RE: [PHP] MySQL and Array's REALLY simple question but I'm not GETTING it .. Ugh..
OK, this works.. This is wonderful.. But I don't get it.. I assigned two
separate variables to that array function.. I don't understand why this
works now, but thanks a lot for the help.
I hope I don't inadvertently run into this again...
-steve
$sql2 = mysql_connect(localhost, eweb, dbfun)
or die(Could not connect using default username and password LINE
14 BR);
mysql_select_db(actionregs, $sql2);
$top_level = mysql_query(SELECT * FROM williams, $sql2)
or die(Could not do query sql1 to find username. Link might not
have been made. What's up? LINE 19 BR);
// $sql2_results = mysql_fetch_array($top_level);
$query_sql2_rows = mysql_num_rows($top_level);
$rows = 0;
echo $query_sql2_rows;
print table style=\font-family:Verdana; font-size:10pt\ border=0
cellpadding=4 width=90%;
print tr bgcolor=\#c0c0c0\th width=50Action ID/thth
width=100Owner/thth width=250Technology/ththSummary/th/tr;
for ($counter=0; $counter $query_sql2_rows; $counter++) {
$tabledata = mysql_fetch_array($top_level);
echo td$tabledata[0]/td;
echo td$tabledata[6]/td;
echo td$tabledata[2]/td;
echo td$tabledata[3]/td;
echo /tr;
}
print /table;
-Original Message-
From: Brad Bonkoski [mailto:[EMAIL PROTECTED]]
Sent: Friday, September 06, 2002 3:34 PM
To: Steve Gaas
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] MySQL and Array's REALLY simple question but I'm not
GETTING it .. Ugh..
Well, when you run the command:
$sql2_results = mysql_fetch_array($top_level);
The first time, it automatically increments the result...
so you fetch the data bu do nothing with it...
so when you get to your loop, you are already at the second entry in
your database..
So, remove the first instance of that and see how it works..
HTH
-Brad
Steve Gaas wrote:
Can anyone tell me what's wrong with my code? All I get output from this
is
the LAST row of data from my Database. There are 2 rows, how do I make it
pull data from all of the rows? It's not going through the loop like it
should I need to be able to tell the mysql_fetch_array which row I
want
in each it iteration of the for loop. The For loop increments counter, but
there is no syntax to add $counter to the result_type. The same goes with
fetch_row...
What am I forgetting!!?? I know it's something simple
***
$sql2 = mysql_connect(localhost, eweb, dbfun)
or die(Could not connect BR);
mysql_select_db(actionregs, $sql2);
$top_level = mysql_query(SELECT * FROM williams, $sql2)
or die(Could not do query BR);
$sql2_results = mysql_fetch_array($top_level);
$query_sql2_rows = mysql_num_rows($top_level);
echo $query_sql2_rows;
// echo's 2 rows
print table style=\font-family:Verdana; font-size:10pt\ border=0
cellpadding=4 width=90%;
print tr bgcolor=\#c0c0c0\th width=50Action ID/thth
width=100Owner/thth width=250Technology/ththSummary/th/tr;
for ($counter=0; $counter $query_sql2_rows; $counter++) {
$tabledata = mysql_fetch_array($top_level);
echo td$tabledata[0]/td;
echo td$tabledata[6]/td;
echo td$tabledata[2]/td;
echo td$tabledata[3]/td;
echo /tr;
}
print /table;
Thanks.
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Re: [PHP] MySQL and Array's REALLY simple question but I'm notGETTING it .. Ugh..
Hello, Steve,
When you call mysql_fetch_array the first time, you are accessing the data
from the first row of your results and then moving the pointer to the next
row. So, when you call mysql_fetch_array the second time, it is already
starting with the second row of your results. Try something like this
instead:
?
$sql2 = mysql_connect(localhost, eweb, dbfun)
or die(Could not connect BR);
mysql_select_db(actionregs, $sql2);
$top_level = mysql_query(SELECT * FROM williams, $sql2)
or die(Could not do query BR);
print table style=\font-family:Verdana; font-size:10pt\ border=0
cellpadding=4 width=90%;
print tr bgcolor=\#c0c0c0\th width=50Action ID/thth
width=100Owner/thth width=250Technology/ththSummary/th/tr;
while ($tabledata = mysql_fetch_array($top_level)) {
echo tr;
echo td$tabledata[0]/td;
echo td$tabledata[6]/td;
echo td$tabledata[2]/td;
echo td$tabledata[3]/td;
echo /tr;
}
print /table;
?
