RE: [PHP-DB] I have a problem to add a new line to the body of my email when I am using 'a href=mailto: ... function
I never knew that Body was supported. I had tried in years ago before it worked, and never looked back. If you look at the source code for this page (http://developer.netscape.com/viewsource/husted_mailto/mailto.html) you'll see that they use the javascript function escape() on the body text. I believe this handles the characters such as spaces and CR/NL. Though it seems you are doing it the same way they are (%0D%0A). If it works on some browsers, but not others, then it's probably a compatibiilty issue, not a scripting issue. You don't indicate which browsers this works/doesn't work on. Or which mail clients you're using. Finally, this is not really a PHP issue, and especially not a PHP-DB issue. --- Martin Rajcok [EMAIL PROTECTED] wrote: I believe the body is supported. I have read everything I could find through Google. People are using it exactly as I am. I can't use '\n' - it is only text in this case and it will be part of the email body. Thanks anyway, Martin -Original Message- From: John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Thursday, 5 June 2003 3:02 p.m. To: 'Martin Rajcok'; [EMAIL PROTECTED] Subject: RE: [PHP-DB] I have a problem to add a new line to the body of my email when I am using 'a href=mailto: ... function I have a problem to add a new line to the body of my email when I am using 'a href=mailto: ... function. This is my code: $details[email] //email of the user from the form $details[contact_name] //name of the user from the form $email_subject = subject text...; $email_form = Welcome .$details[contact_name].,%0D%0A%0D%0A; $email_form .= Thank you for registering with ; a href='mailto:.$details[email].?Body=.$email_form.Subject=.$email_s ub je ct.' class='darkblue'Send the email/a Use \n. email_form = Welcome .$details[contact_name].,\n\n; Are you sure body is even supported? I'll guarantee it's not going to be for every browser/email client. From: http://www.w3.org/TR/WD-html40-970917/htmlweb.html#h-5.1.3.1 -- MAILTO URLs have the following syntax: mailto:email-address User agents may support MAILTO URL extensions that are not yet Internet standards (e.g., appending subject information to a URL with the syntax ?Subject=my%20subject where any space characters are replaced by %20). Some user agents also support ?Cc=email-address. -- ---John W. Holmes... Amazon Wishlist: http://www.amazon.com/o/registry/3BEXC84AB3A5E PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/ -- 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 = Mark Weinstock [EMAIL PROTECTED] *** You can't demand something as a right unless you are willing to fight to death to defend everyone else's right to the same thing. *** __ Do you Yahoo!? Yahoo! Calendar - Free online calendar with sync to Outlook(TM). http://calendar.yahoo.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Datetime help in an INSERT...
I have actually discovered that the $td value is blank. The reason appears to be that the page is reloading when a button is pushed, and that is when the $td value is being lost. My question now is, how do I keep the $td value after the page is reloaded? I would rather keep the value from the original query than perform another database query to populate this value yet again. Thanks yet again for the help. -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Wednesday, June 04, 2003 2:14 PM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Datetime help in an INSERT... I am stumbling across something that I thought I have done before, and I am not having any luck finding an example of this. Basically, I am wanting to timestamp the date and time into new entries in a simple database table. The following section is the actual code for this, and I cannot figure out how to get the date/time stamp to populate into the database. Thanks in advance for the help. I suspect this is a very simple fix. $denylog = INSERT INTO deny (account, td, date) VALUES ($tmp, $td, NOW()); $denylog_result = mysql_query($denylog, $Prod) or die(mysql_error()); The error I am receiving is: You have an error in your SQL syntax near ' NOW())' at line 1 Are you sure $td has a value? If it's blank, you'd get an error like that. ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Cookie issue
We are having complaints on our site about users not being able to log-in
and we feel it maybe related to the setting of cookie. Here is the code
used. Can anyone give me insight.
/* some standard var's or other things like srand */
srand((double)microtime() * 100);
$includes_root = $DOCUMENT_ROOT/includes;
$encoded_url= urlencode($REQUEST_URI);
if($_COOKIE[cookie_check] != Y)
{
setcookie(cookie_check, Y, strtotime(+5 years), /,
$HTTP_HOST, 0);
}
if($_COOKIE[current_surfer_id] == AND $_COOKIE[cookie_check] ==
Y)
{
$current_surfer_id = rand(0, 16777215);
setcookie(current_surfer_id, $current_surfer_id, strtotime(+5
years), /, $HTTP_HOST, 0);
}
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Re: [PHP-DB] Datetime help in an INSERT...
I have actually discovered that the $td value is blank. The reason appears to be that the page is reloading when a button is pushed, and that is when the $td value is being lost. My question now is, how do I keep the $td value after the page is reloaded? I would rather keep the value from the original query than perform another database query to populate this value yet again. Thanks yet again for the help. Stick it in the session? cookie? URL? What is $td? ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] address info, forms, maintanance
John W. Holmes [EMAIL PROTECTED] wrote on 06/04/2003 05:24:22 PM: Thanks to everyone for the suggestions. Got it fixed. Just added a couple lines of code - $sql1 = select * from endusers where name like '$name'; $result1 = mysql_query($sql1); $count1 = mysql_num_rows($result1); if ($count1 == 0 ) $sql = insert into endusers..blah blah blah Now it works fine, no more duplicate entries. But now you're doing two queries for every insert. If you simply made your columns unique and let the database handle it, you'd only have to do one INSERT. Then check affected_rows() or mysql_error() to see if either no rows were affected (no rows inserted) or the error mentions duplicate. If either is the case, the row wasn't inserted because of a unique constraint. If there is an error but it's not duplicate or whatever, then it's another error and you should show it. ---John W. Holmes... Thanks for the tips, I'll work on that. In the meantime, this database is pretty low useage, very specialized information for a small market segment, so for now it will be okay. It'll never have hundreds or thousands of hits per day. But I will look into making the suggested changes. Thanks, Chip -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] while - if - problem
Hey thanks Mike
that surly made a difference.
