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PHP Notice: Undefined index
Posted on 2011-03-06
Okay, I have the following:
<?php
$rqty = $_POST['rqty'];
$unitprice = $_POST['iprice'];
$rtotal = $rqty*$unitprice;
echo $rtotal;?>
I get the error however:
PHP Notice: Undefined index: iprice
Not sure what this means as $rtotal is displaying and is correct.
<?php
$rqty = $_POST['rqty'];
$unitprice = $_POST['iprice'];
$rtotal = $rqty*$unitprice;
echo $rtotal;?>
I get the error however:
PHP Notice: Undefined index: iprice
Not sure what this means as $rtotal is displaying and is correct.
[X]
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7 Comments
Take a look here:
http://php.net/manual/pl/function.error-reporting.php
http://php.net/manual/pl/function.error-reporting.php
Active 2 days ago
How do I get rid of it cos the page isn't going to the next page as it is meant to, it just shows that notice and stays there.
Active 2 days ago
never mind, I read the url you sent me. I used <?php error_reporting(E_ALL ^ E_NOTICE);?>
Is that okay?
Is that okay?
This is a notice that you tried to assign a value that was not set (not passed in by your form). While it does not abort execution, it might indicate an error, for example maybe you intended to POST 'unitprice' instead of 'iprice'?
In your case, it will make $rtotal = 0.
If you want to clean up the notice, you can check if the value is set before you try to use it, and default it to 0 if it is not set:
<?php
if ( isset($_POST['rqty']) ) {
$rqty = $_POST['rqty'];
} else {
$rqty = 0;
}
if ( isset($_POST['iprice']) ) {
$unitprice = $_POST['iprice'];
} else {
$unitprice = 0;
}
$rtotal = $rqty*$unitprice;
echo $rtotal;
?>
In your case, it will make $rtotal = 0.
If you want to clean up the notice, you can check if the value is set before you try to use it, and default it to 0 if it is not set:
<?php
if ( isset($_POST['rqty']) ) {
$rqty = $_POST['rqty'];
} else {
$rqty = 0;
}
if ( isset($_POST['iprice']) ) {
$unitprice = $_POST['iprice'];
} else {
$unitprice = 0;
}
$rtotal = $rqty*$unitprice;
echo $rtotal;
?>
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