variable is undefined PHP

Question

Could please help me understand what I am doing wrong here.

I am able to get the value of $income, but the issue is in the function ovIncome i get this error $income is undefined.

$income = $_REQUEST['income'];

//Gross Income Overview
function ovIncome() {
  //Check if Less Than or More Than
  if ($income == '0') {
    $wageVal = 'Less than € 30.984,- ';
  }
  elseif($income == '1') {
    $wageVal = 'More than € 30.984,- and same as € 61.200,-';
  }
  else {
    $wageVal = 'More than € 30.984,-';
  }
  echo "$wageVal";
}

Alive to die - Anant · Accepted Answer · 2019-08-16T04:55:25.500

1

You need to pass variable in your function as a parameter.

$income = $_REQUEST['income']; 
function ovIncome($income){//pass variable as parameter

And you will good to go.

Sample example: https://3v4l.org/dGYAj

Note: at the time of calling the function you need to pass that variable too( what I did in my code link in last line)

edited Aug 16 '19 at 04:55

answered Aug 14 '19 at 10:49

Alive to die - Anant

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I did, then got this error. Fatal error: Uncaught ArgumentCountError: Too few arguments to function ovIncome(), 0 passed in – Zain Aug 14 '19 at 10:53
That means you didn't pass the variable when you called the function – Andreas Aug 14 '19 at 10:57
Zain you need to pass variable while calling the function. Check my edited answer for more description – Alive to die - Anant Aug 14 '19 at 11:01

variable is undefined PHP

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