Re: [PHP-DB] Undefined Constant
Hi, Thank you all for your quick help, it was just a Youve been up too long working on this and it is the middle of the night mistakes. Once I put quotes around my constant, everything worked. Thanks again, Tyler - Original Message - From: Jason Wong [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Tuesday, November 26, 2002 11:39 PM Subject: Re: [PHP-DB] Undefined Constant On Wednesday 27 November 2002 15:24, Tyler Whitesides wrote: Hi, Although I have been working with PHP for a while, working with Databases is something that I am only so so at. This code is supposed to grab the price from the database given the auto_increment id is inside of $casing. I am trying to get the $row[price] to echo so that Javascript recieves a string, converts it to an integer, and later on in the code, gives the user a total. Any ideas? Use: $row['price'] -- Jason Wong - Gremlins Associates - www.gremlins.biz Open Source Software Systems Integrators * Web Design Hosting * Internet Intranet Applications Development * /* The world really isn't any worse. It's just that the news coverage is so much better. */ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: MySQL/PHP Iterative Tree
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Greetings John On 26 November 2002 at 18:54:04 -0500 (which was 23:54 where I live) John W. Holmes wrote and made these points This is OK for some simple trees, however it does have some limitations. (1) Adding another tree / series of nodes into the middle if the set. Not if you can define what the absolute max size of a node will be. Most of my applications I can not, and this is a limitation that I would never want to impose on on my apps. If you can then good. (2) A child node that belongs to more than one parent. I give a certain company the node numbers of 6-5006. Now assign that left and right pair a unique id. Now you can have multiple people with that same ID. I might have misunderstood you but sorry but this is a big no no in my all my apps. Every item must have a unique ID. This is a fundamental basis of all my systems. - -- See you in Cyber space, ___ David |David Elliott| Software Engineer| _| [EMAIL PROTECTED] | PGP Key ID 0x650F4534 | | Would it help if I got out and pushed? -Princess Leia | -BEGIN PGP SIGNATURE- Version: 6.5.8ckt http://www.ipgpp.com/ iQA/AwUBPeSW6PmK8eZlD0U0EQJ+HwCg4UQSEN5ee3i5kUiiM7aBsDH0cl4AmwbM /l5gv6cPCULmf9KUWJ46LY2a =4Hwb -END PGP SIGNATURE- -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] MySQL: charset
Hi How can i change default charset for mysql database. I want to change default charset on cp1251 just for one database 'dblibrary'. I have seen something like mysql --default-character-set=cp1251, but i dont know does it work? Thanks -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Re: MySQL: charset
Radovan Radic [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... Hi How can i change default charset for mysql database. I want to change default charset on cp1251 just for one database 'dblibrary'. I have seen something like mysql --default-character-set=cp1251, but i dont know does it work? Thanks I thought i could do this somehow when connecting from PHP because i need other databases to be in latin charset. MySQL 3.23.41 -nt PHP 4.2.3 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Connecting to MS SQL 7
A. I think the php3_ isn't a good sign. B. You should remove whatever you found on the web and downloaded. C. Copy all the DLL's in the dll folder (under your PHP folder) to C:\WINNT\SYSTEM32 D. Re-copy the php.ini-recommended from your PHP folder to C:\WINNT\php.ini and see if it stops looking for php3. On Tue, 2002-11-26 at 18:36, Mark Farmer wrote: I have successfully installed PHP 4.2.3 on a Win NT 4 server, running IIS 4, and can execute PHP functions now. However, when I try to connect to an MS SQL 7 database on the same box, I receive this error: PHP Warning: Unable to load dynamic library './php3_mssql70.dll' - %1 is not a valid Windows NT application, and my mssql_connect statment fails. What I had done was to find a copy of php3_mssql70.dll on the web and copy it onto the server -- various faqs told me this needed to happen in order for database calls to work. Why isn't this dll being recognized as a valid Windows NT application, then? Thank you. _ STOP MORE SPAM with the new MSN 8 and get 2 months FREE* http://join.msn.com/?page=features/junkmail -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- Adam Voigt ([EMAIL PROTECTED]) The Cryptocomm Group My GPG Key: http://64.238.252.49:8080/adam_at_cryptocomm.asc signature.asc Description: This is a digitally signed message part
Re: [PHP-DB] Resource id #2
Umm, he is putting them into an array, I quote:
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
See the while condition?
