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"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 9 years ago.
I'm getting an undefined index notice, but can't figure out how to fix it.
Notice: Undefined index: id in /home/jharvard/vhosts/localhost/html/book_details.php on line 10
This is the code that's having problems:
// query for the listing
$listing = query("SELECT name, author, id, edition, price, course, date FROM books WHERE submission = ?", $_GET["submission"]);
// query for the email of the seller
$seller = query("SELECT email FROM users WHERE id = ?", $listing["id"]);
Can anyone help? Thank you very much.
The problem is that the array returned from your first query, $listing doesn't have an element with the key id.
I suspect this would happen if there was no record in the database for whatever was passed in as the submission GET parameter. Hard to say for sure though without knowing how the query() function was written.
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 10 months ago.
I use the section, but this error occurs, in fact, it enters the site and everything is done correctly, but this error occurs behind the scenes. What is the solution?
PHP Notice: Undefined index: walletaddrass in /home/decent/public_html/user/presale.php on line 5
$id_wallet = $_SESSION["walletaddrass"];
The quantification is done as follows:
$_SESSION["walletaddrass"] = $id_wallet;
setcookie('logged-in','true', time() + 286400 , '/');
Please post the whole file where you are getting this error next time.
The issue is because you are trying to use $_SESSION["walletaddrass"] when it's not set, as you are setting it in the line below it.
You must first check if the session variable is set as follows:
if (isset($_SESSION['walletadrass'])) {
$id_wallet = $_SESSION["walletaddrass"];
$_SESSION["walletaddrass"] = $id_wallet; setcookie('logged-in','true', time() + 286400 , '/');
}
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 4 years ago.
I'm very new at php. I get these errors when I write these codes.
check to see if islem was passed in before trying to access it
if (isset($_POST['islem']) {
// your code here
}
Otherwise you won't be able to load the page up initially in order to submit the form. Also notice that I used _POST here, this is because you used the POST method in your form.
You are using the method POST in the form and getting the variables with GET
Either Try method : GET or try $_POST["islem"];
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.
Can you help me with this errors. I am actually trying to generate an invoice. However, my generated invoice having some problems as you can see as the image below.
Here is the image of my coding.
$oid = ( isset($_GET['oid']) ) ? $_GET['oid'] : null;
Also I believe line 38 should come before line 37. bindParam calls a variable defined after it.
You use variable $oid which it's data is $_GET['oid'], in your query before you actually declare/store data to it.
Move line 38 -> $oid = $_GET['oid']; after opening try/catch, that means to line 29, before stating the query. You should implement #fie answer too, to be sure that $oid is not null and that data actually exists.
Also something is wrong with line 73 but we can't see it.
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.
When i press F12 and look at the warnings i see a message that says
Line 18: Undefined index: task_uid
I just dont really understand why it is undefined, considering that i did declare it, and then select it, in here:
<?php
session_start();
require ("../../../main/gerais/DBConn.php");
$task_uid = $_GET['task_uid'];
$proj_id = $_GET['proj_id'];
?>
This here comes three lines before the "Line 18" that is supposed to be the error, according to the console. Here is this Line:
<?php
if(!empty($_POST['proj_id'])||($_POST['task_uid']))
die("Invalid proj_id or task_uid.");
$query = "
SELECT pm.proj_id, pm.task_uid, etc etc etc...
?>
So why is it undefined? And, while i'm not sure if this question is answerable without context, but how much does it matter, considering this "task_uid" shows up in the URL, and considering it only showed up as a warning, not as a real error?
You need to use !empty on both variables:
if(!empty($_POST['proj_id'])||!empty($_POST['task_uid']))
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 7 years ago.
i have session variable associated with user
the first time the following code without the second line is execute it works
on second time it does not giving me undefined offset.
unset($_SESSION['cart_items'][$i]);
this line of code solve the problem but my head is not getting around it
$_SESSION["cart_items"] = array_values($_SESSION["cart_items"]);
The unset() function removes the variable from scope and deletes the data, but it does not re-index the array, so you are left with an array that does not have an index of that number. array_values() just re-indexes the array so it is in order.