You may need to pull the table data array items into variables before
echoing them. Does this work?
HTH!
Jed
On the threshold of genius, Steve Gaas wrote:
Can anyone tell me what's wrong with my code? All I get output from this is
the LAST row of data from my Database. There are 2 rows, how do I make it
pull data from all of the rows? It's not going through the loop like it
should I need to be able to tell the mysql_fetch_array which row I want
in each it iteration of the for loop. The For loop increments counter, but
there is no syntax to add $counter to the result_type. The same goes with
fetch_row...
What am I forgetting!!?? I know it's something simple
***
$sql2 = mysql_connect(localhost, eweb, dbfun)
or die(Could not connect BR);
mysql_select_db(actionregs, $sql2);
$top_level = mysql_query(SELECT * FROM williams, $sql2)
or die(Could not do query BR);
$sql2_results = mysql_fetch_array($top_level);
$query_sql2_rows = mysql_num_rows($top_level);
echo $query_sql2_rows;
// echo's 2 rows
print table style=\font-family:Verdana; font-size:10pt\ border=0
cellpadding=4 width=90%;
print tr bgcolor=\#c0c0c0\th width=50Action ID/thth
width=100Owner/thth width=250Technology/ththSummary/th/tr;
for ($counter=0; $counter $query_sql2_rows; $counter++) {
$tabledata = mysql_fetch_array($top_level);
echo td$tabledata[0]/td;
echo td$tabledata[6]/td;
echo td$tabledata[2]/td;
echo td$tabledata[3]/td;
echo /tr;
}
print /table;
Thanks.
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Re: [PHP] MySQL and Array's REALLY simple question but I'm not GETTING it .. Ugh..
It doesn't matter what variables you assign to fetch the data, one of
the effects of mysql_fetch_array() is to increment the database result
variable. It does this without you even knowing it. check out the
mysql_data_seek() function to see how you can make the adjustments
yourself. i.e. mysql_data_seek($result, 0) would set you back at the
beginning, record one.
-Brad
Steve Gaas wrote:
OK, this works.. This is wonderful.. But I don't get it.. I assigned two
separate variables to that array function.. I don't understand why this
works now, but thanks a lot for the help.
I hope I don't inadvertently run into this again...
-steve
$sql2 = mysql_connect(localhost, eweb, dbfun)
or die(Could not connect using default username and password LINE
14 BR);
mysql_select_db(actionregs, $sql2);
$top_level = mysql_query(SELECT * FROM williams, $sql2)
or die(Could not do query sql1 to find username. Link might not
have been made. What's up? LINE 19 BR);
// $sql2_results = mysql_fetch_array($top_level);
$query_sql2_rows = mysql_num_rows($top_level);
$rows = 0;
echo $query_sql2_rows;
print table style=\font-family:Verdana; font-size:10pt\ border=0
cellpadding=4 width=90%;
print tr bgcolor=\#c0c0c0\th width=50Action ID/thth
width=100Owner/thth width=250Technology/ththSummary/th/tr;
for ($counter=0; $counter $query_sql2_rows; $counter++) {
$tabledata = mysql_fetch_array($top_level);
echo td$tabledata[0]/td;
echo td$tabledata[6]/td;
echo td$tabledata[2]/td;
echo td$tabledata[3]/td;
echo /tr;
}
print /table;
-Original Message-
From: Brad Bonkoski [mailto:[EMAIL PROTECTED]]
Sent: Friday, September 06, 2002 3:34 PM
To: Steve Gaas
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] MySQL and Array's REALLY simple question but I'm not
GETTING it .. Ugh..
Well, when you run the command:
$sql2_results = mysql_fetch_array($top_level);
The first time, it automatically increments the result...
so you fetch the data bu do nothing with it...
so when you get to your loop, you are already at the second entry in
your database..
So, remove the first instance of that and see how it works..
HTH
-Brad
Steve Gaas wrote:
Can anyone tell me what's wrong with my code? All I get output from this
is
the LAST row of data from my Database. There are 2 rows, how do I make it
pull data from all of the rows? It's not going through the loop like it
should I need to be able to tell the mysql_fetch_array which row I
want
in each it iteration of the for loop. The For loop increments counter, but
there is no syntax to add $counter to the result_type. The same goes with
fetch_row...
