On Mon, 2003-12-22 at 12:36, Carey Baird wrote:
Hey,
I have stored the name of a function as a variable. I have then passed the
variable to another function as follows:
//put function name in a variable
$contentfunction = newsadmincontent();
//pass variable to another function
try eval($contentfunction);
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December 2003 19:43
To: Carey Baird
Cc: php-gen
Subject: Re: [PHP] evaluating a variable
On Mon, 2003-12-22 at 12:36, Carey Baird wrote:
Hey,
I have stored the name of a function as a variable. I have then passed the
variable to another function as follows:
//put function name in a variable
[snip]
Eval ($contentfunction); gave me a parse error:
Parse error: parse error in
/home/pickled/public_html/main/inc/html.php(145)
: eval()'d code on line 1
[/snip]
try to add a semicolon to the end of the variable $contentfunction =
newsadmincontent();;
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On Mon, 2003-12-22 at 12:43, Carey Baird wrote:
Eval ($contentfunction); gave me a parse error:
Parse error: parse error in /home/pickled/public_html/main/inc/html.php(145)
: eval()'d code on line 1
$$contentfunction didnt output anything
Any other ideas?
What are you trying to do
Carey Baird wrote:
Hey,
I have stored the name of a function as a variable. I have then passed the
variable to another function as follows:
//put function name in a variable
$contentfunction = newsadmincontent();
Take off the parenthesis...
$contentfunction = 'newsadmincontent';
To call the
On Mon, 2003-12-22 at 13:09, John W. Holmes wrote:
Carey Baird wrote:
Hey,
I have stored the name of a function as a variable. I have then passed the
variable to another function as follows:
//put function name in a variable
$contentfunction = newsadmincontent();
Take off the
[snip]
I'm having trouble evaluating a dynamic variable. I want to check if the
variable $_POST[resolutions$i] is an empty string, as you'll see from
the
code. However, every way I've tried to check the variable so far
(including
empty() and eval()) always returns a null value, even when the
this worked for me.
?
$entries = 20;
for ($i=1; $i=$entries; $i++) {
if(isset($_POST[resolutions$i]) $_POST[resolutions$i] !==
''){
echo Hello $ibr /;
}
}
?
html
body
form method=POST
input type=hidden name=resolutions1 value=etefgtE /
input type=hidden name=resolutions3
Steve Goodman mailto:[EMAIL PROTECTED]
on Tuesday, September 02, 2003 12:54 PM said:
Can someone
recommend a way to reliably evaluate this variable?
1. (not positive on this point so correct me if I'm wrong) you shouldn't
compare with !==. Instead us !=.
2. What does the following do?
Chris W. Parker
on Tuesday, September 02, 2003 1:29 PM said:
1. (not positive on this point so correct me if I'm wrong) you
shouldn't compare with !==. Instead us !=.
Ok, turns out I'm wrong on that. I guess I should go look it up!!
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To
singular, not plural. Thanks for the
help everyone
.
-Original Message-
From: Chris W. Parker [mailto:[EMAIL PROTECTED]
Sent: Tuesday, September 02, 2003 3:29 PM
To: Stephen Goodman; [EMAIL PROTECTED]
Subject: RE: [PHP] evaluating dynamic variable
Steve Goodman mailto:[EMAIL PROTECTED
From: Stephen Goodman [EMAIL PROTECTED]
When I run the code you've attached, with $i iterating up to 3, I get
three 'empty!', even if $resolution1 should be 'not empty!'. It seems
like the $i in $_POST[resolutions.$i] is not getting parsed into a
value, and php is looking for a key
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