On Wed, 2002-11-27 at 06:03, Chris Barnes wrote:
You need to put your $result into an array. you can use:
$result_array = mysql_fetch_array($result);
then, if you know the field names in the array, print them like so:
echo $result_array[field1];
echo $result_array[field2];
or if you dont know their names you can refer to their position numbers
starting from 0 e.g.
echo $result_array[0];
echo $result_array[1];
using the position numbers you could put together a quick script to
crawl through the array and print all the fields with a few lines of
code.
On Wed, 2002-11-27 at 05:09, The Cossins Fam wrote:
Hello.
I am using MySQL as a database for a departmental library. I have written
a quick search script, but keep getting resource id #2 as a result to my
search. I have read the online documentation for the
mysql_fetch_array() function and must say, I don't see that I'm missing
anything. However, I've only started programming, much less working with
PHP, so perhaps someone can help me out. Here's my code:
?
$quickSearch = mcse;
$table1 = Books;
$table2 = BookList;
$table3 = BoxSet;
$table4 = Category;
$table5 = Publisher;
$table6 = AuthUsers;
$table7 = CD;
$connection = mysql_connect(localhost, root) or die(Couldn't connect
to the library database.);
$db_select = mysql_select_db(library, $connection) or die(Couldn't
select the library database.);
$search = SELECT * FROM $table1 LEFT JOIN $table2 ON Books.BookListID =
BookList.BookListID
LEFT JOIN $table3 ON Books.BoxSetID = BoxSet.BoxSetID
LEFT JOIN $table4 ON Books.CategoryID = Category.CategoryID
LEFT JOIN $table5 ON Books.PublisherID = Publisher.PublisherID
LEFT JOIN $table6 ON Books.auID = AuthUsers.auID
LEFT JOIN $table7 ON Books.CD = CD.CD_ID
WHERE Books.Title LIKE \%'$quickSearch'%\
OR Books.Author LIKE \%'$quickSearch'%\
OR Books.ISBN LIKE \%'$quickSearch'%\
OR BookList.dbase LIKE \%'$quickSearch'%\
OR BookList.dbase_user LIKE \%'$quickSearch'%\
OR BoxSet.BoxSet LIKE \%'$quickSearch'%\
OR Category.Category LIKE \%'$quickSearch'%\
OR Category.Sub_category LIKE \%'$quickSearch'%\
OR Publisher.Publisher LIKE \%'$quickSearch'%\;
$result = mysql_query($search, $connection) or die(Couldn't search the
library.);
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
?
I then have some HTML to display the result of the search. I don't
receive any error messages - I just see an empty table from the HTML
code I wrote. I added an echo of the $result to find the resouce id
#2.
Thanks for any help you can provide.
--joel
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[PHP-DB] PHP4, Apache 1.3.x, and mod ssl install document?
Hello all, I have a FreeBSD server which has been updated to 4.7 STABLE. I have updated my ports collection and installed mysql server. Does anyone have an updated doc on how to install apache with PHP4 and modssl support either from ports collection or from source. I would greatly appreciate it. Thanks Ron -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] PHP4, Apache 1.3.x, and mod ssl install document?
