#.
$currentHost = $_SERVER['REMOTE_ADDR'];
/* Obs: this loop runs all the rules *** do not edit */
while($currentRules = $sql-fetchArray($currentRulesQuery,
MYSQL_ASSOC))
{
If you print our $currentRules:
var_dump($currentRules);
is it an array? What's in it?
Hello Chris,
Returned output from the array is:
Array ( [0] = 127.0.0.1 [host] = 127.0.0.1 )
If you think it's the in_array - check it:
if (in_array($currentHost, $currentRules)) {
echo Host: $currentHost is in rulesbr/;
var_dump($currentRules);
}
Strangest thing is:
if i add the excluded IP to the field in the DB it will insert the
visitor info to the db, however if i remove the excluded ip from the
filter, it wont
Then this isn't a copy / paste of your script:
#-- If in array
if(!in_array($currentHost, $currentRules))
Comment says one, the code says another.
--
Postgresql php tutorials
http://www.designmagick.com/
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Problem with app, condition not working.
Chris skrev:
PHP Mailer wrote:
Chris skrev:
PHP Mailer wrote:
Chris skrev:
PHP Mailer wrote:
Hey guys got a problem with an app i'm working on.
App: An app to track visitors and hosts visiting certain pages in
a website.
Situation: the app is to populate a table (db) with IP numbers
which are to be concidered as filtered/excluded in the below
snippet of code from being tracked.
Problem: the below code is used to insert the actual data for the
tracker from the visitor on the page its requested.
However, before the actual information is submitted, there has to
be a condition to see if the current visitor's IP is listed in
the excluded list. My current code returns false to all, and will
not input the data whether the IP is in the list or not in the
table.
Culprit: Seems to be the in_array causing it or an eye-flaw from
my side but cannot find it.
//==
#-- Include db config
require_once(../includes/db.class.phtml);
#-- Connect to db
$sql = new conSQL();
//==
#-- Fetch current rules
$currentRulesQuery = $sql-query(SELECT host FROM filter);
if(!$currentRulesQuery){ print mysql_error(); }
#-- Result to array
#.
$currentHost = $_SERVER['REMOTE_ADDR'];
/* Obs: this loop runs all the rules *** do not edit */
while($currentRules = $sql-fetchArray($currentRulesQuery,
MYSQL_ASSOC))
{
If you print our $currentRules:
var_dump($currentRules);
is it an array? What's in it?
Hello Chris,
Returned output from the array is:
Array ( [0] = 127.0.0.1 [host] = 127.0.0.1 )
If you think it's the in_array - check it:
if (in_array($currentHost, $currentRules)) {
echo Host: $currentHost is in rulesbr/;
var_dump($currentRules);
}
Strangest thing is:
if i add the excluded IP to the field in the DB it will insert the
visitor info to the db, however if i remove the excluded ip from the
filter, it wont
Then this isn't a copy / paste of your script:
#-- If in array
if(!in_array($currentHost, $currentRules))
Comment says one, the code says another.
Hello Chris,
Sorry about that, happens when writing it too fast and changing it too
often.
Fixed the problem now, did a match in the query instead.
Thank you for the excellent help!
Jimmie
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[PHP] array_search function bugged? update
Hi all
Here is an update on my problem reported with the array_search function.
to those who directed me to the bug report page, thanks.
There are a number of reports specific to this function. Maybe event
someone else has written a fix.
here is my solution to my problem.
The following code* was developed and finalized by J.E.Killen 3/2006
* accept of course for array_search() itself.
function word_wise(array $input, int $what) //returns an array
{$str = array();
$stra = array();
$str = str_split($input); //turns string into
array
$stra = $str; // saves a copy for final
reconcilliation
$tmp = '';// used to hold and transfer values
$final = array(); // ***THE WHOLE POINT
$reduced = array(); // reduction array
$formula = array(); // interum array
$output = array(); // test output.
for($i = 0; $i count($str); $i++)
{$formula[$i] = array_search($str[$i],
$str);
if($formula[$i] $i) // use of $formula
{$str[$i] = '';} // repeats eliminated from $str
}; // if $i is greater than formula[$i], formula[$i] is
a repeat.
for($i = 0; $i count($str); $i++)
{if($str[$i] != '') // looks for empty
strings in $str
{
array_push($reduced, $str[$i]); //
fills $reduced with
// non
empty $str items
// to
produce an array of
// all
unique letters
}
};
for($i = 0; $i count($stra); $i++)
{for($j = 0; $j count($reduced); $j++)
{
if($stra[$i] == $reduced[$j])//produces
the final formula for
{$final[$i] = $j; } //
reconstruction of the word.