What am I forgetting!!?? I know it's something simple
***
$sql2 = mysql_connect(localhost, eweb, dbfun)
or die(Could not connect BR);
mysql_select_db(actionregs, $sql2);
$top_level = mysql_query(SELECT * FROM williams, $sql2)
or die(Could not do query BR);
$sql2_results = mysql_fetch_array($top_level);
$query_sql2_rows = mysql_num_rows($top_level);
echo $query_sql2_rows;
// echo's 2 rows
print table style=\font-family:Verdana; font-size:10pt\ border=0
cellpadding=4 width=90%;
print tr bgcolor=\#c0c0c0\th width=50Action ID/thth
width=100Owner/thth width=250Technology/ththSummary/th/tr;
for ($counter=0; $counter $query_sql2_rows; $counter++) {
$tabledata = mysql_fetch_array($top_level);
echo td$tabledata[0]/td;
echo td$tabledata[6]/td;
echo td$tabledata[2]/td;
echo td$tabledata[3]/td;
echo /tr;
}
print /table;
Thanks.
--
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Re: [PHP] MySQL and Array's REALLY simple question but I'm not GETTING it.. Ugh..
$sql2_results = mysql_fetch_array($top_level);
I guess this must be in a loop otherwise your $sq12_results keep getting
overwitten; thus,
the last record will be shown at the end.
$query_sql2_rows = mysql_num_rows($top_level);
echo $query_sql2_rows;
// echo's 2 rows
print table style=\font-family:Verdana; font-size:10pt\ border=0
cellpadding=4 width=90%;
print tr bgcolor=\#c0c0c0\th width=50Action ID/thth
width=100Owner/thth width=250Technology/ththSummary/th/tr;
for ($counter=0; $counter $query_sql2_rows; $counter++) {
$tabledata = mysql_fetch_array($top_level);
echo td$tabledata[0]/td;
echo td$tabledata[6]/td;
echo td$tabledata[2]/td;
echo td$tabledata[3]/td;
echo /tr;
}
print /table;
Thanks.
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--
i18n Engineer
SCO, Inc.
www.sco.com (801) 765-4999 x5647
355 South 520 West, Suit 100, Lindon, Utah 84042 USA
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[PHP] Further Security Clarifications [was: Simple Security Clarifcation]
My main files are located in /var/www/html (the 'DOCUMENT_ROOT' in Apache, according to php.ini). All sensitive files have been moved to '/var/www/secure', but now I can't access them! (According to php.ini, the PHP core 'doc_root=none'). I'm totally confused. If I understand this correctly, I want the files in '/var/www/secure' to be served through php scripts that reside in the individual files that call them up in /var/www/html. So, the problem seems to be that either Apache or PHP doesn't know/can't access them. So, what am I doing wrong here? I've also added a FILes ~\.sht$ directive to refuse all .sht files (they're .inc's). How do get access for php to the secure file directory, and exclude the hackers? At this point, I'm about as confused as I've ever been since beginning PHP. Any clarifications that will guide back into the fold, will be greatly appreciated! Tia, Andre -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Further Security Clarifications [was: Simple Security Clarifcation]
Usually the first thing you want to do here is check your error log. Most of the time, this sort of thing will be a permissions problem, as the apache server runs the PHP scripts as a user (ie you), that user needs to have the ability to execute those files. If you aren't sure, make the file owner you and give the files 777 and work backwards from there. Don't forget to check the directory permissions as well as the file permissions. From a hosting point of view, its different and a little more complicated if you have multiple users on the server, if it's just you though, it makes it a little easier. Take note of what the file permissions/ownership are befor eyou change them (in case this isn't the problem). Another simple things to check - make sure you're using the full path, ie, /var/www/secure/filename.php How are you including them? I use a require(/pathtofile/filename.php); Works for me assuming I have the right permissions. Best Regards Bob Irwin Server Admin Web Programmer Planet Netcom - Original Message - From: Andre Dubuc [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Thursday, August 22, 2002 11:19 AM Subject: [PHP] Further Security Clarifications [was: Simple Security Clarifcation] My main files are located in /var/www/html (the 'DOCUMENT_ROOT' in Apache, according to php.ini). All sensitive files have been moved to '/var/www/secure', but now I can't access them! (According to php.ini, the PHP core 'doc_root=none'). I'm totally confused. If I understand this correctly, I want the files in '/var/www/secure' to be served through php scripts that reside in the individual files that call them up in /var/www/html. So, the problem seems to be that either Apache or PHP doesn't know/can't access them. So, what am I doing wrong here? I've also added a FILes ~\.sht$ directive to refuse all .sht files (they're .inc's). How do get access for php to the secure file directory, and exclude the hackers? At this point, I'm about as confused as I've ever been since beginning PHP. Any clarifications that will guide back into the fold, will be greatly appreciated! Tia, Andre -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Scanned by PeNiCillin http://safe-t-net.pnc.com.au/ Scanned by PeNiCillin http://safe-t-net.pnc.com.au/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Further Security Clarifications [was: Simple Security Clarifcation]