Earl
'
- Original Message -
From: Ford, Mike [LSS] [EMAIL PROTECTED]
To: 'Earl' [EMAIL PROTECTED]; PHP-DB [EMAIL PROTECTED]
Sent: Thursday, June 05, 2003 6:13 AM
Subject: RE: [PHP-DB] while - if - problem
-Original Message-
From: Earl [mailto:[EMAIL PROTECTED]
Sent: 04 June 2003 22:04
To: PHP-DB
FYI
this was beginning to bug me out... so I decided to try the
trim function
and walla... it worked.
Thanks for ya'll assistance.
I was going to say this even before you added the trim() calls in , but
this
really does look like an excellent situation for using the switch
construct
(just look at all those trim() calls and array accesses you save!):
while($cols=ifx_fetch_row($eventQuery))
{
switch (trim($cols['out_type']))
{
case '0':
$r_away['linetype']='L';
$r_home['linetype']='L';
break;
case '1':
$r_away['linetype']='H';
$r_home['linetype']='H';
break;
}
switch (trim($cols[s_acro]))
{
case 'CF':
case 'PF':
$r_away['sport']='1';
$r_home['sport']='1';
$s_lt='PS';
$t_lt='TP';
break;
case 'PB':
case 'CB':
$r_away['sport']='2';
$r_home['sport']='2';
$s_lt='PS';
$t_lt='TP';
break;
case 'B':
$r_away['sport']='3';
$r_home['sport']='3';
$s_lt='ML';
$t_lt='TM';
break;
case 'H':
$r_away['sport']='4';
$r_home['sport']='4';
$s_lt='ML';
$t_lt='TM';
break;
}
Cheers!
Mike
-
Mike Ford, Electronic Information Services Adviser,
Learning Support Services, Learning Information Services,
JG125, James Graham Building, Leeds Metropolitan University,
Beckett Park, LEEDS, LS6 3QS, United Kingdom
Email: [EMAIL PROTECTED]
Tel: +44 113 283 2600 extn 4730 Fax: +44 113 283 3211
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RE: [PHP-DB] Datetime help in an INSERT...
OK. I guess the next question would be what is the best option for performance? This portion of the app currently doesn't have any session functionality built in. I would be setting up the session for this sole purpose. I haven't worked with cookies yet. I suppose the URL method might work, I have just never like it because I think it looks unclean, just personal preference there. As to the $td question... This is a variable that holds the userid of the Technical Director who is the account approver. This portion of the app is simply an approval mechanism to an account request system that I am developing. The TD receives an e-mail with a link back to the approval page to approve the requested accounts. This approval then fires off an e-mail to the SA responsible for user accounts to inform him to create the newly approved account. Well, thanks again for the feedback. Hopefully I can get this going pretty quick. We are currently planning on rolling this app out to our user community on Monday. This is the last issue I have that I am aware of. -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 9:06 AM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Datetime help in an INSERT... I have actually discovered that the $td value is blank. The reason appears to be that the page is reloading when a button is pushed, and that is when the $td value is being lost. My question now is, how do I keep the $td value after the page is reloaded? I would rather keep the value from the original query than perform another database query to populate this value yet again. Thanks yet again for the help. Stick it in the session? cookie? URL? What is $td? ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Datetime help in an INSERT...
Not sure exactly how I would implement this. Currently the reload URL is being provided via the $PHP_SELF variable. I like using this, but suppose I could change this if truly necessary. Also, the $td is not currently being passed into the function that generates the two buttons and hence defines the reload URL. I would rather not pass any additional variables into this function as I just feel this is unclean. Thanks again for the idea, and any additional advice or suggestions are most appreciated. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 9:01 AM To: NIPP, SCOTT V (SBCSI) Subject: RE: [PHP-DB] Datetime help in an INSERT... You could alway place it as a GET echo a href=\yourscript.com/yourpage.php?td=$td\; Gary Every Sr. UNIX Administrator Ingram Entertainment (615) 287-4876 Pay It Forward mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] http://accessingram.com http://accessingram.com -Original Message- From: NIPP, SCOTT V (SBCSI) [ mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] ] Sent: Thursday, June 05, 2003 8:57 AM To: 'CPT John W. Holmes'; [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: RE: [PHP-DB] Datetime help in an INSERT... I have actually discovered that the $td value is blank. The reason appears to be that the page is reloading when a button is pushed, and that is when the $td value is being lost. My question now is, how do I keep the $td value after the page is reloaded? I would rather keep the value from the original query than perform another database query to populate this value yet again. Thanks yet again for the help. -Original Message- From: CPT John W. Holmes [ mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] ] Sent: Wednesday, June 04, 2003 2:14 PM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Datetime help in an INSERT... I am stumbling across something that I thought I have done before, and I am not having any luck finding an example of this. Basically, I am wanting to timestamp the date and time into new entries in a simple database table. The following section is the actual code for this, and I cannot figure out how to get the date/time stamp to populate into the database. Thanks in advance for the help. I suspect this is a very simple fix. $denylog = INSERT INTO deny (account, td, date) VALUES ($tmp, $td, NOW()); $denylog_result = mysql_query($denylog, $Prod) or die(mysql_error()); The error I am receiving is: You have an error in your SQL syntax near ' NOW())' at line 1 Are you sure $td has a value? If it's blank, you'd get an error like that. ---John Holmes... -- PHP Database Mailing List ( http://www.php.net/ http://www.php.net/ ) To unsubscribe, visit: http://www.php.net/unsub.php http://www.php.net/unsub.php
RE: [PHP-DB] Datetime help in an INSERT...
