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 5 years ago.
anyone can help me?
i used this code for managing pages
and it's get
undefined index : page
undefined index : aksi
<?php
$page = $_GET['page'];
$aksi = $_GET['aksi'];
if($page == "") {
if($aksi == "") {
include "page/dashboard.php";
}
} elseif($page == "dashboard") {
if($aksi == "") {
include "page/dashboard.php";
}
}elseif($page == "masuk") {
if($aksi == "") {
include "page/kas_masuk/masuk.php";
}
if($aksi == "hapus") {
include "page/kas_masuk/hapus.php";
}
}
?>
Use isset() function like below,
UPDATE
To get value of $_GET[] use below
$page = (isset($_GET['page'])?$_GET['page']:'');
$aksi = (isset($_GET['aksi'])?$_GET['aksi']:'');
Now we will get value if $_GET[] have value otherwise empty without returning any error or warning
Related
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 5 years ago.
The highlighted line is line 24
Notice: Undefined index: action in C:\xampp\htdocs\authors.php on line 24
**if($_GET['action'] == "list") {**
$current = "members";
$list = "members";
}
else if($_GET["list"] == "authors") $current = "authors";
else $current = "members";
Check first if $_GET['action'] has been set or not
if(isset($_GET['action'])) {
if($_GET['action'] == "list") {
$current = "members";
$list = "members";
} else if($_GET["list"] == "authors") {
$current = "authors";
} else {
$current = "members";
}
}
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 6 years ago.
So i've got a rather complex bit of code that i'm trying to make work (actually; it works fine on my test server but not so much my live server. so i'm trying to make it rework) -- it keeps telling me that a variable i'm asking it to check with isset is not a defined variable. I'm probably over thinking this whole method and there's probably a simplier way. but i need you guys to bash me over the head with it apparently
include('bootstra.php');
include('includes/parts/head.php');
if(!isset($_SESSION['user']['uid']))
{
echo $html;
include('includes/parts/login.php');
if(isset($_GET['mode']))
{
$file = $_GET['mode'];
$path = "includes/parts/{$file}.php";
if (file_exists($path)) {
require_once($path);
} else {
die("The Page REquested could not be found!");
}
}
else
{
include('includes/parts/frontpage.php');
}
}
else
{
if(!isset($_SESSION['x']) || !isset($_SESSION['y']))
{
if(isset($_GET['x']) && isset($_GET['y']))
{
$_SESSION['x'] = $_GET['x'];
$_SESSION['y'] = $_GET['y'];
}
else
{
$_SESSION['x'] = 0;
$_SESSION['y'] = 0;
}
header("location:index.php?x={$_SESSION['x']}&y={$_SESSION['y']}");
}
else
{
if($_SESSION['x'] != $_GET['x'] || $_SESSION['y'] != $_GET['y'] )
{
header("location:index.php?x={$_SESSION['x']}&y={$_SESSION['y']}");
}
else
{
echo $html;
echo $base->displayMap($_GET['x'], $_GET['y']);
include('includes/context_menu.php');
}
}
}
Here is the exact error:
Notice: Undefined index: x in /home/****/public_html/reddactgame/index.php on line 27
Change it
<pre>
if( !isset($_SESSION['x']) || !isset($_SESSION['y']) )
</pre>
to
<pre>
if( !( isset($_SESSION['x']) && isset($_SESSION['y']) ) )
</pre>
There is no error, that is a Warning...
It is saying that $_SESSION['x'] was never setted, so, when you do that isset it tells you that it was never declared.
Example:
This gives you the same warning
empty($x);
This does not give you warning
$x = 0;
empty($x);
EDIT
I see this is a common question, so here is a better explanation.
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.
For some reason the h1 only says Welcome to Arma 3 Life Highway Patrol's
Any ideas?