Thanks Bob, Got a 404: File not Found. Checked the ssl_error_log as suggested, and found a rather interesting entry: No such file: /var/www/html/var/www/secure/test.php Obviously it's goes to DOCUMENT_ROOT (pre-pending the/var/www/html) and adds what I've asked for. So, how do I tell it where to look, and not the default setting? How am I including them? Well, most of the action occurs from the menu so it's: a href=https://localhost/var/www/secure/test.php;Testing for Bugs/a (I've also tried /secure/test.php Any ideas what I'm messing up? Tioa, Andre On Wednesday 21 August 2002 09:26 pm, Bob Irwin wrote: Usually the first thing you want to do here is check your error log. Most of the time, this sort of thing will be a permissions problem, as the apache server runs the PHP scripts as a user (ie you), that user needs to have the ability to execute those files. If you aren't sure, make the file owner you and give the files 777 and work backwards from there. Don't forget to check the directory permissions as well as the file permissions. From a hosting point of view, its different and a little more complicated if you have multiple users on the server, if it's just you though, it makes it a little easier. Take note of what the file permissions/ownership are befor eyou change them (in case this isn't the problem). Another simple things to check - make sure you're using the full path, ie, /var/www/secure/filename.php How are you including them? I use a require(/pathtofile/filename.php); Works for me assuming I have the right permissions. Best Regards Bob Irwin Server Admin Web Programmer Planet Netcom - Original Message - From: Andre Dubuc [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Thursday, August 22, 2002 11:19 AM Subject: [PHP] Further Security Clarifications [was: Simple Security Clarifcation] My main files are located in /var/www/html (the 'DOCUMENT_ROOT' in Apache, according to php.ini). All sensitive files have been moved to '/var/www/secure', but now I can't access them! (According to php.ini, the PHP core 'doc_root=none'). I'm totally confused. If I understand this correctly, I want the files in '/var/www/secure' to be served through php scripts that reside in the individual files that call them up in /var/www/html. So, the problem seems to be that either Apache or PHP doesn't know/can't access them. So, what am I doing wrong here? I've also added a FILes ~\.sht$ directive to refuse all .sht files (they're .inc's). How do get access for php to the secure file directory, and exclude the hackers? At this point, I'm about as confused as I've ever been since beginning PHP. Any clarifications that will guide back into the fold, will be greatly appreciated! Tia, Andre -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Scanned by PeNiCillin http://safe-t-net.pnc.com.au/ Scanned by PeNiCillin http://safe-t-net.pnc.com.au/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Further Security Clarifications [was: Simple Security Clarifcation]
Thanks Bob, Got a 404: File not Found. Checked the ssl_error_log as suggested, and found a rather interesting entry: No such file: /var/www/html/var/www/secure/test.php Ahhh - ok - I thought you were including them internally from PHP. You are actually linking to the file being SERVED by the web server in HTML. IN this case, all you need to do is reference to it as https://secureserveraddress/filename.php First of all, we need to understand this. We have two seperate servers here, the unix server that apache is running on and the apache server (this runs PHP, the secure server etc) itself. So... your normal website (served by the apache server) is at http://mywebsite.com/files.php BUT 'files.php' is located ON THE UNIX SERVER as /var/www/html/files.php The /var/www/html/ is the UNIX path to the file. The users who are using your APACHE server to get file do not see this in anyway. All they see is what is in the root directory, ie, /var/www/html from http://mywebsite.com/, this is exactly the same for the secure server, except the served files are encrypted. Success in this depends on what you are trying to do. Are you trying to secure files that contain information like your database passwords? Or are you just trying to run PHP scripts that produce HTML on a secure server (so that you can take credit card details from the remote users?). If you are trying to hide scripts with important information (ie, passwords) then running a secure server will not work. They will STILL be available from the internet, just at https://mywebsite/myfilewithapassword.php. This is not easily explained and I don't want to spend time going into it if its not what you're after, but if this is what you are doing, let me know and I'll help out. If you are just trying to encrypted the data from the server to the user, then you are doing the right thing, you just need to lose the /var/www/secure/ in the https:// address. Obviously it's goes to DOCUMENT_ROOT (pre-pending the/var/www/html) and adds what I've asked for. So, how do I tell it where to look, and not the default setting? How am I including them? Well, most of the action occurs from the menu so it's: a href=https://localhost/var/www/secure/test.php;Testing for Bugs/a (I've also tried /secure/test.php Any ideas what I'm messing up? Scanned by PeNiCillin http://safe-t-net.pnc.com.au/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] Further Security Clarifications [was: Simple Security Clarifcation]
