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.
So I receive this message when on my category pages:
Notice : Undefined index: text in /home/thespace/public_html/vqmod/vqcache/vq2-catalog_view_theme_spacec_template_product_category.tpl on line 13 ยป Meteorite Jewellery
Though I can't see the problem in that particular line of code:
<?php echo $breadcrumb['separator']; ?><span itemprop="title"><?php echo $breadcrumb['text']; ?></span>
Am I missing something here?
The message is pretty clear: "Undefined index: text", you don't have the index "text" in your variable. In that case $breadcrumb['text'] does not exists.
You can try a var_dump($breadcrumb) to debug this variable.
As you can see the doc, this function will print in the page the parameter given. I suggested to you to use it, just so you could see that the "text" was indeed missing. Now you must see why this index no longer exist or if it even existed before. We can't help you with the current info, though.
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 3 years ago.
I am looking at my 'WP_DEBUG' errors and the following two errors/notices pop up in a number of PHP files:
Notice: Undefined variable: defaultUI in...
Notice: Undefined variable: compileShortcodeUI in...
I have checked all of the the PHP files and lines specifically and every single one of these refers to this same bit of code:
$compileShortcodeUI .= "<div class='whatInsert whatInsert_".$shortcodeName."'>".$defaultUI."</div>";
What do I need to change to remove these errors?
You have to define the variable initially.
$compileShortcodeUI = ''; // null or any default value
$defaultUI = ''; // null or any default value
$compileShortcodeUI .= "<div class='whatInsert whatInsert_".$shortcodeName."'>".$defaultUI."</div>";
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 6 years ago.
Notice: Undefined variable: upid in C:\wamp\www\probeer\opvragen.php on line 112
Call Stack
# Time Memory Function Location
1 0.0007 264840 {main}( ) ..\opvragen.php:0
line 112
echo 'Update</td></tr>';
To simply answer your question the variable $upid is not defined, just like your error says.
Your are trying to echo out the content of the variable $upid right here:
?upid='.$upid.'
So somewhere above line 112 in your file, $upid should be assigned a value, but it isn't.
If you on line 111 do this:
$upid = "Test";
Then your script wont throw the error, but you have to actually assign the value you want for it, so its time to look at your code where you are supposed to define 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.
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:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 7 years ago.
Notice: Undefined index: n in C:\Users\Marseille\Desktop\Activation
W7\UwAmp\www\quezzer\question.php on line 5
this is a code tiped at question.php
$number = (int) $_GET['n'];
why i have this error! i have not understand!
Since this is a get variable you're using, from what I understand is that you maybe trying to run the script questions.php directly. Therefore no get variables are set, try
localhost/yourfolder/questions.php?n=5