Hello again;
I have two variables declared in the global scope of a script.
$string_a = 'stuff $string_b and more stuff';
$string_b = '';
One is a string with a reference for substitution to the other
string which is empty.
In the processing body of the script are if/if else blocks.
In these
On Jan 1, 2008 2:17 PM, jekillen [EMAIL PROTECTED] wrote:
Hello again;
I have two variables declared in the global scope of a script.
$string_a = 'stuff $string_b and more stuff';
$string_b = '';
One is a string with a reference for substitution to the other
string which is empty.
In the
On Tue, January 1, 2008 4:17 pm, jekillen wrote:
Hello again;
I have two variables declared in the global scope of a script.
$string_a = 'stuff $string_b and more stuff';
$string_b = '';
One is a string with a reference for substitution to the other
string which is empty.
In the processing
On Jan 1, 2008 2:17 PM, jekillen [EMAIL PROTECTED] wrote:
Hello again;
I have two variables declared in the global scope of a script.
$string_a = 'stuff $string_b and more stuff';
$string_b = '';
One is a string with a reference for substitution to the other
string which is empty.
In the
On Jan 1, 2008, at 3:31 PM, Richard Lynch wrote:
On Tue, January 1, 2008 4:17 pm, jekillen wrote:
Hello again;
I have two variables declared in the global scope of a script.
$string_a = 'stuff $string_b and more stuff';
$string_b = '';
One is a string with a reference for substitution to the
On Jan 1, 2008, at 3:48 PM, James Ausmus wrote:
On Jan 1, 2008 2:17 PM, jekillen [EMAIL PROTECTED] wrote:
Hello again;
I have two variables declared in the global scope of a script.
$string_a = 'stuff $string_b and more stuff';
$string_b = '';
One is a string with a reference for substitution
I seem to be having a problem in assigning a value to an array where the
array is called dynamically.
e.g. the physical name for the array is my_array, so:
my_array[1] = test;
works fine.
$array_name = my_array;
$array_name[1] = test;
does not work.
I have tried $$array_name[1] = test; but to
Hi,
I seem to be having a problem in assigning a value to an array where the
array is called dynamically.
e.g. the physical name for the array is my_array, so:
my_array[1] = test;
works fine.
$array_name = my_array;
$array_name[1] = test;
does not work.
I have tried $$array_name[1] = test; but
?php
$foo = my_array;
$$foo = array(0 = bar);
var_dump($my_array);
?
array(1) { [0]= string(3) bar }
Regards Frank
-Ursprüngliche Nachricht-
Von: Shaun [mailto:[EMAIL PROTECTED]
Gesendet: Donnerstag, 29. November 2007 13:57
An: php-general@lists.php.net
Betreff: [PHP] Variable Names
?php
$foo = my_array;
$$foo = array(0 = bar);
var_dump($my_array);
?
array(1) { [0]= string(3) bar }
Regards Frank
-Ursprüngliche Nachricht-
Von: Shaun [mailto:[EMAIL PROTECTED]
Gesendet: Donnerstag, 29. November 2007 13:57
An: php-general@lists.php.net
Betreff: [PHP] Variable Names
On Nov 29, 2007 7:47 AM, Mark Head [EMAIL PROTECTED] wrote:
I seem to be having a problem in assigning a value to an array where the
array is called dynamically.
e.g. the physical name for the array is my_array, so:
my_array[1] = test;
works fine.
$array_name = my_array;
$array_name[1] =
Hello Richard,
Wednesday, July 25, 2007, 12:05:22 AM, you wrote:
I think you're just missing $contato_name = $_POST['contato_name'] in
here somewhere...
That was what was missing indeed :-)
That said, if you are using striplashes, you have magic_quotes_gpc on,
and that's something you may
Good afternoon list,
Probably a no-brainer for the most of you but i'm in a pickle: i've
set up a variable to receive in the body of the notification mail
upon sending a form.
The body of the mail should read like this:
Obrigado por nos contatar, $contato_name. Recebemos sua mensagem
[ 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:
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 = $_POST['contato_name'];
Yes!!
