On Mon, 2009-08-24 at 23:16 -0600, mike bode wrote:
I am trying to use PHP on my web site I am developing now. I have installed
Apache 2.2 and PHP 5.2. My problem is that I can execute PHP code embedded
in my HTML code, but I can't execute the same cose when I put it into a
separate .php
To: php-general@lists.php.net
From: mikebo...@hotmail.com
Date: Mon, 24 Aug 2009 23:16:02 -0600
Subject: [PHP] Beginner question
I am trying to use PHP on my web site I am developing now. I have installed
Apache 2.2 and PHP 5.2. My problem is that I can execute PHP code embedded
in
Well, as I said: Beginner ...
I am actually trying to implement some html code that I found on the web,
which uses php (see:
http://www.dynamicdrive.com/dynamicindex4/php-photoalbum.htm). They use the
method I tried to call a function php and then display a photo album.
Well, that doesn't
-Original Message-
From: mike bode [mailto:mikebo...@hotmail.com]
Sent: 25 August 2009 07:16 AM
To: php-general@lists.php.net
Subject: [PHP] Beginner question
I am trying to use PHP on my web site I am developing now. I have installed
Apache 2.2 and PHP 5.2. My problem is that I can
At 11:16 PM -0600 8/24/09, mike bode wrote:
I get a blank page. this is probably something really stupid, but I
have been wrecking my head for days now, and I can't figure it out.
anybody has an idea?
Mike:
Here's an idea -- try this:
http://sperling.com/examples/include-demo/
If you
Thanks.
I will surely do that.
In the meantime I have found out that it really is not a problem with the
code. I uploaded the (non-functioning code) from my PC to a public web
server, and lo and behold, the code works.
It thus appears that there is something wrong with my php-apache setup.
I see. PHP runs on the server and cannot directly interact with my browser,
only through Javascript. So I can use javascript to trigger a php script,
which does something on the server, and returns something to javascript on
my machine, which can then be used to do something on my browser.
At 2:45 PM -0600 8/25/09, mike bode wrote:
I see. PHP runs on the server and cannot directly interact with my
browser, only through Javascript.
Kind of.
PHP runs on the server and can create html, css, javascript et al.
However, PHP will complete it's task before the browser see's
anything.
On Fri, 2007-01-19 at 19:21 +0100, Delta Storm wrote:
Hi,
I have learned as lot about PHP but I still dont know how do they build
PHP based web sites.
Obviously you haven't learned enough ;)
I need to build a complete PHP site with a lot of content and integrate
CSS,javaScript and a lot
Doesn't the Apple Developer Tools disk have all that stuff?
http://developer.apple.com/tools/
-- jon
---
jon roig
web developer
email: [EMAIL PROTECTED]
phone: 888.230.7557
-Original Message-
From: rob [mailto:[EMAIL PROTECTED]
Sent: Thursday, April 08, 2004
On Thu, 2004-04-08 at 22:30, rob wrote:
I am a newcomer to PHP and I am looking for an intsallation package similar
to to FoxPro for MAC OSX 10.2.(Ie intsall appache, MY SQL and Latest version
of PHP) Does anyone know where I may find one?
Thanks RB
Try http://www.serverlogistics.com
They
On 10/4/04 1:37 AM, jon roig [EMAIL PROTECTED] wrote:
Thank you for your help! Php and mysql ship on MAC OSX servers but not on on
the normal machines.
Apple provides support for these two at
http://developer.apple.com/internet/opensource/osdb.html for help on MySQL
on MAC OSX including step by
Paul,
the warning that John refers to does occur - whether you see it
depends on the error reporting level, PHP will create the array for you
if you have not already initialized it but a warning will be generated
in such cases. - errors that may not display on one machine might be
visible
Jochem Maas wrote:
Paul,
the warning that John refers to does occur - whether you see it
depends on the error reporting level, PHP will create the array for you
if you have not already initialized it but a warning will be generated
in such cases. ...
error_reporting = E_ALL
...
At 11:41 AM -0800 2/6/04, Paul Furman wrote:
while ($file = readdir($dh)){
if (strstr ($file, '.jpg')){
$pictures[] = $file;
}
Spotted this problem when staring at your code.
