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 am running a PHP script, and keep getting errors like:
Notice: Undefined index: search in C:\server\data\localweb\easy_ticket\admin\view_tickets.php on line 188
Line 188 looks like this:
if ($_GET["search"] || $_GET["filter"] && $_GET["f"]- && $_GET["sortby"]) {
After googling this issue I got this solution .. Using isset()
The problem is that I cant visualize the way I apply this to the code .. #.#
Yes you should use isset construct separated by commas..
if (isset($_GET["search"],$_GET["filter"],$_GET["f"],$_GET["sortby"])
{
// All variables have been set... do your coding stuff below
} else { echo "Some of the variables are not set"; }
try this
if ((isset($_GET["search"]) || isset($_GET["filter"])) && (isset($_GET["f"]) && isset($_GET["sortby"])) {
Related
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
PHP check value against multiple values with OR-operator
(3 answers)
Closed 4 years ago.
Hi,
I have this code:
if($_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 4 years ago.
I've been getting this error. Any fix??
PHP Notice: Undefined offset: 1 in line 402
if ($argv[1] == '--cover')
{
$Modules = new Modules;
$Modules->cover();
die();
}
The line is if ($argv[1] == '--cover')
An offset is undefined if it doesn't exist in the array.
Try this:
if(isset($argv[1]) && $argv[1] == '--cover'){
$Modules = new Modules;
$Modules->cover();
die();
This error means that the $argv[1] is undefined or null. Make sure this variable is properly assign or have a value before making a condition.
using print_r($argv); before making a condition will help you if the variable have a value or not.
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.
if($_REQUEST['command']=='add' && $_REQUEST['productid']>0){
$pid=$_REQUEST['productid'];
addtocart($pid,1);
header("location:shoppingcart.php");
exit();
}
Notice: Undefined index: command in C:\wamp64\www\WSMshop\products.php on line 5
and that's the error, i do what i know can fix it, but nothing happen. Please help me. im still newbie in this. TIA
Since too little code is provided - my suggestion is to simply make additional check whether the 'command' is being submit by changing the if statement to:
if(isset($_REQUEST['command']) && $_REQUEST['command']=='add' && $_REQUEST['productid']>0){
Cheers!
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 8 years ago.
When I run
if (filter_var($_GET['abc'], FILTER_VALIDATE_URL)) { ... }
and I don't give a ?abc= extension to my .php then I get notice:
Notice: Undefined index: abc in line...
Is there a way to bypass this notice and say if $_GET exists and then is abc then...
Thanks a lot
You have to test if key abc in GET exists, so:
if (isset($_GET['abc']) && filter_var($_GET['abc'], FILTER_VALIDATE_URL)) { ... }
you have to check whether the index is set using isset
if (isset($_GET['abc']) && filter_var($_GET['abc'], FILTER_VALIDATE_URL))
if (isset($_GET['abc']) && filter_var($_GET['abc'], FILTER_VALIDATE_URL)) { ... }
This way it checks if that exist and you won't get a notice , hope this help :)
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.
with this code:
$tstUsername = getValue($_POST['tstUsername']);
$tstPassword = getValue($_POST['tstPassword']);
if ($tstUsername !== false && $tstPassword !== false) {
echo "New string added to database!";
haydayshops_mysql_query($conn,"INSERT INTO table_accounts (username, password) VALUES >('$tstUsername','".sha1($tstPassword)."')");
}
The code works fine, but when i put the error mode on, i see this nasty errors:
Notice: Undefined index: tstUsername in ...
Undefined index: checkPassword in ...
You get these errors because you're trying to access to key of an array which doesn't exists. Here, you access to $_POST['tstUsername'] but it doesn't seems to exists in the $_POST data. Same case with a key named checkPassword somewhere else in your code.
To avoid that, use isset() function to check if they are declared before trying to access to the value.