Oh so simple for a
On Tue, July 24, 2007 3:33 pm, Luc wrote:
if (empty($_POST['altura'])){
$contacter_form_error[] = 'favor preencher altura';
}
if (empty($_POST['largura'])){
$contacter_form_error[] = 'favor preencher largura';
}
else {
On Wed, 2007-03-14 at 12:39 +0100, Jochem Maas wrote:
Richard Lynch wrote:
On Sat, March 10, 2007 6:28 am, Dave Goodchild wrote:
Hi guys, I have just read 'Programming PHP' (O'Reilly) and although I
think
it's a great book, I am confused about variable variables and
references -
not
On Sat, March 10, 2007 6:28 am, Dave Goodchild wrote:
Hi guys, I have just read 'Programming PHP' (O'Reilly) and although I
think
it's a great book, I am confused about variable variables and
references -
not the mechanics, just where you would use them.
The subject of variable variables is
2007/3/10, Dave Goodchild [EMAIL PROTECTED]:
Hi guys, I have just read 'Programming PHP' (O'Reilly) and although I
think
it's a great book, I am confused about variable variables and references -
not the mechanics, just where you would use them.
The subject of variable variables is explained
Hi guys, I have just read 'Programming PHP' (O'Reilly) and although I think
it's a great book, I am confused about variable variables and references -
not the mechanics, just where you would use them.
The subject of variable variables is explained but no examples are given as
to why and where
I must say, in all the years i am programming with PHP (about 5-6 years) i
NEVER used references.
So i don't find it useful, but well, if you want to give your variable
content more than one name, you can :)
I think you just need to start programming now, keeping in mind they are
available, but
: [PHP] Variable variables and references
Hi guys, I have just read 'Programming PHP' (O'Reilly) and although I think
it's a great book, I am confused about variable variables and references -
not the mechanics, just where you would use them.
The subject of variable variables is explained
sometimes you might (think you) need it.
Tijnema
- Original Message
From: Dave Goodchild [EMAIL PROTECTED]
To: PHP-General php-general@lists.php.net
Sent: Saturday, March 10, 2007 5:28:57 AM
Subject: [PHP] Variable variables and references
Hi guys, I have just read 'Programming PHP
]
To: PHP-General php-general@lists.php.net
Sent: Saturday, March 10, 2007 5:28:57 AM
Subject: [PHP] Variable variables and references
Hi guys, I have just read 'Programming PHP' (O'Reilly) and although I
think
it's a great book, I am confused about variable variables and
references
I am using phpmailer for a rich html mailer and I have been using
lines like
this to build up the mailbody
$mail_body .= div align=\center\img
src=\http://www.myurl.org/mylogo.gif\;;
Is there a build in function to assign html code to a php
variable and then
output them? Or can I read
code to a php
variable and then
output them? Or can I read an external php file into a variable?
Have a look at the heredoc syntax for declaring strings:
http://uk.php.net/manual/en/language.types.string.php#language.types.string.
syntax.heredoc
I find it really useful
?php
ob_start();
include( 'someFile.php' );
$content = ob_get_contents();
ob_end_clean();
?
no I think he needs file_get_contents();
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On Mon, 2007-02-12 at 14:52 +0200, clive wrote:
?php
ob_start();
include( 'someFile.php' );
$content = ob_get_contents();
ob_end_clean();
?
no I think he needs file_get_contents();
While that will certainly read PHP into a variable, it won't evaluate
the
I am using phpmailer for a rich html mailer and I have been using
lines like
this to build up the mailbody
$mail_body .= div align=\center\img
src=\http://www.myurl.org/mylogo.gif\;;
Is there a build in function to assign html code to a php
variable and then
output
Robert Cummings wrote:
On Mon, 2007-02-12 at 14:52 +0200, clive wrote:
?php
ob_start();
include( 'someFile.php' );
$content = ob_get_contents();
ob_end_clean();
?
no I think he needs file_get_contents();
While that will certainly read PHP into a variable, it won't evaluate
to assign html code to a php
variable and then
output them? Or can I read an external php file into a variable?
Have a look at the heredoc syntax for declaring strings:
http://uk.php.net/manual/en/language.types.string.php#language.typ
es.string.
syntax.heredoc
I
On Mon, 2007-02-12 at 15:20 +0200, clive wrote:
Robert Cummings wrote:
On Mon, 2007-02-12 at 14:52 +0200, clive wrote:
?php
ob_start();
include( 'someFile.php' );
$content = ob_get_contents();
ob_end_clean();
?
no I think he needs file_get_contents();
Nice work Rob!
You were totally correct as I needed to write the header, mail body and
footer before I dumped the whole thing using OB contents.
Ross
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align=\center\img
src=\http://www.myurl.org/mylogo.gif\;;
Is there a build in function to assign html code to a php variable and
then output them? Or can I read an external php file into a variable?
div id=container
div id=content-topthsi is some content/div
div id=content-middle
div id
While I'm inside a function call, is it possible to know what function
is currently being called?