I just responded to a separate but similar suggestion. The form elements, including the buttons, are all generated in a separate function. I am currently passing about 5 variables into this function and I am a little hesitant to pass any others. I am not sure about complexity, performance, etc. in passing so many variables into a function. Please let me know if anyone has any ideas on this. I doubt that this is happening, but I would really like for the code to be as easily maintainable as possible. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 9:48 AM To: NIPP, SCOTT V (SBCSI) Subject: RE: [PHP-DB] Datetime help in an INSERT... a button is being pressed? so it is processing a form? if its in a form just pass along the value of $td via hidden input data... input type=hidden name=td value=? echo $td; ? jay merritt [EMAIL PROTECTED] [EMAIL PROTECTED] OK. I guess the next question would be what is the best option for performance? This portion of the app currently doesn't have any session functionality built in. I would be setting up the session for this sole purpose. I haven't worked with cookies yet. I suppose the URL method might work, I have just never like it because I think it looks unclean, just personal preference there. As to the $td question... This is a variable that holds the userid of the Technical Director who is the account approver. This portion of the app is simply an approval mechanism to an account request system that I am developing. The TD receives an e-mail with a link back to the approval page to approve the requested accounts. This approval then fires off an e-mail to the SA responsible for user accounts to inform him to create the newly approved account. Well, thanks again for the feedback. Hopefully I can get this going pretty quick. We are currently planning on rolling this app out to our user community on Monday. This is the last issue I have that I am aware of. -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 9:06 AM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Datetime help in an INSERT... I have actually discovered that the $td value is blank. The reason appears to be that the page is reloading when a button is pushed, and that is when the $td value is being lost. My question now is, how do I keep the $td value after the page is reloaded? I would rather keep the value from the original query than perform another database query to populate this value yet again. Thanks yet again for the help. Stick it in the session? cookie? URL? What is $td? ---John Holmes... -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] forms with php/mysql
Hi, Thanks all for your answers. What I mean by I am not a coder is that it will take me 3 month to do it while doing only that, because I code in PHP once a month. And since I help an association I cannot really spend my time doing this. What I can do is to take bits of code and put them toggether without too much trouble. I just found php classes repository, it is not bad at all. I got most of the elements. Philippe Le Jeudi 5 Juin 2003 12:02, Becoming Digital a écrit : If you aren't a coder, you probably cannot do what you ask; there aren't any pre-built simple systems that do exactly what you ask. Yes and no. I agree that someone without coding experience (or a desire to learn) cannot tackle this problem, but what Philippe has requested is a simple user authentication system with sessions, of which there are many. Still, customization to fit his system will require an understanding of the code, which will likely prove problematic. To the poster, check out the PHP Classes Repository for a number of authentication classes. Don't expect them to work if you don't understand the code, though. Edward Dudlik Becoming Digital www.becomingdigital.com - Original Message - From: Peter Beckman [EMAIL PROTECTED] To: Philippe Rousselot [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Sent: Thursday, 05 June, 2003 00:15 Subject: Re: [PHP-DB] forms with php/mysql 1. Search the archives, they are here: http://www.php.net/mailing-lists.php 2. If you aren't a coder, you probably cannot do what you ask; there aren't any pre-built simple systems that do exactly what you ask. You could look on freshmeat.net or sourceforge.com, but I doubt you'd find what you are looking for. Peter On Thu, 5 Jun 2003, Philippe Rousselot wrote: Hi, I need for an association to create a form using php/mysql. If I can understand php, I am not a coder, so if someone had the code I need, I would really be greatful. I need : 1. on a page someone identify himself or register to the service using a form where the person enter name, email(username) and password. 2. once registered, the person can access a secured zone containing a form 3. the persone can fill the form, abandon half way and save the result, then come back and have the previously recorded information appearing again. all data need to be in a Mysql database. Thank you very much for your help. Philippe -- 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 -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Cookie issue
To add to this message:
We are having problems with people using IE 6.0. They appear to sign-in and
then are returned to the sign-in screen over and over. This doesnt happen
to all that use the site.
Thanks in advance
-Original Message-
From: Marie Osypian [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 05, 2003 10:03 AM
To: PHP-DB
Subject: [PHP-DB] Cookie issue
We are having complaints on our site about users not being able to log-in
and we feel it maybe related to the setting of cookie. Here is the code
used. Can anyone give me insight.
/* some standard var's or other things like srand */
srand((double)microtime() * 100);
$includes_root = $DOCUMENT_ROOT/includes;
$encoded_url= urlencode($REQUEST_URI);
if($_COOKIE[cookie_check] != Y)
{
setcookie(cookie_check, Y, strtotime(+5 years), /,
$HTTP_HOST, 0);
}
if($_COOKIE[current_surfer_id] == AND $_COOKIE[cookie_check] ==
Y)
{
$current_surfer_id = rand(0, 16777215);
setcookie(current_surfer_id, $current_surfer_id, strtotime(+5
years), /, $HTTP_HOST, 0);
}
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RE: [PHP-DB] Datetime help in an INSERT...