}
}
for($i = 0; $i count($final); $i++)
{ $tmp = $final[$i];
$output[$i] = $reduced[$tmp]; //test run
reconstructs word from $final and $reduced.
}
switch($what) //options to chose what to return
{case 1:
return $reduced; //reduction to all unique
letters
break;
case 2:
return $stra;//original
break;
case 3:
return $final; // reconstruction formula
break;
case 4:
return $output; //test
break;
}
} // end word_wise()
Just want to return something to the php community
Jeff K
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[PHP] array_search function bugged update errata
sorry all for the mistake in the code; in this line 'array $input' should be string $input. function word_wise(array $input, int $what) //returns an array so it reads function word_wise(string $input, int $what) //returns an array this should be the only mistake. I added this stuff as an afterthought after all the code was working perfectly in my test code. best JK -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] mysql_fecth_array() and function call as parameter
Hi all,
I have wriiten a function that determines whether tomorrows date is
the first of the month or not then returns a SQL string based on
whether its the first of the month or not. According to my apache
error logs I get an error that says:
[Sun Mar 26 21:43:14 2006] [error] [client 192.168.0.2] PHP Warning:
mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in C:\\Program Files\\Apache
Group\\Apache2\\htdocs\\validation.php on line 331
I understand that this is a warning but I believe that it has
something to do with the fact that no output is being displayed. All
other mysql database outputs work fine.
This is the code that sets the query in the mysql_query parameter
$Month_query = mysql_query($this-determineMonth());
this is the code for determineMonth ()
function determineMonth()
{
$return = ;
$query1 = mysql_query(SELECT months FROM Month WHERE m_id =
month(curdate()));
$query2 = mysql_query(SELECT dayNum FROM Days WHERE dayNum =
day(curdate()));
$query3 = mysql_query(SELECT year FROM Year WHERE year =
year(curdate()));
switch ($query1)
{
case January:
case March:
case May:
case July:
case August:
case October:
case December:
if($query2 == 31)
$return = SELECT m_id, months FROM
Month WHERE m_id =
month(curdate())+1;
else
$return = SELECT m_id, months FROM
Month WHERE m_id = month(curdate());
break;
case February:
if ($query2 == 28 || $query2 == 29)
$return = SELECT m_id, months FROM
Month WHERE m_id =
month(curdate())+1;
else
$return = SELECT m_id, months FROM
Month WHERE m_id = month(curdate());
break;
case April:
case June:
case September:
case November:
if ($query2 == 30)
$return = SELECT m_id, months FROM
Month WHERE m_id =
month(curdate())+1;
else
$return = SELECT m_id, months FROM
Month WHERE m_id = month(curdate());
break;
}
return $return;
}
I hope I included everything. If not let me know and I'll post it
Paul
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Re: [PHP] mysql_fecth_array() and function call as parameter
Paul Goepfert wrote: Hi all, I have wriiten a function that determines whether tomorrows date is the first of the month or not then returns a SQL string based on whether its the first of the month or not. According to my apache error logs I get an error that says: [Sun Mar 26 21:43:14 2006] [error] [client 192.168.0.2] PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\\Program Files\\Apache Group\\Apache2\\htdocs\\validation.php on line 331 I understand that this is a warning but I believe that it has something to do with the fact that no output is being displayed. All other mysql database outputs work fine. This is the code that sets the query in the mysql_query parameter $Month_query = mysql_query($this-determineMonth()); Add this after your month_query call: echo mysql_error() . br/; It will tell you what's wrong with the query. I'd probably also print out the query: $qry = $this-determineMonth(); echo $qry . br/; and run it manually (if something does get returned). -- Postgresql php tutorials http://www.designmagick.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] mysql_fecth_array() and function call as parameter