Thanks again Bob, First off, the site is still being debugged off-line, and part of the problem, as you suggested is my confusion over UNIX SERVER and the Apache Server. OK. Got that. What I'm trying to do: Any file that utilizes $_SESSION variables accessed through username/password validation is accessed via https. This includes the original signup screen, logins, etc. The only place for http files are static files that display non-sensitive info, and that do not require db access. If you are trying to hide scripts with important information (ie, passwords) then running a secure server will not work. They will STILL be available from the internet, just at https://mywebsite/myfilewithapassword.php. If the user is not logged in, they get an error message, and can go no further since $_SESSION['authenticate'] must match their username/password. All I'm trying to do is to provide an extra layer of security by shoving all sensitive files into the 'secure' directory, outside of the DOCUMENT_ROOT (but I have no idea what 'doc_root' in PHP is for??). From what I'm trying to accomplish, do I really need to bother setting the 'secure' directory outside of the document_root? Wouldn't the setup I've done so far suffice? At any rate, I've tried just setting : https://localhost/secure/test.php -- it still gives a 404 Tia, Andre On Wednesday 21 August 2002 10:12 pm, Bob Irwin wrote: Thanks Bob, Got a 404: File not Found. Checked the ssl_error_log as suggested, and found a rather interesting entry: No such file: /var/www/html/var/www/secure/test.php Ahhh - ok - I thought you were including them internally from PHP. You are actually linking to the file being SERVED by the web server in HTML. IN this case, all you need to do is reference to it as https://secureserveraddress/filename.php First of all, we need to understand this. We have two seperate servers here, the unix server that apache is running on and the apache server (this runs PHP, the secure server etc) itself. So... your normal website (served by the apache server) is at http://mywebsite.com/files.php BUT 'files.php' is located ON THE UNIX SERVER as /var/www/html/files.php The /var/www/html/ is the UNIX path to the file. The users who are using your APACHE server to get file do not see this in anyway. All they see is what is in the root directory, ie, /var/www/html from http://mywebsite.com/, this is exactly the same for the secure server, except the served files are encrypted. Success in this depends on what you are trying to do. Are you trying to secure files that contain information like your database passwords? Or are you just trying to run PHP scripts that produce HTML on a secure server (so that you can take credit card details from the remote users?). If you are trying to hide scripts with important information (ie, passwords) then running a secure server will not work. They will STILL be available from the internet, just at https://mywebsite/myfilewithapassword.php. This is not easily explained and I don't want to spend time going into it if its not what you're after, but if this is what you are doing, let me know and I'll help out. If you are just trying to encrypted the data from the server to the user, then you are doing the right thing, you just need to lose the /var/www/secure/ in the https:// address. Obviously it's goes to DOCUMENT_ROOT (pre-pending the/var/www/html) and adds what I've asked for. So, how do I tell it where to look, and not the default setting? How am I including them? Well, most of the action occurs from the menu so it's: a href=https://localhost/var/www/secure/test.php;Testing for Bugs/a (I've also tried /secure/test.php Any ideas what I'm messing up? Scanned by PeNiCillin http://safe-t-net.pnc.com.au/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] my PHP scripts hangs over a simple error
Folks: I just setup a Redhat/Apache/PHP box for testing purposes as I develop some PHP websites. The weird thing, is that the server takes FOREVER to respond when I make a very simple error in the code. On other servers it comes up with a 'Parse Error' message immediately, but not so with this box. For example, I made a page with the following code: ? echo I'm making a purposeful mistake; echo The semi-colon is missing at the end of this line echo Now I'm going on; ? When I requested this page with my browser, I started my stopwatch and waited. The little blue progress bar started moving, but ever so slowly. 3 minutes later () I got the error message: Parse error: parse error, expecting `','' or `';'' in /var/www/html/test.php on line 4 Now why didn't that come up immediately?! Any help is appreciated. Thanks, Joseph -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: Increment alphabetical character (simple question)