Brent
On Fri, 2006-09-08 at 14:54 -0400, Brent Meshier wrote:
While I'm inside a function call, is it possible to know what function
is currently being called?
RTFM: http://www.php.net/manual/en/language.constants.predefined.php
Cheers,
Rob.
--
While I'm inside a function call, is it possible to know what
function is currently being called?
Yes.
http://us2.php.net/debug_backtrace
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Hi All
Im trying to store a document template in mysql that has php var names within
it and then find it in the datebase and print it out with the var names
replaced with the var values.
e.g i have this stored
Dear $title $name
we would like to..
I declare the vars
bob pilly wrote:
Hi All
Im trying to store a document template in mysql that has php var names within it and then find it in the datebase and print it out with the var names replaced with the var values.
e.g i have this stored
Dear $title $name
we would like to..
I
.
This is equivalent to scripting:
echo $foo;
I'm using eval() to execute the echo command and interpret the PHP
variable $foo.
___
eval(\$dog = \$bar;);
echo dog = . $dog;
RESULT: dog = cat
Here I'm using eval() to set one PHP variable equal to another
I was wondering if you could create variable variables for objects, please
see examples below, Im having problems getting it to work.
$data['fieldname'] = foo;
// without variable variables
$res = $db-query( SELECT foo FROM table );
while( $res-fetchInto( $db_data ) )
Rob, I thought I had tried everything, except...the following:
$item = $_POST[item$i]; //Where $i increments and loops thru my POST
vars...WOHOO!
This actually works. Why it works, I'll never know. but it does!
Robert Cummings [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
On Sun,
But when I tried your idea, it errors out with a PHP parse error.
I agree with you about it being semantically the same, however it doesn't
work. go figure
Robert Cummings [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
On Sun, 2005-12-11 at 02:03, The.Rock wrote:
Rob, I thought I
On Sun, Dec 11, 2005 at 12:10:22AM -0600, The.Rock wrote:
Stephen,
I'm using a template and so:
input name=item{number} size=60 class=formdata value={item}
becomes:
input name=item1 size=60 class=formdata value={item}
and so on.
You seemed to miss the point Stephen was
Is there a way to do this?
$var = $_POST[$var];
I have a form that submits data, except the name of each input field is
incrementing. So I need to increment the $_POST var. Any ideas?
I can't seem to get anything to work
Chris
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PHP General Mailing List (http://www.php.net/)
To unsubscribe,
On 12/10/05 9:06 PM, The.Rock [EMAIL PROTECTED] wrote:
Is there a way to do this?
$var = $_POST[$var];
I have a form that submits data, except the name of each input field is
incrementing. So I need to increment the $_POST var. Any ideas?
I can't seem to get anything to work
Chris
Here is an example of one of the fields:
input name=item{number} size=60 class=formdata value={item}
I'm looping thru this form several times, so each time the name gets
incremented. Do you have an example of what your talking about?
Stephen Johnson [EMAIL PROTECTED] wrote in message
Here is an example of one of the fields:
input name=item{number} size=60 class=formdata value={item}
I'm looping thru this form several times, so each time the name gets
incremented. Do you have an example of what your talking about?
Stephen Johnson [EMAIL PROTECTED] wrote in message
Here is an example of one of the fields:
input name=item{number} size=60 class=formdata value={item}
I'm looping thru this form several times, so each time the name gets
incremented. Do you have an example of what your talking about?
Stephen Johnson [EMAIL PROTECTED] wrote in message
I would do this instead
input name=item[$var] size =60 class=formdata value=item
Increment the $var with the {number} that you are using.
In the following PHP page
You would do
$var = $_POST['item'];
$var would then be an array accessed the same way you would access it
normally.
Does that
Here is an example of one of the fields:
input name=item{number} size=60 class=formdata value={item}
I'm looping thru this form several times, so each time the name gets
incremented. Do you have an example of what your talking about?
Stephen Johnson [EMAIL PROTECTED] wrote in message
On Sun, 2005-12-11 at 00:27, The.Rock wrote:
Here is an example of one of the fields:
input name=item{number} size=60 class=formdata value={item}
I'm looping thru this form several times, so each time the name gets
incremented. Do you have an example of what your talking about?
?php
$i =
Stephen,
I'm using a template and so:
input name=item{number} size=60 class=formdata value={item}
becomes:
input name=item1 size=60 class=formdata value={item}
and so on.