I placed the echo statements in my code and I found out that my query was empty. I also went a step further and tested if I was getting the correct outputs from the queries that initially do at the beginning of the method. Instead of getting an output like March I got an output that says Resource id#9. I don't get it. I tested the sql statement in MySQL that I have on my computer. It worked, Why wouldn't it give the same output through mysql_query()? Paul On 3/26/06, Chris [EMAIL PROTECTED] wrote: Paul Goepfert wrote: Hi all, I have wriiten a function that determines whether tomorrows date is the first of the month or not then returns a SQL string based on whether its the first of the month or not. According to my apache error logs I get an error that says: [Sun Mar 26 21:43:14 2006] [error] [client 192.168.0.2] PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\\Program Files\\Apache Group\\Apache2\\htdocs\\validation.php on line 331 I understand that this is a warning but I believe that it has something to do with the fact that no output is being displayed. All other mysql database outputs work fine. This is the code that sets the query in the mysql_query parameter $Month_query = mysql_query($this-determineMonth()); Add this after your month_query call: echo mysql_error() . br/; It will tell you what's wrong with the query. I'd probably also print out the query: $qry = $this-determineMonth(); echo $qry . br/; and run it manually (if something does get returned). -- Postgresql php tutorials http://www.designmagick.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] mysql_fecth_array() and function call as parameter
Paul Goepfert wrote:
I placed the echo statements in my code and I found out that my query
was empty. I also went a step further and tested if I was getting the
correct outputs from the queries that initially do at the beginning of
the method. Instead of getting an output like March I got an output
that says Resource id#9. I don't get it. I tested the sql
statement in MySQL that I have on my computer. It worked, Why
wouldn't it give the same output through mysql_query()?
Ahh, didn't notice that.
$query1 = mysql_query(SELECT months FROM Month WHERE m_id =
month(curdate()));
$query2 = mysql_query(SELECT dayNum FROM Days WHERE dayNum =
day(curdate()));
$query3 = mysql_query(SELECT year FROM Year WHERE year = year(curdate()));
These return result resources (kinda like '$fp = fopen' doesn't return
the file - it returns a handle only), not the results. You need to do:
$query1_data = mysql_fetch_assoc($query1, 0, 0);
then
switch($query1_data) {
}
see http://www.php.net/mysql_query and
http://www.php.net/mysql_fetch_assoc for more info.
On 3/26/06, Chris [EMAIL PROTECTED] wrote:
Paul Goepfert wrote:
Hi all,
I have wriiten a function that determines whether tomorrows date is
the first of the month or not then returns a SQL string based on
whether its the first of the month or not. According to my apache
error logs I get an error that says:
[Sun Mar 26 21:43:14 2006] [error] [client 192.168.0.2] PHP Warning:
mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in C:\\Program Files\\Apache
Group\\Apache2\\htdocs\\validation.php on line 331
I understand that this is a warning but I believe that it has
something to do with the fact that no output is being displayed. All
other mysql database outputs work fine.
This is the code that sets the query in the mysql_query parameter
$Month_query = mysql_query($this-determineMonth());
Add this after your month_query call:
echo mysql_error() . br/;
It will tell you what's wrong with the query. I'd probably also print
out the query:
$qry = $this-determineMonth();
echo $qry . br/;
and run it manually (if something does get returned).
--
Postgresql php tutorials
http://www.designmagick.com/
--
Postgresql php tutorials
http://www.designmagick.com/
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Re: [PHP] opening .Z archive with gzopen
Chris a écrit : nicolas figaro wrote: Hi I'm using php 4.3.11 on linux RH9. I'd like to open a .Z archive with gzopen. I looked at the options of gzopen, but I'm not sure there is a way to specify the compression used by compress/uncompress. Has anyone ever tried and done this yet ? (without a system(uncompress $file) Since a compressed file isn't the same as a gzipp'ed file I'd guess this won't work. This might: http://pear.php.net/package/File_Archive but it doesn't mention .Z files specifically. Hi and thanks for the answer. The package contains an uncompress function, but the goal was to open the file withuout uncompressing it (like the zcat command or gzopen function). Or I can ask the guys who generates .Z files to generate .gz instead. N F -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
