PHP: Notice Undefined index even using isset

问题

i'm getting this even using 'isset':

Notice: Undefined index

it's giving the error at:

returnifisset($_COOKIE["miceusername"], ' value="', '"');

even though i am checking if the cookie isset or not. The function is:

function returnifisset($variable, $first = '', $last = ''){
    if(isset($variable) and !empty($variable)){ return $first.$variable.$last; }
}

how i should modify this function to make it work and not give that error!

回答1:

You are actually accessing the variable by passing it with your function, before the isset is ever called. You can't solve this problem.

回答2:

You use different function names printifisset and returnifisset. Also you can use only !empty() statement

来源：https://stackoverflow.com/questions/7096914/php-notice-undefined-index-even-using-isset