This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Reference - What does this error mean in PHP?
(38 answers)
How to use store and use session variables across pages?
(8 answers)
Closed 3 years ago.
I have a login page when clicked on defines the variable $_session[username] = $name;
this $_session['username'] is displayed on my main page and it works fine if someone does login. however when someone does not login, the error notice:undefined index pops up. if i use
<?php session_start(); if($_SESSION['username']==""|| isset($_SESSION['username'])){echo "Login";}else{echo "Logout";}?>
i thought it would fix the problem. but it has the same error
if $_SESSION['username] is not set, you'll receive an undefined error message.
I'd recommend using empty(). It's the same thing as saying if( isset($_SESSION['username']) && $_SESSION['username] != '' ). This is more resource intensive (minimally in your case), as it has to check if the variable isset and that it isn't empty (blank, false, 0, null, etc).
if( empty($_SESSION['username']) ){
echo 'Login';
}else{
echo 'Logout';
}
Related
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
PHP check value against multiple values with OR-operator
(3 answers)
Closed 4 years ago.
Hi,
I have this code:
if($_GET['s']=="page1" || $_GET['s']=="page2" || $_GET['s']=="page3") {
dosomething();
}
and I get this error: : Undefined index: s in
Which I can dismiss only by adding this line:
$_GET['s']="";
but then this wont execute the code correctly since $_GET['s'] is not supposed to have any initial value. How do I fix this other than disabling the notices and errors?
Thank you.
You can check your $_GET['s']
if(isset($_GET['s'])) {
// your code goes here...
}
isset() is used to check if the index exists.
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 6 years ago.
I am aware that I can use sessions and GET/ POST method but I would like to achieve this using cookies. My code on page1.php is :
$_COOKIE['varname'] = $id;
and on page2.php is:
$id = $_COOKIE['varname'];
I get the following notice on my browser: Undefined index $id
What is the problem with my code?
Try using setcookie('varname', $id) then
if (isset($_COOKIE['varname']){ echo $_COOKIE['varname']; }
To set a cookie you need to use setcookie(). And it has to be done prior to any output.
setcookie("mycookie", "myvalue" , $validtime); // validtime is a integer.
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 6 years ago.
I am trying to pass a variable from one page to another using a URL but I get a NOTICE. I have read that NOTICE or WARNINGS can be suppressed, by I want to learn how to fix this if possible. Thanks!
Notice: Undefined index: status in /view_dept.php on line 139
this is the URL and where the query happens - (technically the page1 - where the variable comes from):
header('location:view_dept.php?status=success');
This is the "page2", where I need to pass the variable so I can echo a success message. This is the MAIN PAGE.
<?php
if( $_GET['status'] == 'success'):
echo 'SUCCESS';
endif;
?>
To avoid the Notice, try isset() to check if the index status is present
<?php
if( isset($_GET['status']) && $_GET['status'] == 'success'):
echo 'SUCCESS';
endif;
?>
This question already has answers here:
Reference - What does this error mean in PHP?
(38 answers)
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 8 years ago.
Okay so I have a simple script that is supposed to route traffic on my site however, I am have a hard time figuring out how to fix the "Undefined Index Error" that appears at the top of the page when ever I use the URL "localhost/sitename/"
if(isset($_GET['p'])){ $notify = $_GET['p']; }
if ($page == "home"){
include(__DIR__.'/_includes/pages/home.php');
} elseif($page == "maintenance") {
include(__DIR__.'/_includes/pages/maintenance.php');
} else {
include(__DIR__.'/_includes/pages/home.php');
}
?>
What am I doing wrong in regards towards this? Now the question was flagged, but I simply trying to redirect users to the home page even if they just type localhost/site name. And I still get the following error: Notice: Undefined index: p in C:\xampp\htdocs\BayingHound\index.php on line 3
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Reference - What does this error mean in PHP?
(38 answers)
Closed 8 years ago.
recently i just configure my script with common script for entry data
while im trying to submit the data , the data is successful to submit . but there is something notice that bothering me , they say Notice: Undefined index: type in D:\xampp\htdocs\project\submit.php on line 7
and the line is
<?php
include 'includes/config.php';
if($_SERVER["REQUEST_METHOD"] == "POST")
{
$type=addslashes($_POST['type']); // this is line 7
$nama_barang=addslashes($_POST['nama_barang']);
$kategori=addslashes($_POST['kategori']);
$deskripsi=addslashes($_POST['deskripsi']);
im using xampp v.3.2.1 , it is possible the notice is from the xampp ?
thanks you guys im so glad for your answer :))
type (and other $_POST members) may not always be set so you should try and code to detect that.
e.g:
$type = (isset($_POST['type'])) ? addslashes($_POST['type']) : false;
The notice mentions that your $_POST array has no index type.
So you should check for it before you try to access it:
$type = ""; //you could set a default here
if(array_key_exists("type", $_POST))
$type = addslashes($_POST['type']);