Take a look at the ord() and chr() functions. They'll do what you want. On Tue, 12 Feb 2002, phantom wrote: How do I get the ASCII value of a character increment it by one... and convert back to a character. For example: I want to grab a char such as C and increment it to D ??? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] Re: Increment alphabetical character (simple question)
In article [EMAIL PROTECTED], [EMAIL PROTECTED] says... How do I get the ASCII value of a character increment it by one... and convert back to a character. For example: I want to grab a char such as C and increment it to D ??? ord() and chr() will do this. $oldchar = C; $newchar = chr(ord($oldchar) + 1); should be close to the mark. NB untested!! -- David Robley Temporary Kiwi! -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] PHP GuestBook - KISGB (Keep It Simple Guestbook) Version 3.0 Released
I have released version 3.0 of my KISGB. KISGB is a PHP guestbook program that does not require sessions, cookies, or an rdbms. Can be Public or Private through HTTP Authentication. Automated install script, fully customizable, clean, and fast. Separate multiple logging capabilty for tracking anything! Includes web-based password protected Admin functionality, along with email notification, greeting, ip logging, ip banning, bad word filter, smileys, allowable html tags in comments, next/previous, etc. Themes for controlling appearance that allow for background colors, images, animations, etc. Language support for Dutch, English, French, German, Polish, Portuguese, Spanish, and Surinam. -- Gaylen [EMAIL PROTECTED] Home http://www.gaylenandmargie.com/ PHP KISGB v3.0 Guest Book http://www.gaylenandmargie.com/phpwebsite/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] Regular Expressions - A relatively simple search...
Hello, I'm trying to replace a couple of lines of code: $dotpos = 1 - (strlen($userfile_name) - strpos($userfile_name, '.')); $extension = substr($userfile_name, $dotpos); with a simpler regular expression: $extension = eregi_replace( /.*, , $userfile_name); However it isn't working.. What I'd like to do is to find the extension of a file name and place that in a variable. So in '/home/mike/test.txt', I want to have the statement return 'txt' Any help would be appreciated.. Mike -- Mike Gifford, OpenConcept Consulting, http://openconcept.ca Offering everything your organization needs for an effective web site. Abolish Nuclear Weapons Now!: http://pgs.ca/petition/ It is a miracle that curiosity survives formal education. - A Einstein -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP] *IF* it was that simple
Ok here is a simple IF statement I am messing around with, yet the parser is
sending back an error message of which I cannot see the cause:
Code:
?php
if ($name == tarrant $username == costtar $password ==
password);
{
print(Your are now logged in $name, thank you.);
}
else
{
print(Access Denied! IP Logged);
}
?
Error:
Parse error: parse error in C:\locahost\parser.php on line 16
Line 16 being the line which contains else (line 6).
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Re: [PHP] *IF* it was that simple
Loose the ; after the if
Jeroen
Tarrant Costelloe [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
Ok here is a simple IF statement I am messing around with, yet the parser
is
sending back an error message of which I cannot see the cause:
Code:
?php
if ($name == tarrant $username == costtar $password ==
password);
{
print(Your are now logged in $name, thank you.);
}
else
{
print(Access Denied! IP Logged);
}
?
Error:
Parse error: parse error in C:\locahost\parser.php on line 16
Line 16 being the line which contains else (line 6).
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[PHP] Re: [PHP-DB] super simple....... but!
try
HTMLBODY
?php
echo "starting...";
if ( mysql_connect("localhost","php","php") )
{ echo "ok"; }
else {
echo "error!";
}
?
/BODY/HTML
- Original Message -
From: "Francois Boucher" [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Saturday, April 21, 2001 10:34 PM
Subject: [PHP-DB] super simple... but!
I wrote this code but nothing append. If i look de code in the browser it's
stopping to starting!
What is the problme?
HTMLBODY
?php
echo "starting...";
if ( mysql_connect("localhost","php","php") )
{ echo "ok"; }
else
echo "error!";
?
/BODY/HTML
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-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
-= Franois Boucher =-
-= [EMAIL PROTECTED]=-
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
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[PHP] Re: [PHP-DB] super simple....... but!
try
HTMLBODY
?php
echo "starting...";
if ( mysql_connect("localhost","php","php") )
{ echo "ok"; }
else {
echo "error!";
}
?
/BODY/HTML
- Original Message -
From: "Francois Boucher" [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Saturday, April 21, 2001 10:34 PM
Subject: [PHP-DB] super simple... but!