Thought I was on track, but stumped again... I am now getting a completely different error that I cannot figure out how it is related. I have passed the $td variable into the function and then get it back as a POST variable from a hidden field within the function. Here is the error I am now seeing; Unknown column 'sn4265' in 'field list' This one is really throwing me at the moment as the only place the value sn4265 is stored is in the $sbcuid variable. This $sbcuid variable is not a part of the INSERT that I am performing which I think is generating this error. I am guessing that you guys will need more info to help out with this now. Just let me know what you want. Thanks again. -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 10:07 AM To: NIPP, SCOTT V (SBCSI) Subject: Re: [PHP-DB] Datetime help in an INSERT... Just pass the variable to your function. If you're using a form, then make it a hidden element. ---John Holmes... - Original Message - From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED] To: 'CPT John W. Holmes' [EMAIL PROTECTED] Sent: Thursday, June 05, 2003 10:53 AM Subject: RE: [PHP-DB] Datetime help in an INSERT... Do you think URL propagation is a better option than passing another variable into the function and then passing it back out through a hidden form element? Here was another snippet from another suggestion on this. a button is being pressed? so it is processing a form? if its in a form just pass along the value of $td via hidden input data... input type=hidden name=td value=? echo $td; ? jay merritt [EMAIL PROTECTED] [EMAIL PROTECTED] -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 9:53 AM To: NIPP, SCOTT V (SBCSI) Subject: Re: [PHP-DB] Datetime help in an INSERT... Propagating it in the URL is probably the best method. ---John Holmes... - Original Message - From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED] To: 'CPT John W. Holmes' [EMAIL PROTECTED]; [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Sent: Thursday, June 05, 2003 10:44 AM Subject: RE: [PHP-DB] Datetime help in an INSERT... OK. I guess the next question would be what is the best option for performance? This portion of the app currently doesn't have any session functionality built in. I would be setting up the session for this sole purpose. I haven't worked with cookies yet. I suppose the URL method might work, I have just never like it because I think it looks unclean, just personal preference there. As to the $td question... This is a variable that holds the userid of the Technical Director who is the account approver. This portion of the app is simply an approval mechanism to an account request system that I am developing. The TD receives an e-mail with a link back to the approval page to approve the requested accounts. This approval then fires off an e-mail to the SA responsible for user accounts to inform him to create the newly approved account. Well, thanks again for the feedback. Hopefully I can get this going pretty quick. We are currently planning on rolling this app out to our user community on Monday. This is the last issue I have that I am aware of. -Original Message- From: CPT John W. Holmes [mailto:[EMAIL PROTECTED] Sent: Thursday, June 05, 2003 9:06 AM To: NIPP, SCOTT V (SBCSI); [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: Re: [PHP-DB] Datetime help in an INSERT... I have actually discovered that the $td value is blank. The reason appears to be that the page is reloading when a button is pushed, and that is when the $td value is being lost. My question now is, how do I keep the $td value after the page is reloaded? I would rather keep the value from the original query than perform another database query to populate this value yet again. Thanks yet again for the help. Stick it in the session? cookie? URL? What is $td? ---John Holmes... -- 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
RE: [PHP-DB] PHP/Mysql Script help.
Thanks for all you wonderful help. With a little tweaking, and a new
debugging program, I have been able to get the results I want. Now I'm
going to begin adding some security features. Again, thanks for you help.
Yours in Service,
Jason Britton
Scar's Legion
http://st.fulco.net
[EMAIL PROTECTED]
Vincit imitationem veritas.
-Original Message-
From: Becoming Digital [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 05, 2003 4:59 AM
To: 'PHP-DB'
Subject: Re: [PHP-DB] PHP/Mysql Script help.
I debugged the code and there were a slew of errors. I've fixed all but
one,
which I can't correct bcs I don't know the intention. Most of the errors
came
from improper syntax, a problem which could have been avoided by using
consistent code formatting. I chose to do that for you as well and have
included the code below.
As for the error I could not correct, it involves the final elseif()
statement.
The way your code was written, you had two 'else' statements, which is an
obvious violation. I corrected that by changing the first (before the
select
statement) to an elseif(), but I do not know the condition. Once you
enter it,
the code should work smoothly.
?php
require(racesetup.php);
$con = mysql_connect($host, $user, $pass) or die(mysql_error());
mysql_select_db($db, $con) or die(mysql_error());
// display individual record
if ($id)
{
error_reporting(E_ALL);
$result = mysql_query(SELECT * FROM nuke_race WHERE id=$id,$con) or
die(mysql_error());
$myrow = mysql_fetch_array($result) or die(mysql_error());
printf (bRace:/b %s\nbr, $myrow[race]);
echo br;
printf(bDescription:/b %s\nbr, $myrow[racetxt]);
echo br;
echo brh3a href=\javascript:history.go(-1)\Back/a/h3;
}
elseif ($_Post['letter'])
{
if ($_Post['letter'])
{
$result = mysql_query(SELECT * FROM nuke_race WHERE race LIKE
$_Post['letter']%,$con) or die(mysql_error());
if ($myrow = mysql_fetch_array($result))
{ // display list if there are records to display
do {
printf(a href=\%s?id=%s\%s/abr\n, $PHP_SELF,
$myrow[id], $myrow[race]);
}
while ($myrow = mysql_fetch_array($result));
}
}
}
//NO CONDITION LISTED!!
elseif ()
{
echo form name=\letter\ method=\post\ action=\races.php\
. pPick a letter to begin your search/p
. p
. select name=\select\
. option value=\a\A/option
. option value=\b\B/option
. option value=\c\C/option
. option value=\d\D/option
. option value=\e\E/option
. option value=\f\F/option
. option value=\g\G/option
. option value=\h\H/option
. option value=\i\I/option
. option value=\j\J/option
. option value=\k\K/option
. option value=\L\L/option
. option value=\m\M/option
. option value=\n\N/option
. option value=\o\O/option
. option value=\p\P/option
. option value=\q\Q/option
. option value=\r\R/option
. option value=\s\S/option
. option value=\t\T/option
. option value=\u\U/option
. option value=\v\V/option
. option value=\w\W/option
. option value=\x\X/option
. option value=\y\Y/option
. option value=\z\Z/option
. /select
. /p
. p
. input type=\Submit\ name=\submit\ value=\Enter
information\
. /p
. /form
.;
}
else
// no records to display
echo Sorry, no records were found!;
?
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Fulco of Scarborough [EMAIL PROTECTED]
To: 'PHP-DB' [EMAIL PROTECTED]
Sent: Thursday, 05 June, 2003 00:25
Subject: Re: [PHP-DB] PHP/Mysql Script help.
I have still been getting errors, so here goes again.
I am attempting to design a script that presents the user with a form with
26 letters to choose from. When they pick a letter I want it to take them
to a list of all the entries in my database that begin with the letter
they
selected in link form. When they click on the term, I want it to pull up
the info for that entry.