Off the top of my head (general guidelines, don't use word for word): mkdir /usr/local/apache cd /usr/local/apache wget http://www.apache.org/dist/httpd/apache_1.3.27.tar.gz tar -zxf * cd apache_1.3.27 ./configure --prefix=/usr/local/apache_1.3.27 mkdir /usr/local/modssl cd /usr/local/modssl wget http://www.modssl.org/source/mod_ssl-2.8.12-1.3.27.tar.gz tar -zxf * cd mod_ssl-2.8.12-1.3.27 ./configure --with-apache=/usr/local/apache/apache_1.3.27 cd /usr/local/apache/apache_1.3.27 ./configure --prefix=/usr/local/apache_1.3.27 --enable-module=ssl --enable-module=so make make certificate type=test (OPTIONAL) make install mkdir /usr/local/php cd /usr/local/php lynx http://www.php.net/do_download.php?mr=http%3A%2F%2Fwww.php.net%2Fdf=php-4.2.3.tar.gz (Save the file with lynx) tar -zxf * cd php_4.2.3 ./configure --with-apxs=/usr/local/apache_1.3.27/bin/apxs --with-mysql=/usr make make install Add the line: application/x-httpd-php .php To somewhere near the bottom of your httpd.conf and while your in httpd.conf configure the values you like. On Wed, 2002-11-27 at 10:38, RClark wrote: Hello all, I have a FreeBSD server which has been updated to 4.7 STABLE. I have updated my ports collection and installed mysql server. Does anyone have an updated doc on how to install apache with PHP4 and modssl support either from ports collection or from source. I would greatly appreciate it. Thanks Ron -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- Adam Voigt ([EMAIL PROTECTED]) The Cryptocomm Group My GPG Key: http://64.238.252.49:8080/adam_at_cryptocomm.asc signature.asc Description: This is a digitally signed message part
Re: [PHP-DB] PHP4, Apache 1.3.x, and mod ssl install document?
Actually, it's easier: cd /usr/ports/www/apache13-modssl make install cd /usr/ports/www/mod_php4 make install apachectl start (or restart) Peter On Wed, 27 Nov 2002, RClark wrote: Hello all, I have a FreeBSD server which has been updated to 4.7 STABLE. I have updated my ports collection and installed mysql server. Does anyone have an updated doc on how to install apache with PHP4 and modssl support either from ports collection or from source. I would greatly appreciate it. Thanks Ron -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php --- Peter Beckman Internet Guy [EMAIL PROTECTED] http://www.purplecow.com/ --- -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Resource id #2
But what do all those $row['fieldname'} rows do? Call me ignorant
(you wouldn't be the first), but a statement that simply has a
variable name doesn't DO anything. Should these have echos in front
of them?
--- Adam Voigt [EMAIL PROTECTED] wrote:
Umm, he is putting them into an array, I quote:
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
See the while condition?
On Wed, 2002-11-27 at 06:03, Chris Barnes wrote:
You need to put your $result into an array. you can use:
$result_array = mysql_fetch_array($result);
then, if you know the field names in the array, print them like
so:
echo $result_array[field1];
echo $result_array[field2];
or if you dont know their names you can refer to their position
numbers
starting from 0 e.g.
echo $result_array[0];
echo $result_array[1];
using the position numbers you could put together a quick script
to
crawl through the array and print all the fields with a few lines
of
code.
On Wed, 2002-11-27 at 05:09, The Cossins Fam wrote:
Hello.
I am using MySQL as a database for a departmental library. I
have written
a quick search script, but keep getting resource id #2 as a
result to my
search. I have read the online documentation for the
mysql_fetch_array() function and must say, I don't see that I'm
missing
anything. However, I've only started programming, much less
working with
PHP, so perhaps someone can help me out. Here's my code:
?
$quickSearch = mcse;
$table1 = Books;
$table2 = BookList;
$table3 = BoxSet;
$table4 = Category;
$table5 = Publisher;
$table6 = AuthUsers;
$table7 = CD;
$connection = mysql_connect(localhost, root) or
die(Couldn't connect
to the library database.);
$db_select = mysql_select_db(library, $connection) or
die(Couldn't
select the library database.);
$search = SELECT * FROM $table1 LEFT JOIN $table2 ON
Books.BookListID =
BookList.BookListID
LEFT JOIN $table3 ON Books.BoxSetID = BoxSet.BoxSetID
LEFT JOIN $table4 ON Books.CategoryID =
Category.CategoryID
LEFT JOIN $table5 ON Books.PublisherID =
Publisher.PublisherID
LEFT JOIN $table6 ON Books.auID = AuthUsers.auID
LEFT JOIN $table7 ON Books.CD = CD.CD_ID
WHERE Books.Title LIKE \%'$quickSearch'%\
OR Books.Author LIKE \%'$quickSearch'%\
OR Books.ISBN LIKE \%'$quickSearch'%\
OR BookList.dbase LIKE \%'$quickSearch'%\
OR BookList.dbase_user LIKE \%'$quickSearch'%\
OR BoxSet.BoxSet LIKE \%'$quickSearch'%\
OR Category.Category LIKE \%'$quickSearch'%\
OR Category.Sub_category LIKE \%'$quickSearch'%\
OR Publisher.Publisher LIKE \%'$quickSearch'%\;
$result = mysql_query($search, $connection) or die(Couldn't
search the
library.);
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
?