I wrote this code but nothing append. If i look de code in the browser it's
stopping to starting!
What is the problme?
HTMLBODY
?php
echo "starting...";
if ( mysql_connect("localhost","php","php") )
{ echo "ok"; }
else
echo "error!";
?
/BODY/HTML
--
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
-= Franois Boucher =-
-= [EMAIL PROTECTED]=-
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
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Re: [PHP] please help with this simple problem
i tryed it and it ended up having an error that was caused originally by a lack of a $ on the 3rd line variable... after i fixed that it said wrong perameter count for fopen() on the third line, and "Warning: Supplied argument is not a valid File-Handle resource" for the remaining lines below that in that block of code. i'm sorry, i'm kinda new to this : ( "Stewart Taylor" [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... You need to copy the contents of the current file. Then recreate the file by writing the new message then writing back the original contents. e.g. $fname = basename($PHP_SELF). ".comment"; $fsize = filesize($fname); fp = fopen(basename($fname)); $data = fread($fp,fsize); fwrite($fp,$message); fwrite($fp,$data); fclose($fp); -Stewart -Original Message- From: adam [mailto:[EMAIL PROTECTED]] Sent: 22 March 2001 11:17 To: [EMAIL PROTECTED] Subject: [PHP] please help with this simple problem i am coding a simple script to post a text area into a file. it works, but it posts it at the bottom and i wanted to have it post to the top of the text already there.. here's a snip of the important part of the script: $fp = fopen (basename($PHP_SELF) . ".comment", "a"); fwrite ($fp, $message); fclose ($fp); } any help would be much appreciated -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] please help with this simple problem
you haven't specified the open type (a = append, w = write, r = read only , etc etc) solution $fp=fopen(basename($fname), "a"); just to match your example. On 22-Mar-2001 adam wrote: i tryed it and it ended up having an error that was caused originally by a lack of a $ on the 3rd line variable... after i fixed that it said wrong perameter count for fopen() on the third line, and "Warning: Supplied argument is not a valid File-Handle resource" for the remaining lines below that in that block of code. i'm sorry, i'm kinda new to this : ( "Stewart Taylor" [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... You need to copy the contents of the current file. Then recreate the file by writing the new message then writing back the original contents. e.g. $fname = basename($PHP_SELF). ".comment"; $fsize = filesize($fname); fp = fopen(basename($fname)); $data = fread($fp,fsize); fwrite($fp,$message); fwrite($fp,$data); fclose($fp); -Stewart -Original Message- From: adam [mailto:[EMAIL PROTECTED]] Sent: 22 March 2001 11:17 To: [EMAIL PROTECTED] Subject: [PHP] please help with this simple problem i am coding a simple script to post a text area into a file. it works, but it posts it at the bottom and i wanted to have it post to the top of the text already there.. here's a snip of the important part of the script: $fp = fopen (basename($PHP_SELF) . ".comment", "a"); fwrite ($fp, $message); fclose ($fp); } any help would be much appreciated -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] Rui Barreiros Software Developer WEBSOLUT - Soluções Internet Emailto: [EMAIL PROTECTED] Personal Info: http://websolut.net/people/rui.html As informações contidas neste email são confidenciais e destinam-se apenas à(s) pessoa(s) a quem foi enviado: http://websolut.net/confidencialidade-responsabilidade.html -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
RE: [PHP] please help with this simple problem
Sorry I missed "w+" from the fopen function $fname = basename($PHP_SELF). ".comment"; $fsize = filesize($fname); $fp = fopen(basename($fname),"w+"); --- added "w+" $data = fread($fp,$fsize); fwrite($fp,$message); fwrite($fp,$data); fclose($fp) -Stewart -Original Message- From: adam [mailto:[EMAIL PROTECTED]] Sent: 22 March 2001 12:00 To: [EMAIL PROTECTED] Subject: Re: [PHP] please help with this simple problem i tryed it and it ended up having an error that was caused originally by a lack of a $ on the 3rd line variable... after i fixed that it said wrong perameter count for fopen() on the third line, and "Warning: Supplied argument is not a valid File-Handle resource" for the remaining lines below that in that block of code. i'm sorry, i'm kinda new to this : ( "Stewart Taylor" [EMAIL PROTECTED] wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... You need to copy the contents of the current file. Then recreate the file by writing the new message then writing back the original contents. e.g. $fname = basename($PHP_SELF). ".comment"; $fsize = filesize($fname); fp = fopen(basename($fname)); $data = fread($fp,fsize); fwrite($fp,$message); fwrite($fp,$data); fclose($fp); -Stewart -Original Message- From: adam [mailto:[EMAIL PROTECTED]] Sent: 22 March 2001 11:17 To: [EMAIL PROTECTED] Subject: [PHP] please help with this simple problem i am coding a simple script to post a text area into a file. it works, but it posts it at the bottom and i wanted to have it post to the top of the text already there.. here's a snip of the important part of the script: $fp = fopen (basename($PHP_SELF) . ".comment", "a"); fwrite ($fp, $message); fclose ($fp); } any help would be much appreciated -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
Re: [PHP] please help with this simple problem
it works now, only it's earasing everything and then writing to the file.