I have made some changes and added some error checking:
?php
require(racesetup.php);
$con = mysql_connect($host, $user, $pass) or die(mysql_error());
mysql_select_db($db, $con) or die(mysql_error());
// display individual record
if ($id) {
error_reporting(E_ALL);
$result = mysql_query(SELECT * FROM nuke_race WHERE id=$id,$con) or
die(mysql_error());
$myrow = mysql_fetch_array($result) or die(mysql_error());
printf(bRace:/b %s\nbr, $myrow[race]);
echo br;
Re: [PHP-DB] forms with php/mysql
done you can find the files at www.alcatorda.com/jcp/test/files.html and try it at www.alctorda.com/jcp/test/enregistrer.php I am sure it is full of bad codding but it works. I just have to make it nice now. Philippe Le Jeudi 5 Juin 2003 06:15, Peter Beckman a écrit : 1. Search the archives, they are here: http://www.php.net/mailing-lists.php 2. If you aren't a coder, you probably cannot do what you ask; there aren't any pre-built simple systems that do exactly what you ask. You could look on freshmeat.net or sourceforge.com, but I doubt you'd find what you are looking for. Peter On Thu, 5 Jun 2003, Philippe Rousselot wrote: Hi, I need for an association to create a form using php/mysql. If I can understand php, I am not a coder, so if someone had the code I need, I would really be greatful. I need : 1. on a page someone identify himself or register to the service using a form where the person enter name, email(username) and password. 2. once registered, the person can access a secured zone containing a form 3. the persone can fill the form, abandon half way and save the result, then come back and have the previously recorded information appearing again. all data need to be in a Mysql database. Thank you very much for your help. Philippe -- 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] PHP sort from .... best solution?
Hi!
Thanks! But I got a problem with the code:
The PHP file as it is with the code pasted into it will i'll past at the end
of this post. The problem is I get the error:
Warning: Invalid argument supplied for foreach() in
/home/amotor/www/list_ingresser_artikler.php on line 43
If I put a // in front of line 32 the script feeds correct, but not sorted.
I think there is a very easy solution to this, but I dont know so mutch PHP
:( I'm reading though.
Thanx alot for the help I got so far. I've learned alot!
Regards,
A. Lyse
The PHP file
?php
//Dette er en mal for å lette implementering av pusyset til de enkelte
sider.
//Denne malen viser ingressene til de 50 siste artikler denne siden skal ha.
include ../lib/ingress.lib;
$nettstedid=11; //Id til nettstedet. Dersom man ikke husker sin nettstedid
finnes den i admindelen av pubsyset.
$sprakid=1; //Språkid. 1=Norsk
$offset=0; //Offset, dersom man ikke skal vise fra artikkel 0, men f.eks.
fra artikkel 51.
$offset3ingr=2; //Offset, dersom man ikke skal vise fra artikkel 0, men
f.eks. fra artikkel 51.
$offsetannet=1; //Offset, dersom man ikke skal vise fra artikkel 0, men
f.eks. fra artikkel 51.
$antall=50; //Hvor mange artikler skal vises
$antallmotor=25; //Hvor mange artikler skal vises på motorsport
$antallmotor3ingresser=3; //Hvor mange artikler skal vises videre på
motorsport
$antallmc=1; //Hvor mange artikler skal vises på MC
$antallmc3ingresser=3; //Hvor mange artikler skal vises videre på MC
$antallstyling=1; //Hvor mange artikler skal vises på MC
$antallstyling3ingresser=3; //Hvor mange artikler skal vises videre på MC
$typeid=0; //Hva slags ingresser skal vises? 0 viser alle typer.
$i1 = hentIngresser($nettstedid,38,$sprakid,$offset,$antallmotor);
$i2 = hentIngresser($nettstedid,39,$sprakid,$offset,$antallmotor);
function obj_date_compare($a, $b) {
return strcmp($a-publisertLang, $b-publisertLang);
}
if($ingresser = array_merge($i1 ,$i2))
{
$ingresser = usort($ingresser, 'obj_date_compare');
listIngresser($ingresser);
}
else
echo 'Det oppstod en feil ved henting av ingresser.';
function listIngresser($ingresser){
foreach($ingresser as $i){
echo '
tr
td width=100% align=left bgcolor=#ebe8e4 a
href='.$i-artikkellink.' img class=lesmer
src=/bilder/design/lesmer.gif width=61 height=19 align=right
b a class=forsidelinker
href='.$i-artikkellink.''.$i-overskrift.'/a /b
nbsp;nbsp;b class=dato('.$i-publisert.')/b
/tr /td
tr
td height=121 class=ingresshoved
a href='.$i-artikkellink.'img class=bilde src=' . $i-bilde . '
width=100 height=85 border=1 align=right/a
/font'.$i-ingress.'
tr td align=left width=100% bgcolor=#f4f4f4 p
/p /td '
;
}
}
?
** END PHP file ***
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RE: [PHP-DB] Datetime help in an INSERT...
