Re: [PHP] Variable problem?

2007-07-24 Thread Daniel Brown
[ For those of you who are going to get pissed for top-posting, find something better to bitch about. I'm in a rush, but trying to help. ;-P ] Luc, You're not defining $contato_name. Do this: $contato_name = $_POST['contato_name']; On 7/24/07, Luc [EMAIL PROTECTED] wrote:

Re: [PHP] Variable problem?

2007-07-24 Thread Luc
Good evening Daniel, It was foretold that on 24/7/2007 @ 17:47:46 GMT-0400 (which was 18:47:46 where I live) Daniel Brown would write: snipped a bit You're not defining $contato_name. Do this: $contato_name = $_POST['contato_name']; Yes!! Oh so simple for a

Re: [PHP] Variable problem?

2007-07-24 Thread Richard Lynch
On Tue, July 24, 2007 3:33 pm, Luc wrote: if (empty($_POST['altura'])){ $contacter_form_error[] = 'favor preencher altura'; } if (empty($_POST['largura'])){ $contacter_form_error[] = 'favor preencher largura'; } else {

Re: Re: RE: [PHP] Variable Problem

2003-02-06 Thread Sunfire
Wong [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Thursday, February 06, 2003 12:25 AM Subject: Re: Re: RE: [PHP] Variable Problem On Thursday 06 February 2003 07:48, Sunfire wrote: on any server i ever dealt with if i put: input type=sent value=?php\$sent\? in the edit box for the value

Re: Re: RE: [PHP] Variable Problem

2003-02-06 Thread Jason Wong
On Thursday 06 February 2003 22:50, Sunfire wrote: no that doesnt work either what you get when you do that one in the edit box is: using: input type=text name=sent value=?php echo \$sent\;? or any different with echo in the value field gives me in the box: ?php \\;?} ...rest of the script

Re: [PHP] Variable Problem

2003-02-05 Thread Jason Wong
On Thursday 06 February 2003 04:41, Sierra Times.com wrote: I have a variable that get's chopped off at the %20 character. In the form page I have: bFrom: ?= $name;? brE-mail address: ?= $email;?br ?= $sent;?/b/font/td $sent and $name shows up fine, but immedieately I added

RE: [PHP] Variable Problem

2003-02-05 Thread Leonard Burton
You need to put quotes around the vars. If I were you I would do this: print input type=text name=sent value=\$sent\; print input type=text name=name value=\$name\; You need the quotes. Leonard. www.phpna.com -Original Message- From: Sierra Times.com [mailto:[EMAIL PROTECTED]] Sent:

Re: Re: [PHP] Variable Problem

2003-02-05 Thread Sunfire
PM Subject: Spam: Re: [PHP] Variable Problem On Thursday 06 February 2003 04:41, Sierra Times.com wrote: I have a variable that get's chopped off at the %20 character. In the form page I have: bFrom: ?= $name;? brE-mail address: ?= $email;?br ?= $sent;?/b/font/td $sent

Re: RE: [PHP] Variable Problem

2003-02-05 Thread Sunfire
need to echo...specially if there are 300 lines of code a block... - Original Message - From: Leonard Burton [EMAIL PROTECTED] To: Sierra Times.com [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Wednesday, February 05, 2003 4:16 PM Subject: Spam: RE: [PHP] Variable Problem You need to put

Re: RE: [PHP] Variable Problem

2003-02-05 Thread Chris Shiflett
--- Sunfire [EMAIL PROTECTED] wrote: you can do: echo BLOCK //any valid html code here input type=sent value=\sent\ BLOCK; No, he needed to output $sent, which is dynamic. Also, it is a lot cleaner (and likely faster) just to use ? (or ?php) and ? to switch in/out of PHP as needed. If

Re: Re: RE: [PHP] Variable Problem

2003-02-05 Thread Sunfire
Shiflett [EMAIL PROTECTED] To: Sunfire [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Wednesday, February 05, 2003 5:44 PM Subject: Spam: Re: RE: [PHP] Variable Problem --- Sunfire [EMAIL PROTECTED] wrote: you can do: echo BLOCK //any valid html code here input type=sent value=\sent\ BLOCK

Re: Re: RE: [PHP] Variable Problem

2003-02-05 Thread Chris Shiflett
--- Sunfire [EMAIL PROTECTED] wrote: on any server i ever dealt with if i put: input type=sent value=?php\$sent\? in the edit box for the value i always always end up with ?php instead of what $sent stands for..the only way i found it to work is with echo but maybe im missing something You

Re: Re: RE: [PHP] Variable Problem

2003-02-05 Thread Jason Wong
On Thursday 06 February 2003 07:48, Sunfire wrote: on any server i ever dealt with if i put: input type=sent value=?php\$sent\? in the edit box for the value i always always end up with ?php instead of what $sent stands for..the only way i found it to work is with echo but maybe im missing

RE: [PHP] Variable problem

2002-03-22 Thread Rick Emery
show your code -Original Message- From: Leif K-Brooks [mailto:[EMAIL PROTECTED]] Sent: Friday, March 22, 2002 5:18 PM To: [EMAIL PROTECTED] Subject: [PHP] Variable problem I have a variable name in a print END and then some text after it. The thing is, php thinks that the text is

RE: [PHP] Variable problem

2002-03-22 Thread Rick Emery
${varable}ABC -Original Message- From: Leif K-Brooks [mailto:[EMAIL PROTECTED]] Sent: Friday, March 22, 2002 5:21 PM To: Rick Emery Subject: Re: [PHP] Variable problem on 3/22/02 6:18 PM, Rick Emery at [EMAIL PROTECTED] wrote: show your code -Original Message- From