But when I need to get the post data, I cannot access in the way I'm used to
like:
$var1 = $_POST['item1'];
I tried that, and every combination thereafter. All I get is the following:
PHP Parse error: parse error, unexpected
I tried!
Robert Cummings [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
On Sun, 2005-12-11 at 00:27, The.Rock wrote:
Here is an example of one of the fields:
input
On Sun, 2005-12-11 at 00:41, Robert Cummings wrote:
On Sun, 2005-12-11 at 00:27, The.Rock wrote:
Here is an example of one of the fields:
input name=item{number} size=60 class=formdata value={item}
I'm looping thru this form several times, so each time the name gets
incremented. Do you
On Sun, 2005-12-11 at 01:15, The.Rock wrote:
I tried that, and every combination thereafter. All I get is the following:
PHP Parse error: parse error, unexpected
I tried!
Hmmm, the following works for me verbatim in a shell script... make sure
you don't include the opening and closing PHP
Rob, I thought I had tried everything, except...the following:
$item = $_POST[item$i]; //Where $i increments and loops thru my POST
vars...WOHOO!
This actually works. Why it works, I'll never know. but it does!
Robert Cummings [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
On
Rob, I thought I had tried everything, except...the following:
$item = $_POST[item$i]; //Where $i increments and loops thru my POST
vars...WOHOO!
This actually works. Why it works, I'll never know. but it does!
Robert Cummings [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
On Sun,
Rob, I thought I had tried everything, except...the following:
$item = $_POST[item$i]; //Where $i increments and loops thru my POST
vars...WOHOO!
This actually works. Why it works, I'll never know. but it does!
Robert Cummings [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
On Sun,
On Sun, 2005-12-11 at 02:03, The.Rock wrote:
Rob, I thought I had tried everything, except...the following:
$item = $_POST[item$i]; //Where $i increments and loops thru my POST
vars...WOHOO!
This actually works. Why it works, I'll never know. but it does!
*lol* It's semantically the same
Hi all, i have a question :), i need to display the instant value of a
variable (think at a variable that stores the interface trafic at one
moment) in browser, without having to refresh browser. I was wondering
if i can do that with php (not perl, cgi, etc).
If i missed something in my
FSA wrote:
Hi all, i have a question :), i need to display the instant value of a
variable (think at a variable that stores the interface trafic at one
moment) in browser, without having to refresh browser. I was wondering
if i can do that with php (not perl, cgi, etc).
If i missed something
Hi,
Try something like:
$classname = MyClass;
$construct_params = array(param1,param2,param3);
$return =null;
if(class_exists($classname)){
$param=explode( ',' , $construct_param);
# You must add here some security checks
eval($retrun = new $className($param));
var_dump($return);
}
olivier wrote:
Hi,
I answered this also, but apparently I only replied to the OP...
(reproduced here - minus the typos in my first post on this topic :-/)
if (class_exists($className, false)) {
$obj = new $className($construct_params);
} else {
die(Hack off mate.);
}
Try something
Sorry for typo error, just need my cup of cofee...
Here is a good post for:
http://fr.php.net/manual/fr/function.call-user-func-array.php
from taylor
08-May-2005 12:04
?php
/**
* Create an object of a specified type using
-- Message transmis --
Subject: Re: [PHP] variable object creating
Date: Mercredi 22 Juin 2005 14:19
From: olivier [EMAIL PROTECTED]
To: php-general@lists.php.net
Sorry for typo error, just need my cup of cofee...
Here is a good post for:
http://fr.php.net/manual/fr
Sorry for typo error, just need my cup of cofee...
Here is a good post for:
http://fr.php.net/manual/fr/function.call-user-func-array.php
- see:
from taylor
08-May-2005 12:04
- using eval.
I dont like eval too but i think that is depending on the pb we want to
solve... if you can change contrutor
olivier wrote:
Sorry for typo error, just need my cup of cofee...
:-) no probs ... I think we gave the OP plenty to think about!
liked you 'Register' idea - seems like it could work well.
also nice one for pointing out my php5isms - I forget that alot
of stuff I use is php5 only (e.g. second
On Tue, June 21, 2005 8:33 pm, Eli said:
Hi,
I want to create an object in a form that I get the class name and its
parameters, and I need to create that object...
How can this been done?
i.e:
$classname = MyClass;
$construct_params = array(param1,param2,param3);
/* Now create the
Hi,
I want to create an object in a form that I get the class name and its
parameters, and I need to create that object...
