我不明白为什么这个 php 函数不起作用

Question

I am currently learning php. I just coded this function and I was hoping for the function to take a name and check if the name is in the multi array. If the name is contained in it, it will give back the name and the salary and job.

I used a foreach to loop through the list. I created the list outside of the function in a global scope. But am getting this error when I run the code. It is saying "multy is undefined". I don't understand why because is defined.

The following message occurs:

Notice: Undefined variable: multy in C:\xampp\htdocs\projects\index.php on line 19

Warning: Invalid argument supplied for foreach() in C:\xampp\htdocs\projects\index.php on line 19

Here's my code

<?php

$multy = [
    [
        'staff' => 'ben', 'job' => 'cooking', 'salary' => 1500,
    ],
    [
        'staff' => 'cy', 'job' => 'chef', 'salary' => 2000,
    ],
    [
        'staff' => 'sylva', 'job' => 'software engineer', 'salary' => 15000,
    ],
];

function checkingList($name){
  foreach($multy as $mult){
    if($mult['staff'] === $name){
      echo $mult['staff'] .', You are hired. Your job is ' . $mult['job'].' and your salary is'. $mult['salary'];
    }
  }
}
checkingList('sylva')
?>

Answer 1

The Problem is Variable Scopes.
Variables that are used and defined inside any function in PHP are by default, local variables; which means that they can only be interpreted inside the functions by the values that are given to them.
More on Variable Scope
More on PHP Variables

There are 2 ways to use a variable inside a function that is defined outside of its scope:
(1) [Common Way] Pass the variable to the function as an input.
(2) Using "global":

<?php

$multy = [
    [
        'staff' => 'ben', 'job' => 'cooking', 'salary' => 1500,
    ],
    [
        'staff' => 'cy', 'job' => 'chef', 'salary' => 2000,
    ],
    [
        'staff' => 'sylva', 'job' => 'software engineer', 'salary' => 15000,
    ],
];

function checkingList($name){

  //changed code here:
  global $multy;
  //end of change

  foreach($multy as $mult){
    if($mult['staff'] === $name){
      echo $mult['staff'] .', You are hired. Your job is ' . $mult['job'].' and your salary is'. $mult['salary'];
    }
  }
}
checkingList('sylva')
?>

Answer 2

You defined $multy variable outside of the function, so you need to pass it as a second parameter. So your code will be like this :

$multy = [
    [
        'staff' => 'ben', 'job' => 'cooking', 'salary' => 1500,
    ],
    [
        'staff' => 'cy', 'job' => 'chef', 'salary' => 2000,
    ],
    [
        'staff' => 'sylva', 'job' => 'software engineer', 'salary' => 15000,
    ],
];

function checkingList($name, $multy){
  foreach($multy as $mult){
    if($mult['staff'] === $name){
      echo $mult['staff'] .', You are hired. Your job is ' . $mult['job'].' and your salary is '. $mult['salary'];
    }
  }
}
checkingList('cy', $multy); 
// cy, You are hired. Your job is chef and your salary is 2000

Above code tested here