I then have some HTML to display the result of the search. I
don't
receive any error messages - I just see an empty table from the
HTML
code I wrote. I added an echo of the $result to find the
resouce id
#2.
Thanks for any help you can provide.
--joel
_
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[PHP-DB] is innodb faster than MyISAM???
i just read a benchmark on innodb.org http://www.innodb.com/bench.html that says under the innodb vs MyISAM section... MyISAM is the default table type used in MySQL and InnoDB is a table type supporting transactions in MySQL. I wrote a Perl program which inserts 100 000 rows to a table with 3 integer columns and two indexes. Then another Perl program fetches each row either through a secondary index or the primary key. The Perl programs for each test are at the end of this web page. I ran the tests on a Linux 2-CPU Xeon 450 MHz. The times below are wall clock times. The results were: InnoDBMyISAM 100 000 inserts25 s. 40 s. 100 000 selects on primary key57 s. 58 s. 100 000 selects on secondary key 68 s. 95 s. Is this true? For regular inserts that seems cool - obviously transactions would take a little more, but hey, why bother with MyISAM then? dave -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Resource id #2
They don't do anything, but my point was, he said that what he pulled
from the DB needed to be put into an array, and I was pointing out, it
already was.
On Wed, 2002-11-27 at 12:23, Mark wrote:
But what do all those $row['fieldname'} rows do? Call me ignorant
(you wouldn't be the first), but a statement that simply has a
variable name doesn't DO anything. Should these have echos in front
of them?
--- Adam Voigt [EMAIL PROTECTED] wrote:
Umm, he is putting them into an array, I quote:
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
See the while condition?
On Wed, 2002-11-27 at 06:03, Chris Barnes wrote:
You need to put your $result into an array. you can use:
$result_array = mysql_fetch_array($result);
then, if you know the field names in the array, print them like
so:
echo $result_array[field1];
echo $result_array[field2];
or if you dont know their names you can refer to their position
numbers
starting from 0 e.g.
echo $result_array[0];
echo $result_array[1];
using the position numbers you could put together a quick script
to
crawl through the array and print all the fields with a few lines
of
code.
On Wed, 2002-11-27 at 05:09, The Cossins Fam wrote:
Hello.
I am using MySQL as a database for a departmental library. I
have written
a quick search script, but keep getting resource id #2 as a
result to my
search. I have read the online documentation for the
mysql_fetch_array() function and must say, I don't see that I'm
missing
anything. However, I've only started programming, much less
working with
PHP, so perhaps someone can help me out. Here's my code:
?