How can this been done?
i.e:
$classname = MyClass;
$construct_params = array(param1,param2,param3);
/* Now create the object with the given classname and params... how?
* Brad Brevet [EMAIL PROTECTED]:
This seems to be what I was looking for, but I am curious, will the / be
included in the variable? Will I have to do a stripslashes() command on it?
If you echo out $_SERVER['PATH_INFO'] for the URL shown below, it will
give you:
/321
Usually what you do
What I do to control it only by PHP without using the mod_rewrite for
apache is to use URL with this format:
http://sample.com/script.php/param1/param2/param3
Then, work in the script looking at the variable
$_SERVER['REQUEST_URI'] wich will contain, in this sample:
* Jordi Canals [EMAIL PROTECTED]:
What I do to control it only by PHP without using the mod_rewrite for
apache is to use URL with this format:
http://sample.com/script.php/param1/param2/param3
Then, work in the script looking at the variable
$_SERVER['REQUEST_URI'] wich will contain, in
Hi, I am curious how to pass a variable without using something like id=321.
I have seen sites that have something like
http://www.website.com/something/321 and the variable is passed how exactly
is that done? And is it called something specific so I know how to refer to
it in the future?
On Apr 8, 2005 4:11 PM, Brad Brevet [EMAIL PROTECTED] wrote:
Hi, I am curious how to pass a variable without using something like id=321.
I have seen sites that have something like
http://www.website.com/something/321 and the variable is passed how exactly
is that done? And is it called
Brad Brevet:
Hi, I am curious how to pass a variable without using something like id=321.
I have seen sites that have something like
http://www.website.com/something/321 and the variable is passed how exactly
is that done? And is it called something specific so I know how to refer to
it
Brad Brevet wrote:
Hi, I am curious how to pass a variable without using something like id=321.
I have seen sites that have something like
http://www.website.com/something/321 and the variable is passed how exactly
is that done? And is it called something specific so I know how to refer to
it in
This seems to be what I was looking for, but I am curious, will the / be
included in the variable? Will I have to do a stripslashes() command on it?
Brad
Hans Juergen von Lengerke [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Brad Brevet:
Hi, I am curious how to pass a variable
James Williams wrote:
Hey guys, I'm coding a forum right now and I just decided that I needed
James,
I'll assume that you're coding a forum because you feel like it. :-)
(i.e. there are lots of forum 'packages' out there)
to code some variable replacement to create a simple template system.
Hey guys, I'm coding a forum right now and I just decided that I
needed to code some variable replacement to create a simple template
system. This is the code I have but it doesn't work
|?php
$tpl = file
Hi all
I have few variables in this format:
$isproductssorttext = 150;
$isofferssorttext = 250;
$isnewproductssorttext = 350;
$isproductscount = 50;
$isofferscount = 30;
$isnewproductscount = 20;
etc
What I want to do is have a variable
e.g. $x = products;
and create from that, the
From: mario [mailto:[EMAIL PROTECTED]
Hi all
I have few variables in this format:
$isproductssorttext = 150;
$isofferssorttext = 250;
$isnewproductssorttext = 350;
$isproductscount = 50;
$isofferscount = 30;
$isnewproductscount = 20;
etc
What I want to do is have a
On Tue, 4 Jan 2005, mario wrote:
Hi all
I have few variables in this format:
$isproductssorttext = 150;
$isofferssorttext = 250;
$isnewproductssorttext = 350;
$isproductscount = 50;
$isofferscount = 30;
$isnewproductscount = 20;
etc
What I want to do is have a variable
mario wrote:
Hi all
I have few variables in this format:
$isproductssorttext = 150;
$isofferssorttext = 250;
$isnewproductssorttext = 350;
$isproductscount = 50;
$isofferscount = 30;
$isnewproductscount = 20;
etc
What I want to do is have a variable
e.g. $x = products;
and create from that,
On Tue, 2005-01-04 at 16:54 +0200, mario wrote:
Hi all
I have few variables in this format:
$isproductssorttext = 150;
$isofferssorttext = 250;
$isnewproductssorttext = 350;
$isproductscount = 50;
$isofferscount = 30;
$isnewproductscount = 20;
etc
What I want to do is have
Hi there, I have some code where I am using the $_SERVER['PHP_SELF']
array to reference my script for a form action... now for some reason
the file name is not being rendered out. I am including only small
snippets of my code to see where my error is...