Number of open braces: 2
Number of close braces: 1
You need to close off your while loop.
Should I
You didn't close your loop;
while ($file = readdir($dh)){
if (strstr ($file, '.jpg')){
$pictures[] = $file;
}
#vardump ($pictures[]);
} = here
?
instead of;
while ($file = readdir($dh)){
if (strstr ($file, '.jpg')){
$pictures[] =
Eric Gorr wrote:
the while function knows to just go through those and fills in array
numbers accordingly?
The while function has nothing to do with it. Using the syntax $array[]
simply adds an element onto the _end_ of the array and PHP picks the
next logical, numerical index.
OK thanks
Paul Furman [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Eric Gorr wrote:
[snip]
while ($file = readdir($fh)){
if (strstr ($file, '.jpg')){
$pictures[] = $file;
#print $pictures[]; #Fatal error: Cannot use [] for reading
Which element are you trying
:[EMAIL PROTECTED]
Sent: Friday, February 06, 2004 3:09 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP] beginner question about while loops
Eric Gorr wrote:
the while function knows to just go through those and fills in array
numbers accordingly?
The while function has nothing to do
- Original Message -
From: Paul Furman [EMAIL PROTECTED]
Totally ignorant need for clarification... Should I set up a counter for
loops $integer++ or if I'm going through something, the while function
knows to just go through those and fills in array numbers accordingly?
[snip]
OK thanks again for helping through the stumbling blocks... I'm rolling
again now.
John W. Holmes wrote:
Just make sure $pictures is defined as an array before you try to push a
value onto it (even using the method you have now), otherwise you'll get a
warning.
It seems to be working fine this
From: Paul Furman [EMAIL PROTECTED]
So when assigning values to an array inside a loop, it knows to advance
to the next but then if I want to print those out at the same time, it's
complaining
while ($file = readdir($fh)){
if (strstr ($file, '.jpg')){
$pictures[] = $file;
you could start with something like:
$old_string=44 55 99 111;
$new_string=ereg_replace( ,,$old_string);
echo $new_string;
should display
445599111
There are other regular expression aids listed on the page at:
At 01:59 AM 12/21/2001 -0500, jtjohnston wrote:
Michael,
$fp = fopen (./users.txt, r);
while (!feof ($fp)) {
$buffer = fgets($fp, 4096);
echo $buffer;
}
fclose($fp);
Ok. But $buffer is not an array, is it? Why/what is 4096? What is !feof ?
Don't you have access to the WWW? You need to
: [PHP] Beginner question?
Michael,
$fp = fopen (./users.txt, r);
while (!feof ($fp)) {
$buffer = fgets($fp, 4096);
echo $buffer;
}
fclose($fp);
Ok. But $buffer is not an array, is it? Why/what is 4096? What is !feof ?
Can anyone show me how to split() each line and parse to see if $user
?php
//$fp = fopen (C:/Program Files/localhost/info.txt, r);
//$fp = fopen (http://compcanlit.ca/;, r);
$fp = fopen (./info.txt, r);
while (!foef($fp))
{
$line = fgets($fp, 1024);
echo $line;
}
fclose($fp);
?
what you were doing was opening a file ready for reading and then printing
the
At 12:13 AM 12/21/2001 -0500, jtjohnston wrote:
This seems newbie. So what am I doing wrong? I want to print the
contents of the file. I get resource id #1 instead. Do I need to add a
header or what?
?php
//$fp = fopen (C:/Program Files/localhost/info.txt, r);
//$fp = fopen
Michael,
$fp = fopen (./users.txt, r);
while (!feof ($fp)) {
$buffer = fgets($fp, 4096);
echo $buffer;
}
fclose($fp);
Ok. But $buffer is not an array, is it? Why/what is 4096? What is !feof ?
Can anyone show me how to split() each line and parse to see if $user exists?
Here's what I'm
:) Again why 1024? or 4096 or ... ?
while (!foef($fp))
{
$line = fgets($fp, 1024);
echo $line;
}
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