The answer to the interpolation problem is to use double quotes:
$dsn =
"{$servers["$id"]["type"]}://{$servers["$id"]["username"]}:{$servers["$id"]["password"[EMAIL PROTECTED]"$id"]["hostName"]}/{$servers["$id"]["databaseName"]}";
John Ellingsworth wrote:
Hi.
I am sure there is an easy answe
Hi.
I am sure there is an easy answer to this, but I do not know it and have
not found the answer anywhere. I want to be able to dynamically set my
database from an array of options:
$database = array(
array(
"type" => "mssql",
"name" => "database1",
"hostName" =>
-3-2004 12:42
Subject: Re: [PHP-WIN] variable passing problem
then it says...
Notice: Undefined index: fileToUpload in e:\viraj\office\inventory\2.php
on line 3
(fileToUpload is the variable name)
any suggestions?
Viraj.
"Svensson, B.A.T. (HKG)" wrote:
&g
then it says...
Notice: Undefined index: fileToUpload in e:\viraj\office\inventory\2.php
on line 3
(fileToUpload is the variable name)
any suggestions?
Viraj.
"Svensson, B.A.T. (HKG)" wrote:
>
> > (on windows: Notice: Undefined variable: fileToUpload in
> > e:\viraj\office\inventory\2.ph
> (on windows: Notice: Undefined variable: fileToUpload in
> e:\viraj\office\inventory\2.php on line 3) (on Linux:
> Resource id #1 )
Using $_POST['fileToUpload'], will probably solve that
problem for you.
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Hi all,
I have a script to upload a file into MySQL as BLOB.
this script works nicely on a LAMP server, but i have to implement it
on a Windows2000 server running PHP, Apache and MySQL.
-- here goes the "gather.php" content, which allow us to browse for a
specific
file in the system, and a "Up
Hi all,
I'm having trouble using variables.
If I create a PHP file and define the variables within it (e.g. $name =
"Bob";) and then ask the same file to ouput the variable as part of an echo
command (e.g. echo( "Hello, $name"); ), it works fine.
The problem comes when I try to link to a PHP fi
Hi, is there anyway to set the $document_root customized for each site? so
a user can take as doc root the root of their homesite? im win2k adv serv
with php 4.3.0, danke :)
--
SSS. Lic. Ádrian Enrique Labastida Cañizares
Gerencia de Sistemas & Administración Web
Host Depot S.A. de C.V.
[EMAIL P
I am new to php and I am having problems passing variables from one page to
another. I wrote a simple script as such:
name
then pass to this page:
I get this for the response:
Notice: Undefined variable: test in E:\websites\girls\cgi-bin\insert2.php on
line 8
Please help!!
e this helps
Stephen
- Original Message -
From: "suhailkaleem" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, November 18, 2002 1:05 PM
Subject: [PHP-WIN] Variable problem
> hi !
>
> yes you were right there is space between server name but
hi !
yes you were right there is space between server name but still i am unable
to remove it
$username ,$password,$server are comming from file and they have correct
values
$username1 = $username;
$password1 = $password;
$server1 = $server;
echo $server."test" ;
echo $password."test" ;
echo $
I have an file application that works for some data but not other?? The
error I'm getting causes an error on line 64
for (reset($t1); $ti=current($tl)
any thoughts??
System;
$connID = connect_to_np_system($sysid);
// get list of steps for the ENTIRE
I have register_globals = off and whenever I try to access a variable I MUST
use $_SERVER["var"], $_POST["var"], etc... Is there a way to just simply
access the variable as $var instead? I don't want to turn on
register_globals but I like to have short variable names... Someone has
told me that
Hi Tom,
First problem, is that the variable $incident_array isn't declared
beforehand outside the loop.
Do:
var $incident_array = array();
before the loop. That way PHP doesn't assume that it's scoping is only for
within the loop, since that's where the variable is created, not having
been p
Whoops, my mistake. I read your code completely wrong. Ignore my
comments. You are calling the first variable but I didn't expect it to
be called like that. I expected that the array would be appended to by
calling it like this:
$incident_array[]=$value;
Nevermind, sorry 'bout that.
Paul
Well, first of all, to get the first value of the array you should call
it as $incidnet_array[0]. Now doing that *might* fix your problem if you
were only recieving one value for the array. So if you were getting
output inside the array, how many outputs were you getting? One or two
or more?
~Pau
Hi folks.
I have an interesting problem with regards to variable scope in loops.
Now I understand how this operates in functions but its the first time I
have seen variable scope in a loop.
Below is a code snippet of what I have:
// execute SQL query
OCIExecute($sql_statement) or die("Couldn't
Hi,
>global $titel;
> $titel='html'
missing colon;
> $db = mysql_connect("localhost", "root") or die ("Kan geen connectie
> maken met het database programma. De Database server kan uit staan.");;
> mysql_select_db("setf",$db) or die ("Kan geen connectie maken met de
> database");
>
> $r
Hi, I've got a problem with variables.
I am trying to make a select staement for mysql using a variable in the
where clause. If i don't use the variable the result comes out fine but
with it there are parser errors on other lines. Can anybody tell me how
to insert a variable in such a script.
Hi.
I've sort of programmed myself into a corner on a recent projekt i've been working on.
This is the last part of the projekt, and i sure hope someone can help me on this one.
Scenario is:
I want to read a HTML file containing PHP variables into a string, then replace the
variables with tex
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