$quickSearch = mcse;
$table1 = Books;
$table2 = BookList;
$table3 = BoxSet;
$table4 = Category;
$table5 = Publisher;
$table6 = AuthUsers;
$table7 = CD;
$connection = mysql_connect(localhost, root) or
die(Couldn't connect
to the library database.);
$db_select = mysql_select_db(library, $connection) or
die(Couldn't
select the library database.);
$search = SELECT * FROM $table1 LEFT JOIN $table2 ON
Books.BookListID =
BookList.BookListID
LEFT JOIN $table3 ON Books.BoxSetID = BoxSet.BoxSetID
LEFT JOIN $table4 ON Books.CategoryID =
Category.CategoryID
LEFT JOIN $table5 ON Books.PublisherID =
Publisher.PublisherID
LEFT JOIN $table6 ON Books.auID = AuthUsers.auID
LEFT JOIN $table7 ON Books.CD = CD.CD_ID
WHERE Books.Title LIKE \%'$quickSearch'%\
OR Books.Author LIKE \%'$quickSearch'%\
OR Books.ISBN LIKE \%'$quickSearch'%\
OR BookList.dbase LIKE \%'$quickSearch'%\
OR BookList.dbase_user LIKE \%'$quickSearch'%\
OR BoxSet.BoxSet LIKE \%'$quickSearch'%\
OR Category.Category LIKE \%'$quickSearch'%\
OR Category.Sub_category LIKE \%'$quickSearch'%\
OR Publisher.Publisher LIKE \%'$quickSearch'%\;
$result = mysql_query($search, $connection) or die(Couldn't
search the
library.);
while ($row = mysql_fetch_array($result)) {
$row['Books.Title'];
$row['Books.Author'];
$row['Books.ISBN'];
$row['BookList.dbase'];
$row['BookList.dbase_user'];
$row['BoxSet.BoxSet'];
$row['Category.Category'];
$row['Category.Sub_category'];
$row['Publisher.Publisher'];
$row['AuthUsers.email'];
}
?
I then have some HTML to display the result of the search. I
don't
receive any error messages - I just see an empty table from the
HTML
code I wrote. I added an echo of the $result to find the
resouce id
#2.
Thanks for any help you can provide.
--joel
_
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***
You can't demand something as a
[PHP-DB] imap_search problems
I'm writing a command line script to go through a mailbox on the file
system. PHP the binary is compiled with imap, I've confirmed by using
imap_headers to get the array of the 10,051 messages and printed the array
out.
Here's my code:
#!/usr/local/bin/php -q
?php
$id = $_SERVER[argv][1];
ini_set(max_execution_time,9000);
list($usec, $sec) = explode( ,microtime());
$start = ((float)$usec + (float)$sec);
$mbox = imap_open(/home/beckman/public_html/spamtracker/a-copy-of-spam, , )
or die(can't open);
print Finding '$id' (Executing imap_search($mbox,'$id'))\n;
$matches = imap_search($mbox, '$id');
print_r($matches);
#while(list($key,$val)=each($matches)) {
# echo $key: $val\n;
#}
imap_close($mbox);
?
Here's what I get when searching for a variety of different possibilities (no results
being returned in $matches):
Finding 'ALL fat' (Executing imap_search(Resource id #1,'ALL fat'))
Finding 'ALL fat' (Executing imap_search(Resource id #1,'ALL fat'))
Finding 'BODY fat' (Executing imap_search(Resource id #1,'BODY fat'))
Then I put SE_UID as the last option in imap_search:
Finding 'BODY fat' (Executing imap_search(Resource id #1,'BODY fat',SE_UID))
Finding 'TO [EMAIL PROTECTED]' (Executing imap_search(Resource id #1,'TO
[EMAIL PROTECTED]',SE_UID))
I get nothing in $matches ever. What is going on? Am I doing it wrong?
I've read the documentation and the contributed notes and copied and pasted
replacing the search strings and continue to fail. Argh.
I know the word fat is in the body of more than 50 messages, as well as
in the headers, as there are 10,051 pieces of spam I'm trying to parse.
Peter
---
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Re: [PHP-DB] imap_search problems
Don't you hate it when, right after you post to the list, you figure out
your problem? :-)
I removed the quotes around $id after using:
$err = imap_errors();
print_r($err);
to determine what the problem was.
Anyway, Happy Thanksgiving all, happy troubleshooting and never forget to use
the debugging tools the nice people at PHP gave you for Christmas.
Beckman
On Wed, 27 Nov 2002, Peter Beckman wrote:
I'm writing a command line script to go through a mailbox on the file
system. PHP the binary is compiled with imap, I've confirmed by using
imap_headers to get the array of the 10,051 messages and printed the array
out.