?php
$editFormAction =
On Saturday 04 December 2004 15:36, Dustin Krysak wrote:
Hi there, I have some code where I am using the $_SERVER['PHP_SELF']
array to reference my script for a form action... now for some reason
the file name is not being rendered out. I am including only small
snippets of my code to see
I have a form element which allows a user to go back
x number of days in the records.
$sql.=sprintf(SELECT * FROM records WHERE
Date_Sub(Curdate(), interval day) $%s =
PostStart)
This formula works when I actually hardcode the number
of days in where this mess -- $%s is right now.
I'm not
, September 16, 2004 6:02 PM
To: revDAVE
Cc: PHP
Subject: Re: [PHP] Novice PHP Variable/Link Question
On Thu, 16 Sep 2004 14:54:34 -0700, revDAVE
[EMAIL PROTECTED] wrote:
How can I use a PHP variable as the destination for a link?
? $mylink = 'thispage.htm'
a href=thispage.htmgo
On Friday 17 September 2004 21:49, Gryffyn, Trevor wrote:
Alternately, you can do it the lazy way like me:
a href=?=$url??=$url?/a
? Echo $url; ?
Is the same as...
?=$url?
I also think that's a little easier to read. But that's my preference
in style.
The use of this syntax is
How can I use a PHP variable as the destination for a link?
? $mylink = 'thispage.htm'
a href=thispage.htmgo here/a
With var...? How do I write this?
a href=??? $mylink ???go here/a
?
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Thanks - RevDave
[EMAIL PROTECTED]
[db-lists]
Check out some great Domain Names at:
http
On Thu, 16 Sep 2004 14:54:34 -0700, revDAVE [EMAIL PROTECTED] wrote:
How can I use a PHP variable as the destination for a link?
? $mylink = 'thispage.htm'
a href=thispage.htmgo here/a
With var...? How do I write this?
a href=??? $mylink ???go here/a
a href=?php echo $url;??php
On 9/16/04 3:01 PM, Greg Donald [EMAIL PROTECTED] wrote:
a href=?php echo $url;??php echo $url;?/a
It worked great - thanks a lot.
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Thanks - RevDave
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[db-lists]
Check out some great Domain Names at:
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PHP General Mailing List
Hi List,
I'm using PHP5 with global variables off. I've got around 20 dynamically
generated forms with a total of 300 different variables sent thru these
forms. I'd like to use variable variables, but according to the manual this
is not possible.
Now comes my Questoin, how do I receive my
JanBro wrote:
Hi List,
I'm using PHP5 with global variables off. I've got around 20 dynamically
generated forms with a total of 300 different variables sent thru these
forms. I'd like to use variable variables, but according to the manual this
is not possible.
Now comes my Questoin, how do I
On Sat, 28 Aug 2004 11:43:13 +0200, JanBro [EMAIL PROTECTED] wrote:
Hi List,
I'm using PHP5 with global variables off. I've got around 20 dynamically
generated forms with a total of 300 different variables sent thru these
forms. I'd like to use variable variables, but according to the
I have the following:
[SNIP]
...
$Emp_Status_Rqmt=$row[Emp_Status_Rqmt];
}
if( $Emp_Status_Rqmt == 'Permanent' )
{
$UserStatus = 'Permanent';
}
if( $Emp_Status_Rqmt == 'Contractor' )
{
$UserStatus = 'Contractor';
}
else
{
$UserStatus == 'Flexible';
}
[SNIP]
On Wed, 11 Aug 2004 00:36:05 +0100, Harlequin
[EMAIL PROTECTED] wrote:
I have the following:
[SNIP]
$Emp_Status_Rqmt=$row[Emp_Status_Rqmt];
}
if( $Emp_Status_Rqmt == 'Permanent' )
{
$UserStatus = 'Permanent';
}
if( $Emp_Status_Rqmt == 'Contractor' )
{
Hey,
I was looking at your code and should that not be an If - ElseIf - Else
structure? The way I read this is it first checks to see if the var is set
to 'Permanent' and sets $UserStatus. Then regardless of the previous test it
does another test on the $Emp_Status_Rqmt to see if it is
On Tue, 2004-08-10 at 16:36, Harlequin wrote:
I have the following:
[SNIP]
...
$Emp_Status_Rqmt=$row[Emp_Status_Rqmt];
}
if( $Emp_Status_Rqmt == 'Permanent' )
{
$UserStatus = 'Permanent';
}
if( $Emp_Status_Rqmt == 'Contractor' )
{
$UserStatus = 'Contractor';
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