I am now getting a completely different error that I cannot figure out how it is related. I have passed the $td variable into the function and then get it back as a POST variable from a hidden field within the function. Here is the error I am now seeing; Unknown column 'sn4265' in 'field list' This one is really throwing me at the moment as the only place the value sn4265 is stored is in the $sbcuid variable. This $sbcuid variable is not a part of the INSERT that I am performing which I think is generating this error. I am guessing that you guys will need more info to help out with this now. Just let me know what you want. Thanks again. Well, if $sbcuid isn't being used, something with its value is. What does the query look like? Common sense would say that if you're having trouble with a query, you'd at least post that bit of code, right? ---John W. Holmes... Amazon Wishlist: http://www.amazon.com/o/registry/3BEXC84AB3A5E PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Problem when using browder back button
I am a bit confused. I have built a web site from dynamic pages using PHP Mysql. When a user selects an item from a drop down list that takes then to a detail page. When they click back on the browser button they get a message about the page has expired. Is this an issue of session or cookie? How do I correct this ? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Problem when using browder back button
I am a bit confused. I have built a web site from dynamic pages using PHP Mysql. When a user selects an item from a drop down list that takes then to a detail page. When they click back on the browser button they get a message about the page has expired. Is this an issue of session or cookie? How do I correct this ? It's an issue of using POST instead of GET and has nothing to do with PHP. Most browsers won't re-post POST data for you for security reasons. GET, however, is just a URL with a query string, so it simply requests it. ---John W. Holmes... Amazon Wishlist: http://www.amazon.com/o/registry/3BEXC84AB3A5E PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Problem when using browder back button
It's an issue of using POST instead of GET and has nothing to do with PHP. Most browsers won't re-post POST data for you for security reasons. GET, however, is just a URL with a query string, so it simply requests it. Assuming you can't re-write the POST inputs as GETs, is there a way around this so users don't have to see that error? matt -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Problem when using browder back button
It's an issue of using POST instead of GET and has nothing to do with PHP. Most browsers won't re-post POST data for you for security reasons. GET, however, is just a URL with a query string, so it simply requests it. Assuming you can't re-write the POST inputs as GETs, is there a way around this so users don't have to see that error? Don't show any page that's created by a POST operation. If you must use POST in your forms, use a middle-man page that processes the form, inserts the data into the database or whatever, and redirects using header() back to another page. Then the user can go back and forth without issues. ---John W. Holmes... Amazon Wishlist: http://www.amazon.com/o/registry/3BEXC84AB3A5E PHP Architect - A monthly magazine for PHP Professionals. Get your copy today. http://www.phparch.com/ -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Calling Stored Procedures of Oracle in php
Hi,
I want to call a oracle stored procedure from php and retrieve the values.
My procedure has 2 parameters which i have to pass and based on that it will
return values. I am using the below code. I am getting errors like wrong
arguements passed. Can anybody tell me the reason?
$conn = ora_logon(abc,xyz);
$curs = ora_open($conn);
$query = begin prcoedurename('[EMAIL PROTECTED]','3');end;;
ora_parse($curs, $query);
ora_exec($curs);
ora_fetch($curs);
$nrows = ora_numrows($curs);
echo $nrows;
exit;
--
Thanks,
Ketan Parekh
-
Cybage Software Pvt Ltd.
6686359 Ext 235
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RE: [PHP-DB] Calling Stored Procedures of Oracle in php
If this is a cut-n-paste of the code, check the line below for spelling:
begin prcoedurename('[EMAIL PROTECTED]','3');end;;
^^^
Gary Every
Sr. UNIX Administrator
Ingram Entertainment
(615) 287-4876
Pay It Forward
mailto:[EMAIL PROTECTED]
http://accessingram.com
-Original Message-
From: Ketan Parekh [mailto:[EMAIL PROTECTED]
Sent: Friday, June 06, 2003 9:24 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Calling Stored Procedures of Oracle in php
Hi,
I want to call a oracle stored procedure from php and
retrieve the values.
My procedure has 2 parameters which i have to pass and based
on that it will
return values. I am using the below code. I am getting errors
like wrong
arguements passed. Can anybody tell me the reason?
$conn = ora_logon(abc,xyz);
$curs = ora_open($conn);
$query = begin prcoedurename('[EMAIL PROTECTED]','3');end;;
ora_parse($curs, $query);
ora_exec($curs);
ora_fetch($curs);
$nrows = ora_numrows($curs);
echo $nrows;
exit;
--
Thanks,
Ketan Parekh
-
Cybage Software Pvt Ltd.
6686359 Ext 235
--
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[PHP-DB] Transfering variables
Hi everybody! I'm wondering how to transfer variables in a php-document to the same script after the form has been submitted!? Thanks. André -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] MYSQL_CLIENT_COMPRESS
Hi, Is there any body who have experience with the flags in mysql_connect() in php = 4.3 . Especially the MYSQL_CLIENT_COMPRESS flag. Give it sense, to use this feature if both, apache/php and the mysql server reside on the same machine? Thanks, Armand -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] In Addition to 'Transfering variables'
I'm not sure if its get/post I need, beacause the variables are not a part of the form itself! -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] In Addition to 'Transfering variables'
We all might benefit from an explanation of what you're trying to do here. Where does the data in your variables come from? If the variables are not part of the form itself, what _are_ they part of? What will be done with the data (e.g., used as query parameters, page display parameters, etc.)? Have you researched Sessions? Rich -Original Message- From: Andri Sannerholt [mailto:[EMAIL PROTECTED] Sent: Friday, June 06, 2003 2:46 PM To: [EMAIL PROTECTED] Subject: [PHP-DB] In Addition to 'Transfering variables' I'm not sure if its get/post I need, beacause the variables are not a part of the form itself! -- 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] PHP Upgrade question...
I know there is another mailing list for installation questions, but I am hoping for some help here. I get enough mail from this list alone without having to join another. Anyway, while attempting to compile version 4.3.2 I am running into the following error during the configure... checking whether byte ordering is bigendian... configure: error: can not run tes t program while cross compiling At this point the configure simply dies. I have no idea on how to proceed from this. Scott Nipp Phone: (214) 858-1289 E-mail: [EMAIL PROTECTED] Web: http:\\ldsa.sbcld.sbc.com -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Is it worth upgrading Mysql on linux?
Hi there everyone, I run a server from a great company which gives everything you need from $99 a month (450 gigs bandwidth, PHP, MySQL, SSL etc ...) Anyway, I currently run one of the latest builds of PHP but MySQL is only version 3.23.56 (Or something similar) - does anyone see any advantages of upgrading to 4.0 ? Will I need to recompile PHP to use MySQL 4? This kind of thing is new to me (Linux as I used to use Windows). Thanks for everything. Chris
Re: [PHP-DB] Is it worth upgrading Mysql on linux?
hi! if you upgrade i'd say upgrade to 4.1 which provides the IN()-method which allows komplex subselects. But may be this is not a very good idea cause 4.1 is only alpha afaik .ma e: [EMAIL PROTECTED] w: http://www.abendstille.at Life would be easier if i knew the source code. @06.06.03 [21:31] Chris Payne wrote: Hi there everyone, I run a server from a great company which gives everything you need from $99 a month (450 gigs bandwidth, PHP, MySQL, SSL etc ...) Anyway, I currently run one of the latest builds of PHP but MySQL is only version 3.23.56 (Or something similar) - does anyone see any advantages of upgrading to 4.0 ? Will I need to recompile PHP to use MySQL 4? This kind of thing is new to me (Linux as I used to use Windows). Thanks for everything. Chris -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
RE: [PHP-DB] Is it worth upgrading Mysql on linux?
