This question already has answers here:
Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors?
(3 answers)
Closed 8 years ago.
I have one page where i need to send some results to another PHP file.
include("test.php");
but i want to pass variable (var) also so that which can be processed by test.php using $_GET['var'] command. How is it possible in PHP?
$var1="hello";
include ("test.php");
the test file contains below code
echo $var1." world";
Related
This question already has answers here:
Reference - What does this error mean in PHP?
(38 answers)
Closed 2 years ago.
I have 2 php files inside joomla and I want to make a variable from php-file 1 available in php-file 2.
php-file_1.php
$a_variable = $input->getString('text');
php-file_2.php
include(php-file_1.php);
echo $a_variable;
I get Notice: Undefined variable: a_variable in /var/www/vhosts/a_domain.de/httpdocs/php-file_2.php on line 2
what's wrong here?
include and require are not functions, they are language constructs. Therefore they need to be used without brackets.
Either way, you missed quotes around php-file_1.php.
include 'php-file_1.php'; should work fine.
This question already has answers here:
Reference - What does this error mean in PHP?
(38 answers)
Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors?
(3 answers)
Closed 5 years ago.
In first.php i have $name = "Josh". How to echo $name in secound.php. Both are in same folder.
In first.php
$name = "Josh";
In secound.php
echo $name;
Notice: Undefined variable: name in D:\XAMPP\htdocs\Folder\secound.php on line 2
Edit
include 'first.php';
echo $name;
To access the data in the second file, you'll need to include it.
At the top of your secound.php script, add this code:
include 'first.php';
You could also use require, like this:
require 'first.php';
The difference between the two can be found in the PHP manual:
require is identical to include except upon failure it will also
produce a fatal E_COMPILE_ERROR level error. In other words, it will
halt the script whereas include only emits a warning (E_WARNING) which
allows the script to continue.
This question already has answers here:
Reference Guide: What does this symbol mean in PHP? (PHP Syntax)
(24 answers)
Closed 7 years ago.
I want to know about meaning of $$val; what is the actual meaning is?i tried to find meaning of this in google but not understand properly. Please help me in this situations.
For example: suppose i have one variable which has $$value;
meaning of $$value?
You didn't put the language, but I'll assume you mean PHP
That's a variable variable.
That means you ware asking for the value of the variable whose name.is the first variable.
Here's an example, since that's quite confusing:
$foo = "Hi";
$bar = "world";
$world = "Hello!";
echo $$bar; // "Hello!"
php fiddle: http://ideone.com/Ve4YOO
Reference: https://secure.php.net/manual/en/language.variables.variable.php
This question already has answers here:
Redefining constants in PHP
(5 answers)
Closed 8 years ago.
I have a php file like this.
define('TEXT_ONE', 'testvalue1');
define('TEXT_TWO', 'testvalue12');
define('TEXT_THREE', 'testvalue13');
define('TEXT_FOUR', 'testvalue14');
define('TEXT_FIVE', 'testvalue15');
define('TEXT_SIX', 'testvalue16');
define('TEXT_SEVEN', 'testvalue17');
define('TEXT_EIGHT', 'testvalue18');
define('TEXT_NINE', 'testvalue19');
define('TEXT_TEN', 'testvalue10);
define('TEXT_ELEVEN', 'testvalue11');
I want to change some of the defined value through php code.
for ex:- I Want to change above file to
define('TEXT_ONE', 'newtext1');
define('TEXT_TWO', 'newtext12');
define('TEXT_THREE', 'newtext13');
define('TEXT_FOUR', 'newtext14');
define('TEXT_FIVE', 'newtext15');
define('TEXT_SIX', 'newtext16');
define('TEXT_SEVEN', 'newtext17');
define('TEXT_EIGHT', 'newtext18');
define('TEXT_NINE', 'testvalue19');
define('TEXT_TEN', 'newtext10);
define('TEXT_ELEVEN', 'testvalue11');
Can any one help me?
Thanks
define — Defines a named constant.
As the name suggests, that value cannot change during the execution of the script
This question already has answers here:
Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors?
(3 answers)
Closed 8 years ago.
MySQLiconfig.php:
<?php
$MySQLi = new mysqli('localhost','root','123','Database') or die('ERROR');
Other document:
<?php
require 'MySQLiconfig.php';
function DoAQuery($Query){
$MySQLi->query($Query);
}
The error is:
Undefined variable: MySQLi
You are wrong with variable scope. In the function you need to use like global $MySQLi or you can pass as function parameter.