RE: [PHP] Variable problem

2002-03-22 Thread Chris
On 22 Mar 2002 at 17:27, Rick Emery wrote: ${varable}ABC in quotes you can help php with {$(varabl)}ABC -Original Message- From: Leif K-Brooks [mailto:[EMAIL PROTECTED]] Sent: Friday, March 22, 2002 5:21 PM To: Rick Emery Subject: Re: [PHP] Variable problem on 3/22

Re: [PHP] Variable problem

2002-02-04 Thread Randy Johnson
:26 PM Subject: RE: [PHP] Variable problem Use an array !?, I mean : $result[$i] = test; How do I combine the following so it is treated as one variable $i=10 $result$i=test; I want this to be: $result10=test; $i changes so I cannot just put in 10 instead of I

Re: [PHP] Variable Problem

2002-01-18 Thread val petruchek
?php $i = 10; eval (\$result$i=\test\;); echo $result10; ? Valentin Petruchek (aki Zliy Pes) http://zliypes.com.ua mailto:[EMAIL PROTECTED] - Original Message - From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Saturday, January 19, 2002 2:47 AM Subject: [PHP] Variable Problem

RE: [PHP] Variable problem

2002-01-18 Thread Yoel Benitez Fonseca
Use an array !?, I mean : $result[$i] = test; How do I combine the following so it is treated as one variable $i=10 $result$i=test; I want this to be: $result10=test; $i changes so I cannot just put in 10 instead of I. anybody know how i can do that? -- Yoel Benitez

Re: [PHP] Variable Problem

2002-01-18 Thread $BM'Ln8&8c(B
Hello. How do I combine the following so it is treated as one variable $i=10 $result$i="test"; I want this to be: $result10="test"; $i changes so I cannot just put in 10 instead of I. anybody know how i can do that? TIA Randy How about the below. ? $i=10; ${"result$i"}="test"; echo

Re: [PHP] Variable Problem

2002-01-18 Thread Philip Olson
How do I combine the following so it is treated as one variable. It's a good question, but why? Most likely an array will work best for this job, arrays are good: http://www.php.net/manual/en/language.types.array.php The man page on foreach is nice too, and has many examples which include

RE: [PHP] Variable Problem

2002-01-18 Thread K.Tomono
To: —F–쌤Œá Cc: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: Re: [PHP] Variable Problem How do I combine the following so it is treated as one variable. It's a good question, but why? Most likely an array will work best for this job, arrays are good: http://www.php.net/manual/en

RE: [PHP] Variable Problem

2002-01-18 Thread Steven Maroney
-Original Message- From: Philip Olson [mailto:[EMAIL PROTECTED]] Sent: Saturday, January 19, 2002 2:30 PM To: —F–쌤Œá Cc: [EMAIL PROTECTED]; [EMAIL PROTECTED] Subject: Re: [PHP] Variable Problem How do I combine the following so it is treated as one variable. It's a good

Re: [PHP] variable problem - help!

2002-01-16 Thread Nick Wilson
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 * On 16-01-02 at 12:45 * Dani said Hi! I 'm trying not to hard code my php coding and I'm trying to pass a variable name into the $row[coloumname_$variable]; I get an error message for this code. I'm just wondering if I could do

Re: [PHP] variable problem - help!

2002-01-16 Thread Stefan Rusterholz
Hi! I 'm trying not to hard code my php coding and I'm trying to pass a variable name into the $row[coloumname_$variable]; $row[columname_$variable] should do it. $row[constant] is threaten as constant which is normally not the programmers intention (set_error_reporting(0) and you'll see

RE: [PHP] variable problem - help!

2002-01-16 Thread Rick Emery
$row[coloumname_${v$ariable}] -Original Message- From: Dani [mailto:[EMAIL PROTECTED]] Sent: Wednesday, January 16, 2002 8:36 AM To: [EMAIL PROTECTED] Subject: [PHP] variable problem - help! Importance: High Hi! I 'm trying not to hard code my php coding and I'm trying to pass a

Re: [PHP] Variable Problem when UPGRADING...

2001-01-30 Thread Philip Olson
hi again james, btw, consider doing this : if (empty($submit)) { or if (!empty($submit)) { as it doesn't run into such problems. or use isset although i tend to almost always use empty, i LOVE empty() !!! each have their uses : http://www.php.net/manual/en/function.empty.php

Re: [PHP] Variable Problem when UPGRADING...

2001-01-30 Thread Teodor Cimpoesu
James Smith wrote: Alright, when i was programming with PHP3, I would use if statements like this: if(!$submit) { // display form } else { // display signup complete } to make multiple pages. Or I would do this: if($action == "signup") { if(!$submit) { //display

Re: [PHP] Variable Problem when UPGRADING...

2001-01-30 Thread Richard Lynch
You set error_reporting to 15 (E_ALL) instead of 7. Turn it back down to 7 or fix your scripts to use if (isset($submit)){ } -- Visit the Zend Store at http://www.zend.com/store/ Wanna help me out? Like Music? Buy a CD: http://l-i-e.com/artists.htm Volunteer a little time:

Re: [PHP] Variable Problem when UPGRADING...

2001-01-29 Thread Philip Olson
this has to do with error reporting, using the function : http://www.php.net/manual/en/function.error-reporting.php definition of types : http://www.php.net/manual/en/phpdevel-errors.php you may have E_ALL or E_NOTICE on within php.ini philip olson http://www.cornado.com/ On Mon, 29 Jan