Here's my code:
#!/usr/local/bin/php -q
?php
$id = $_SERVER[argv][1];
ini_set(max_execution_time,9000);
list($usec, $sec) = explode( ,microtime());
$start = ((float)$usec + (float)$sec);
$mbox = imap_open(/home/beckman/public_html/spamtracker/a-copy-of-spam, ,
) or die(can't open);
print Finding '$id' (Executing imap_search($mbox,'$id'))\n;
$matches = imap_search($mbox, '$id');
print_r($matches);
#while(list($key,$val)=each($matches)) {
# echo $key: $val\n;
#}
imap_close($mbox);
?
Here's what I get when searching for a variety of different possibilities (no
results being returned in $matches):
Finding 'ALL fat' (Executing imap_search(Resource id #1,'ALL fat'))
Finding 'ALL fat' (Executing imap_search(Resource id #1,'ALL fat'))
Finding 'BODY fat' (Executing imap_search(Resource id #1,'BODY fat'))
Then I put SE_UID as the last option in imap_search:
Finding 'BODY fat' (Executing imap_search(Resource id #1,'BODY fat',SE_UID))
Finding 'TO [EMAIL PROTECTED]' (Executing imap_search(Resource id #1,'TO
[EMAIL PROTECTED]',SE_UID))
I get nothing in $matches ever. What is going on? Am I doing it wrong?
I've read the documentation and the contributed notes and copied and pasted
replacing the search strings and continue to fail. Argh.
I know the word fat is in the body of more than 50 messages, as well as
in the headers, as there are 10,051 pieces of spam I'm trying to parse.
Peter
---
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[EMAIL PROTECTED] http://www.purplecow.com/
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Re: FW: [PHP-DB] PHP and PostrgeSQL tutorial
I have tried the designmagick tutorial and cant get ti to post data to the DB. It tells me I have posted data and a new row appears, but only th eid field has any data. Are there any other tutorials or am I doing something wrong? I am cutting and pasting from the tutorial. I do have to make adjustments to the web page talking to the DB server (the connection). But everything else is the same. On Wed, 10 Jan 2001 17:25:28 +0800, Ronnie Esguerra wrote: http://designmagick.50megs.com Ronnie .php.net -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] 'Remote' data
Hi I need to create function which would read a data (mysql) from the table from the other page, put this data in my MySQL table and return them on my page. My problem is that I have no idea how to get that data. Please give me a hint or maybe you know some articles about that, Thanks Bartosz -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Session questions
Hi there everyone, I have a login system which uses sessions, and when people login it puts the date they logged in in their profile and moves their last login date over to another field, that works fine, but how can I check if someone has logged out via code? In other words, if someone does what they want and then just closes the browser without logging out no record is saved, how can I tell whether they have logged out / gone from the website ? I need to know this for security. Which brings me to my next question :-) How would I make it so only 1 person can login at any time with any given username/password combo? I know I have to track the sessions, but not sure how :-( Any help would really be appreciated. Chris -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Session questions
I have a login system which uses sessions, and when people login it puts the date they logged in in their profile and moves their last login date over to another field, that works fine, but how can I check if someone has logged out via code? In other words, if someone does what they want and then just closes the browser without logging out no record is saved, how can I tell whether they have logged out / gone from the website ? I need to know this for security. You can't know for sure. It's a limitation of HTTP. What you do is assume that if after X minutes, the user hasn't done anything, then they have left. If you're using sessions, the files will be cleaned up after X minutes, anyhow, and they will lose their login. Which brings me to my next question :-) How would I make it so only 1 person can login at any time with any given username/password combo? I know I have to track the sessions, but not sure how :-( Set a flag in the database that says whether that user is logged in. If that username tries to log in again, set an error. The problem you'll run into with this, though, is you'll have to implement a cleanup system to unset the flag after X minutes of inactivity. Also, if I accidentally close my window and then go back to log in, it won't let me because the flag is still set until your cleanup program unsets it. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Session questions
Here's a couple of suggestions: 1) You can use javascript to trap the window's closure and create a new window that simply calls one of your scripts that closes the session. This is not 100% foolproof, however, so you need to come up with a backup plan, like closing sessions automatically with a batch process in the backend. 2) That's easy enough--just store a unique token in their session and in the database when they log in. If, when they come to you with a particular user id they do not have the right token, then that means that either (a) somebody is trying to steal a session or (b) two people have signed on with the same username, in which case you can decide who gets the boot. This, too, is not a completely foolproof method (particularly if you don't use SSL), but it's a good starting point. Hope this helps. Cheers, Marco -- php|architect - The magazine for PHP Professionals The first monthly worldwide magazine dedicated to PHP programmers Come visit us at http://www.phparch.com! On Wed, 2002-11-27 at 20:19, Chris Payne wrote: Hi there everyone, I have a login system which uses sessions, and when people login it puts the date they logged in in their profile and moves their last login date over to another field, that works fine, but how can I check if someone has logged out via code? In other words, if someone does what they want and then just closes the browser without logging out no record is saved, how can I tell whether they have logged out / gone from the website ? I need to know this for security. Which brings me to my next question :-) How would I make it so only 1 person can login at any time with any given username/password combo? I know I have to track the sessions, but not sure how :-( Any help would really be appreciated. Chris -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] foreach loop from MySQL select query to HTML select list
Hi,
I'm trying to pick up data from my MySQL database, use a foreach loop to
extact the data from the result put that data into a select list.