The following link may be of assistance: http://www.mysql.com/doc/en/Upgrading-from-3.23.html -Original Message- From: .ma [mailto:[EMAIL PROTECTED] Sent: Friday, June 06, 2003 3:40 PM To: PHP-DB Subject: Re: [PHP-DB] Is it worth upgrading Mysql on linux? hi! if you upgrade i'd say upgrade to 4.1 which provides the IN()-method which allows komplex subselects. But may be this is not a very good idea cause 4.1 is only alpha afaik .ma e: [EMAIL PROTECTED] w: http://www.abendstille.at Life would be easier if i knew the source code. @06.06.03 [21:31] Chris Payne wrote: Hi there everyone, I run a server from a great company which gives everything you need from $99 a month (450 gigs bandwidth, PHP, MySQL, SSL etc ...) Anyway, I currently run one of the latest builds of PHP but MySQL is only version 3.23.56 (Or something similar) - does anyone see any advantages of upgrading to 4.0 ? Will I need to recompile PHP to use MySQL 4? This kind of thing is new to me (Linux as I used to use Windows). Thanks for everything. Chris -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] How to find the last ID?
Hi there everyone,
I'm creating a new entry using the following:
mysql_query (INSERT INTO agents (agent_name)
VALUES ('$agentname')
but I need to find out the ID value it created, how can I do this easily?
Thanks
Chris
Re: [PHP-DB] How to find the last ID?
hi
you can easily do this by using
mysql_insert_id();
see
http://www.php.net/manual/en/function.mysql-insert-id.php
.ma
e: [EMAIL PROTECTED]
w: http://www.abendstille.at
/***/
life would be easier if i knew the source code.
Am Freitag, 06.06.03 um 23:16 Uhr schrieb Chris Payne:
Hi there everyone,
I'm creating a new entry using the following:
mysql_query (INSERT INTO agents (agent_name)
VALUES ('$agentname')
but I need to find out the ID value it created, how can I do this
easily?
Thanks
Chris
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Re: [PHP-DB] How to find the last ID?
Hi there,
Thanks for your quick reply, very appreciated :-)
Chris
hi
you can easily do this by using
mysql_insert_id();
see
http://www.php.net/manual/en/function.mysql-insert-id.php
.ma
e: [EMAIL PROTECTED]
w: http://www.abendstille.at
/***/
life would be easier if i knew the source code.
Am Freitag, 06.06.03 um 23:16 Uhr schrieb Chris Payne:
Hi there everyone,
I'm creating a new entry using the following:
mysql_query (INSERT INTO agents (agent_name)
VALUES ('$agentname')
but I need to find out the ID value it created, how can I do this
easily?
Thanks
Chris
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Re: [PHP-DB] How to find the last ID?
$last_id = mysql_insert_id();
Check out this page for details.
http://www.php.net/manual/en/function.mysql-insert-id.php
On Fri, 6 Jun 2003 17:16:01 -0400, Chris Payne wrote:
Hi there everyone,
I'm creating a new entry using the following:
mysql_query (INSERT INTO agents (agent_name)
VALUES ('$agentname')
but I need to find out the ID value it created, how can I do this easily?
Thanks
Chris
---
Listserv only address.
Jeff Shapiro
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[PHP-DB] php4.3+apache2+interbase 7 problem
hello, i am installing Apache 2.0.40 and Php 4.3.1 and Interbase 7 and for some reason i have problem when my code try to select from a database if the query is update or insert delete etc. all works well until i try a select. if i install interbase 6 or lower version then everything is ok. also i tryed running the interbase as: --with-interbase=shared and load it as extension but that failed too, also tryed to use an interbase.so of version 6 when the actual installed interbase is 7 and that worked but failes when doing transaction and blob stuff - which means - no good any of you guys out there encountered that kind of problem? any idea ? thanks ahead gmb. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP-DB] Seccond addition: 'Transfering variables'
Ok, the thing I accually want to do is to keep all the entries in one specific selection-option-box after the form has been submitted. I hope that is enough of explanation, because I cannot expect from you to go through my whole project... Regards, André -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] Seccond addition: 'Transfering variables'
Entries from what? Do you have a code example of what you're working with now? Believe it or not, most of us would rather read through it than guess what you're trying to accomplish. Edward Dudlik Becoming Digital www.becomingdigital.com - Original Message - From: Andr Sannerholt [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Friday, 06 June, 2003 17:51 Subject: [PHP-DB] Seccond addition: 'Transfering variables' Ok, the thing I accually want to do is to keep all the entries in one specific selection-option-box after the form has been submitted. I hope that is enough of explanation, because I cannot expect from you to go through my whole project... Regards, Andr -- 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
Re: [PHP-DB] Displaying groups from SELECT
On Fri, 2003-06-06 at 21:49, Becoming Digital wrote:
I'm wearing the stupid hat today, so please pardon this. I know I must be
overlooking something.
I have a small catalogue with two tables (categories, products) from which I'm
trying to display items. I'm trying to print the contents as below without
using two queries, but I'm having a difficult time with it.
cat1
prod1
prod2
cat2
prod1
prod2
etc.
I think this came up fairly recently, but I cannot for the life of me figure out
what search terms would answer this question. As you can see from the message
subject, I don't even know how to refer to my problem. Thanks a lot for all
your help.