<?php
//This file will contain the page name and the credits!
function thisPage($page){
if ($page == 'sops.php'){
$Npage = 'SOPs';
return $Npage;
}elseif ($page == 'changelog.php'){
$Npage = 'Website Changelog';
return $Npage;
}elseif ($page == 'ftodocuments.php'){
$Npage = 'FTO Documents';
return $Npage;
}elseif ($page == 'index.php'){
$Npage = 'Homepage';
return $Npage;
}elseif ($page == 'login.php'){
$Npage = 'FTO Login Page';
return $Npage;
}elseif ($page == 'socialstuff.php'){
$NPage = 'Social Stuff';
return $Npage;
}else{
return false;
}
}
?>
<?php
$findPage = (basename($_SERVER['PHP_SELF']));
$thisPageName = thisPage($findPage);
?>
<h1>Welcome to the Arma 3 Life Highway Patrol's <?php echo $thisPageName ?></h1>
<p> Website created by Austin Sharp and assisted by Greyson Henley.</p>
the problem is your variables. You have two different variables:
$NPage = 'Social Stuff';
return $Npage;
$NPage and $Npage they do not match.
You can use this $findPage = basename($_SERVER['SCRIPT_FILENAME']);
It will give the filename and print the proper output.
I actually just tested the code on MAMP and it worked as it should (saved the file as index.php). What server are you testing this on?
Shot in the dark, but try changing $_SERVER['PHP_SELF'] to $HTTP_SERVER_VARS['PHP_SELF'].
Maybe you could try phpinfo() to see if global php variables are empty.
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 7 years ago.
I have made this for loop to get a variable amount of columns selected from the database. As I'm running this, I get an error saying :
Notice: Undefined variable: kolom_1
Notice: Undefined variable: kolom_2
Notice: Undefined variable: kolom_3
Notice: Undefined variable: kolom_4
Notice: Undefined variable: kolom_5
Notice: Undefined variable: kolom_6
But I have it all placed in a for loop, why does he not recognize them? I am not getting what I'm doing wrong.
function lijst_ophalen($data, $from){
$totaal = count($data);
for ($i=1; $i<$totaal; $i++){
$kolom_[$i] = $this->mysqli->real_escape_string($data['kolom_' . $i . '']);
if($kolom_[$i]!="") $kolom_[$i] = "{$kolom_[$i]},"; else $kolom_[$i]="";
if($kolom_[$i]==$totaal) $kolom_[$i] = "{$kolom_[$i]}";
}
$from_table = "";
$categorie = "";
if($from == "bv"){
$from_table = "klanten_algemene_gegevens_bv";
$categorie = "";
}
if(($from == "1manszaak") || ($from == "vof")){
$from_table = "klanten_algemene_gegevens_vof_1manszaak";
if($from == "1manszaak"){
$categorie = "1manszaak";
}
if($from == "vof"){
$categorie = "vof";
}
$categorie = "WHERE soort_onderneming = '{$categorie}'";
}
if($from == "ib"){
$from_table = "klanten_ib";
$categorie = "";
}
$result = $this->mysqli->query(
<<<EOT
SELECT
{$kolom_1}
{$kolom_2}
{$kolom_3}
{$kolom_4}
{$kolom_5}
{$kolom_6}
FROM {$from_table}
{$categorie}
EOT
);
if($result){
$waardes = array();
while ($row = $result->fetch_assoc()) {
$waardes[]=$row;
}
return $waardes;
}
}
When you are initializing / filling data into the $kolom_ you are creating an' array instead of your (properly) intended series of different variables.
$kolom_[$i] <--- ARRAY
And lower down you are calling a series of variables $kolom_1, which doesn't exist because you created an' array and not a series of variables.
To avoid the error you can simply change your
$kolom_1
$kolom_2 (and so on)
calls into
$kolom_[1]
$kolom_[2] (and so on)
and you should be set to go.
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 8 years ago.
I was trying to print the value of a variable from a function, but it is showing undefined variable value. Please help me and tell me how to solve this problem; I'm pretty new
to PHP coding. I tried the function as public, but it shows some other error.
<?php
if(isset($_POST['btn_submit']))
{
$num1 = $_POST['cards1'];
$num2 = $_POST['cards2'];
$num1=substr($num1,0,(strlen($num1)-1));
$num2=substr($num2,0,(strlen($num2)-1));
}
function get_value() {
switch ($num2) {
case 'A' :
$value = 11;
break;
case 'J':
case 'Q':
case 'K':
$value = 10;
break;
default:
$value = $num2;
break;
}
//return $value;
}
echo '<script type="text/javascript">alert("Out Put is'.$value.' ");</script>';
//$total_value = $num1+$num2;
?>
$num2 is local to your function. You need to pass it in as a parameter:
function get_value($num2) {
*** YOUR CODE
}