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 want to check in my If-Statement if the status is 1(true) or 0(false). The value is stored in a mySQL DB with tinyint.
The following code returns an error:
if($zeile['STATUS']=='1')
{
echo "<td align='center'>true;</td>";
}
else{
echo "<td align='center'>false;</td>";
}
The error is:
Notice: Undefined index: STATUS in /home/u123210707/public_html/index.php on line 49'. line 49
is my If-statement
if($zeile['STATUS'] > 0)
{
/* whatever */
}
In your $zeile array there is no element like STATUS, that's why its give this error,
i think before if condition you must have to check array using print_r and check there is an element called STATUS.
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($_POST["place_id"]) )
{
// your request
} else {
echo("error");
}
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
PHP check value against multiple values with OR-operator
(3 answers)
Closed 4 years ago.
Hi,
I have this code:
if($_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 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.