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 2 years ago.
So here is the code. I get error such as "Undefined index: fad in C:\wamp\www*imdone.php* on line ..."
But the link is - http://localhost/exercises.php?page=htmlex&get=imdone#container.
Hope i could explain barely
$folder = 'test/';
$uploadingfile = $folder . basename($_FILES['fad']['name']);
if (move_uploaded_file($_FILES['fad']['tmp_name'], $uploadingfile))
{
echo "Upload successfully.<br>";
} else {
echo "Upload unsuccesfully\n";
}
The 'fad' in the $_FILES['fad']['name'] is suppose to be the name of your input type=file field in the form. You should first double check with your form to confirm the name.
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 4 years ago.
i had a problem in this line, because the error message tell me theres an undefined index:
$strSQL = " UPDATE place SET
rating_value = '".$_POST["rating_value"]."'
WHERE place_id = '".$_POST["place_id"]."'
";
pls help me if you have a solution. thanks
Do you check your variables before do the request like :
if( !empty($_POST["rating_value"]) && !empty($_POST["place_id"]) )
{
// your request
} else {
echo("error");
}
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.
What I want to do is echo the value of $test on my page, through the uri. I know I'm doing something wrong, but I can't figure out what.
Code:
<?php echo $test?>
URL:
localhost/testMap/test.php?test=hallo
You can get the value of url with a get, and store it in the variable $test and then echo that out.
$test = $_GET['test'];
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 8 years ago.
Im getting the following error: Notice:undefined index:done
please advise what is wrong
<?php
if($_Get['done']==1 $msg = "Account details saved";
?>
Check first if the key exists, then the comparison to 1.
if (isset($_GET['done']) && $_GET['done'] == 1) {
$msg = '...';
}
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 9 years ago.
Can someone tell me what is wrong with this line of code? I am getting an error (yet the the page still functions properly despite the error)
Error:
"Notice: Undefined index: picture in C:\wamp\www\pplogin\members.php on line 180"
Here is the code:
179 $target = "user_images/fs/";
180 $target = $target . basename($_FILES['picture']['name']);
Like I said earlier despite the error the image gets uploaded to the destination and the image name is saved to my MySQL database.
I've been scratching my head for a while now...
$target = !empty($_FILES['picture']['name']) ? $target.basename($_FILES['picture']['name']) : false;
A file with id
picture
doesn't exist.
Use
isset()
to make sure It's set.
You can also use the ternary operator:
$img = isset($_FILES['picture']) ? $_FILES['picture'] : null;
Or something like that :)