将变量传递给PHP中的另一个函数[重复]

Question

This question already has an answer here:

• “Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP 28 answers

In following function:

function dateSelect($monthname,$dayname,$yearname) {
   echo "rs: " . $monthname;    
  dropdownDateSelect($monthName, $dayName, $yearName);    
}

$monthname gets echoed out correctly. However, it is empty inside function dropdownDateSelect. I am not sure why. I am getting "Variable $monthName seems to be uninitialized" error message in Netbeans IDE.

function dropdownDateSelect($monthName, $dayName, $yearName){
  echo "rss: " . $monthName;
}

Now in"rss: $monthName", $monthName is empty.

Could you elaborate why this is happening and how I can pass $monthName to dropdownDateSelect function?

Thank you.

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Answer 1

You have to be aware of upper- and lowercase as they are not the same in PHP (see also the comments above). So this is how your function will work (with $monthName):

function dateSelect($monthName, $dayName, $yearName) {
    echo "rs: " . $monthName;    
    dropdownDateSelect($monthName, $dayName, $yearName);    
}