i think we have almost got this figured out. here's what the code looks like
for the entire tag...
---
?
if ($message)
{
/* uncomment the next two lines to strip out html from input */
$name = strip_tags($name);
/* $message = strip_tags($message); */
$message = ereg_replace("\r\n\r\n", "\nP", $message);
$date = date("l, F j Y, h:i a");
$message = "font size=2 face=verdanaba href=mailto:$email$name/a
/bfont size=1 -- $date/font\n
blockquote\n
$message\n
/blockquote/font\nhr noshade color=white size=1 width=100%\n";
$fname = basename($PHP_SELF) . ".comment";
$fsize = filesize($fname);
$fp = fopen(basename($fname),"w+");
$data = fread($fp,$fsize);
fwrite($fp,$message);
fwrite($fp,$data);
fclose($fp);
}
@readfile(basename(($PHP_SELF . ".comment")));
?
""adam"" [EMAIL PROTECTED] wrote in message
99cmfj$mai$[EMAIL PROTECTED]">news:99cmfj$mai$[EMAIL PROTECTED]...
i am coding a simple script to post a text area into a file. it works, but
it posts it at the bottom and i wanted to have it post to the top of the
text already there..
here's a snip of the important part of the script:
$fp = fopen (basename($PHP_SELF) . ".comment", "a");
fwrite ($fp, $message);
fclose ($fp);
}
any help would be much appreciated
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Re: [PHP] please help with this simple problem
it works now, only it's earasing everything and then writing to the file.
i think we have almost got this figured out. here's what the code looks like
for the entire tag...
---
?
if ($message)
{
/* uncomment the next two lines to strip out html from input */
$name = strip_tags($name);
/* $message = strip_tags($message); */
$message = ereg_replace("\r\n\r\n", "\nP", $message);
$date = date("l, F j Y, h:i a");
$message = "font size=2 face=verdanaba href=mailto:$email$name/a
/bfont size=1 -- $date/font\n
blockquote\n
$message\n
/blockquote/font\nhr noshade color=white size=1 width=100%\n";
$fname = basename($PHP_SELF) . ".comment";
$fsize = filesize($fname);
$fp = fopen(basename($fname),"w+");
$data = fread($fp,$fsize);
fwrite($fp,$message);
fwrite($fp,$data);
fclose($fp);
}
@readfile(basename(($PHP_SELF . ".comment")));
?
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[PHP] Walking Through Array Values, Simple?
Well... I think I probably owe most of the people on this list some serious cash in consulting fees by now. But we're opensource, so I guess I'm off the hook, right? =) Anyways, another question for you people. I'm on the last leg of designing this e-commerce site. All I have left to design is the shopping cart. How I've set it up to add items is quite simple. The data is posted add to cart and stored like this: $cart[$prodid] = $my_qty; cart is the registered variable in my sessions. prodid is the product id, and my_qty is the quantity. now say a user wants 2 items with the id of 111 and 1 item with the id of 898. it should have the data stored as so: $cart[111] = 2 $cart[898] = 1 now.. had i been more experienced with arrays, this would not be hard. but.. i'm not sure when the user goes to check out how to walk through the data. basically i need to pull out each index (which is the product id number) as it's own variable and the qty that corresponds with it as it's own variable. From there, although irrelevant to what the array, there variables will be passed in a mysql query to get the info to display in the basket (ie- name, description, price). so, i figure i am going to have to loop the array to extract the info. this would probably be best since i'm going to be making seperate queries on it as well. loop until all indexes have been accounted for--- get prod id and qty from next instance in array, into $prodid and $qty query the db with the prodid display results end loop -- so.. i'm fine with the query and display, i just need to know how to get each instance into a variable in a loop. thanks a lot