I have tried the following code with without the $row = ... line, both
unsucessfully:
?PHP
$db = mysql_connect(myhost, username, );
mysql_select_db(mydatabase, $db);
## FETCH INFO
$sql_dept = SELECT dept FROM content GROUP BY dept;
$sql_deptsub = SELECT deptsub FROM content GROUP BY deptsub;
mysql_close($db);
?
select name=dept
option-select-/option
?PHP
$result = mysql_query($sql_dept, $db);
$row = mysql_fetch_array($result);
foreach ($row as $value) {echo option
value=\$value\$value/option\n;}
?
/select
I get the following error message in my HTML output code:
bWarning/b: 1 is not a valid MySQL-Link resource in
b/home/.../public_html/content.php/b on line b48/bbr
br
bWarning/b: Invalid argument supplied for foreach() in
b/home/.../public_html/content.php/b on line b49/bbr
Line 48 is the line begining with: foreach...
Any help would be appreciated.
Thanks,
Gav
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Re: [PHP-DB] foreach loop from MySQL select query to HTML selectlist
On Wed, 2002-11-27 at 23:33, Gavin Amm wrote:
Hi,
I'm trying to pick up data from my MySQL database, use a foreach loop to
extact the data from the result put that data into a select list.
I have tried the following code with without the $row = ... line, both
unsucessfully:
?PHP
$db = mysql_connect(myhost, username, );
mysql_select_db(mydatabase, $db);
## FETCH INFO
$sql_dept = SELECT dept FROM content GROUP BY dept;
$sql_deptsub = SELECT deptsub FROM content GROUP BY deptsub;
mysql_close($db);
How can run a query when it's closed? move this to after the select and
see if it works.
?
select name=dept
option-select-/option
?PHP
$result = mysql_query($sql_dept, $db);
$row = mysql_fetch_array($result);
foreach ($row as $value) {echo option
value=\$value\$value/option\n;}
?
/select
Put your mysql_close here if you want to close it.
John Coder
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RE: [PHP-DB] foreach loop from MySQL select query to HTML select list
good point! whoops.
ok, i've moved the $result_dept $result_deptsub queries,
now i'm getting 2 of the same entries in my html code:
select name=dept
option-select-/option
option value=auditaudit/option
option value=auditaudit/option
/select
any thoughts?
Gav
-Original Message-
From: John Coder [mailto:[EMAIL PROTECTED]]
Sent: Thursday, 28 November 2002 3:44 PM
To: Gavin Amm
Cc: Php-Db (E-mail)
Subject: Re: [PHP-DB] foreach loop from MySQL select query to HTML
select list
On Wed, 2002-11-27 at 23:33, Gavin Amm wrote:
Hi,
I'm trying to pick up data from my MySQL database, use a foreach loop to
extact the data from the result put that data into a select list.