Edward Dudlik
Becoming Digital
www.becomingdigital.com
Hi Ed,
The magic word is DISTINCT :)
$query=SELECT DISTINCT(category) AS cat_name FROM table_name;
$result=mysql_query($query);
print ul;
while ($row=mysql_fetch_array($result)) {
print li.($row[cat_name]).;
$query1=SELECT productname FROM tablename WHERE
category=.($row[cat_name]).;
$result1=mysql_query($query1);
while ($row1=mysql_fetch_array($result1)) {
print li.($row1[productname]).;
}
print /ul;
}
print /ul;
I hope this helps..
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Re: [PHP-DB] Displaying groups from SELECT
Thanks, Grhan, but I think I needed to explain things better. As is generally
a good idea, the categories are referenced in the product table by ID, not name.
Additionally, this is something along the lines of what I already had. I was
trying to use only one query and make PHP do the remaining work.
Here's the code I'm currently using, which I should have posted in the first
place.
?
$query = SELECT * FROM categories;
$result = mysql_query( $query );
$rows = mysql_num_rows( $result );
print ul\n;
while ( $cats = mysql_fetch_array( $result ) )
{
print li .$cats[cat_name] ./li;
$queryB = SELECT prod_name FROM products
WHERE prod_cat= .$cats[cat_id];
$resultB = mysql_query( $queryB );
print ul;
while ( $items = mysql_fetch_array( $resultB ) )
{
print li .$items[prod_name] ./li;
}
print /ul;
}
print /ul;
?
The more I think about it, the more it seems like I'll just have to use two
queries. I just didn't want to do so bcs the second query will run at least
four times and it seemed inefficient.
Edward Dudlik
Becoming Digital
www.becomingdigital.com
- Original Message -
From: Grhan zen [EMAIL PROTECTED]
To: Becoming Digital [EMAIL PROTECTED]
Cc: PHP-DB [EMAIL PROTECTED]
Sent: Friday, 06 June, 2003 23:07
Subject: Re: [PHP-DB] Displaying groups from SELECT
On Fri, 2003-06-06 at 21:49, Becoming Digital wrote:
I'm wearing the stupid hat today, so please pardon this. I know I must be
overlooking something.
I have a small catalogue with two tables (categories, products) from which I'm
trying to display items. I'm trying to print the contents as below without
using two queries, but I'm having a difficult time with it.
cat1
prod1
prod2
cat2
prod1
prod2
etc.
I think this came up fairly recently, but I cannot for the life of me figure
out
what search terms would answer this question. As you can see from the message
subject, I don't even know how to refer to my problem. Thanks a lot for all
your help.
Edward Dudlik
Becoming Digital
www.becomingdigital.com
Hi Ed,
The magic word is DISTINCT :)
$query=SELECT DISTINCT(category) AS cat_name FROM table_name;
$result=mysql_query($query);
print ul;
while ($row=mysql_fetch_array($result)) {
print li.($row[cat_name]).;
$query1=SELECT productname FROM tablename WHERE
category=.($row[cat_name]).;
$result1=mysql_query($query1);
while ($row1=mysql_fetch_array($result1)) {
print li.($row1[productname]).;
}
print /ul;
}
print /ul;
I hope this helps..
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Re: [PHP-DB] Calling Stored Procedures of Oracle in php
RE: [PHP-DB] Calling Stored Procedures of Oracle in phpYou did not get it. Actually
the procedure name means Any procedure Name say getName. It was just an example :-)
- Original Message -
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED] ; [EMAIL PROTECTED]
Sent: Friday, June 06, 2003 9:00 PM
Subject: RE: [PHP-DB] Calling Stored Procedures of Oracle in php
If this is a cut-n-paste of the code, check the line below for spelling:
begin prcoedurename('[EMAIL PROTECTED]','3');end;;
^^^
Gary Every
Sr. UNIX Administrator
Ingram Entertainment
(615) 287-4876
Pay It Forward
mailto:[EMAIL PROTECTED]
http://accessingram.com
-Original Message-
From: Ketan Parekh [mailto:[EMAIL PROTECTED]
Sent: Friday, June 06, 2003 9:24 AM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Calling Stored Procedures of Oracle in php
Hi,
I want to call a oracle stored procedure from php and
retrieve the values.
My procedure has 2 parameters which i have to pass and based
on that it will
return values. I am using the below code. I am getting errors
like wrong
arguements passed. Can anybody tell me the reason?
$conn = ora_logon(abc,xyz);
$curs = ora_open($conn);
$query = begin prcoedurename('[EMAIL PROTECTED]','3');end;;
ora_parse($curs, $query);
ora_exec($curs);
ora_fetch($curs);
$nrows = ora_numrows($curs);
echo $nrows;
exit;
--
Thanks,
Ketan Parekh
-
Cybage Software Pvt Ltd.
6686359 Ext 235
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[PHP-DB] PHP Problem
Hi All Hi i install PHP4.3 and apache2 now i create one page for PHP Info i use ?php phpinfo();? i show all info on web page . But Problem is here When i create one page for Connect Database on PostgreSQL.it not work. mean Postgresql work well i test it. i thing PHP Command problem. Please help where is Problem. in PHP Configuration or Apache Configuration. Ramesh Patel [EMAIL PROTECTED] -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP-DB] PHP Problem
Please post the code you are trying to use so that we can try to find the problem. Without data, we can't tell you where things went wrong. Edward Dudlik Becoming Digital www.becomingdigital.com - Original Message - From: Ramesh PAtel [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Saturday, 07 June, 2003 01:12 Subject: [PHP-DB] PHP Problem Hi All Hi i install PHP4.3 and apache2 now i create one page for PHP Info i use ?php phpinfo();? i show all info on web page . But Problem is here When i create one page for Connect Database on PostgreSQL.it not work. mean Postgresql work well i test it. i thing PHP Command problem. Please help where is Problem. in PHP Configuration or Apache Configuration. Ramesh Patel [EMAIL PROTECTED] -- 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