I have tried the following code with without the $row = ... line, both
unsucessfully:
?PHP
$db = mysql_connect(myhost, username, );
mysql_select_db(mydatabase, $db);
## FETCH INFO
$sql_dept = SELECT dept FROM content GROUP BY dept;
$sql_deptsub = SELECT deptsub FROM content GROUP BY deptsub;
mysql_close($db);
How can run a query when it's closed? move this to after the select and
see if it works.
?
select name=dept
option-select-/option
?PHP
$result = mysql_query($sql_dept, $db);
$row = mysql_fetch_array($result);
foreach ($row as $value) {echo option
value=\$value\$value/option\n;}
?
/select
Put your mysql_close here if you want to close it.
John Coder
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9413 2944 and destroy the original message. Thank you.
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should be scanned by an up to-date Anti Virus programme automatically by
your system. It is the responsibility of the recipient to ensure that all
Emails and any attachments are cleared of Viruses before opening. KSG can
not accept any responsibility for viruses that maybe contained here in.
Please advise KSG by return Email if you believe any Email sent by our
system may contain a virus. It should be noted that most Anti Virus
programmes can not scan encrypted file attachments (example - documents
saved with a password). Thus extra care should be taken when opening these
files.
Liability limited by the Accountants Scheme, approved under the Professional
Standards Act 1994 (NSW).
Level 4
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Ph: +61 2 9413 2944 Fax: +61 2 9413 9901
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RE: [PHP-DB] foreach loop from MySQL select query to HTML select list
ok,
i've looked around the web a bit more it would appear you can't just use a
foreach function, apparently you need to use a while{foreach{}} structure. i
was hoping you could just use the foreach function?
so this is the code now:
select name=deptsub_select
option-select-/option
?PHP
while ($row = mysql_fetch_array($result_deptsub)){
foreach ($row as $value) {
echo option value=\$value\$value/option\n;
}
}
?
/select
but now the output HTML code is doubling up:
select name=dept
option-select-/option
option value=auditaudit/option
option value=auditaudit/option
option value=consultingconsulting/option
option value=consultingconsulting/option
...
/select
cheers,
Gav
This e-mail and any attachments are intended solely for the named addressee,
are confidential and may contain legally privileged information.
The copying or distribution of them or of any information they contain, by
anyone other than the addressee, is prohibited. If you received this e-mail
in error, please notify us immediately by return e-mail or telephone +61 2
9413 2944 and destroy the original message. Thank you.
As Email is subject to viruses we advise that all Emails and any attachments
should be scanned by an up to-date Anti Virus programme automatically by
your system. It is the responsibility of the recipient to ensure that all
Emails and any attachments are cleared of Viruses before opening. KSG can
not accept any responsibility for viruses that maybe contained here in.
Please advise KSG by return Email if you believe any Email sent by our
system may contain a virus. It should be noted that most Anti Virus
programmes can not scan encrypted file attachments (example - documents
saved with a password). Thus extra care should be taken when opening these
files.
Liability limited by the Accountants Scheme, approved under the Professional
Standards Act 1994 (NSW).
Level 4
54 Neridah StreetPO Box 1290
CHATSWOOD NSW 2067 CHATSWOOD NSW 2057
Ph: +61 2 9413 2944 Fax: +61 2 9413 9901
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Re: [PHP-DB] foreach loop from MySQL select query to HTML select list
On Thursday 28 November 2002 14:21, Gavin Amm wrote:
ok,
i've looked around the web a bit more it would appear you can't just use
a foreach function, apparently you need to use a while{foreach{}}
structure. i was hoping you could just use the foreach function?
so this is the code now:
select name=deptsub_select
option-select-/option
?PHP
while ($row = mysql_fetch_array($result_deptsub)){
foreach ($row as $value) {
echo option value=\$value\$value/option\n;
}
}
?
/select
but now the output HTML code is doubling up:
select name=dept
option-select-/option
option value=auditaudit/option
option value=auditaudit/option
option value=consultingconsulting/option
option value=consultingconsulting/option
...
/select
mysql_fetch_array() by default returns an array that's associative AND numeric
(print_r($row) to see what it looks like).
Use mysql_fetch_assoc() instead.
--
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Open Source Software Systems Integrators
* Web Design Hosting * Internet Intranet Applications Development *
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