How to use foreach() properly in PHP?

Question

It is my first time coding in PHP. I am trying to return an array which contains of the amount of each type of cash

public function test($a){
    $a = [];
        $cash = array(
            '20'   => 0,
            '30'   => 0,
            '40'   => 0,
            '50'  => 0,
            '60'  => 0,
            '70'  => 0,
            '80' => 0,
        );

        foreach($a as $k => $v) {
            if (array_key_exists("$k",$cash) && $v!= null ) {
                $cash[$k] += 1;
            }
        }

        return  $cash;
    }

When I try testing the code I get the error: Invalid argument supplied for foreach()

I test it with PHPÙnit where I call `$this->assertEquals($expected, $this->TestClass->findCashPayment($a));` — Bab, Oct 16 '18 at 23:28
I don't see how foreach can fail here since you redefine $a to be an empty array. Are you sure this is the line in your unit test that is failing? Or did you change anything after you ran your test? — Cave Johnson, Oct 16 '18 at 23:46
I get the error: `"Failed asserting that two arrays are equal"`. The expected array doesn't match the one I get — Bab, Oct 16 '18 at 23:48
Please edit your question with your _real_ issue, the input that you provide to that function, and the output that you expect from it — Federkun, Oct 16 '18 at 23:50
if `$a` is empty, `foreach` (which loops over each of its elements) will do nothing. That's probably why the test fails ;) — Stratadox, Oct 17 '18 at 00:32

score 0 · Answer 1 · answered Oct 16 '18 at 23:28

0

the 1st argument for foreach() must be an array. $a is not an array Also note that you are trying to use $count, which was not previously defined as an array. Either that "$cash" assignment should be "$count" or vice-versa.

answered Oct 16 '18 at 23:28

Yoshimitsu

370
2
9

My bad :) It's fixed now, but I still get the same error :/ – Bab Oct 16 '18 at 23:30
you must pass an array to your test() function. – Yoshimitsu Oct 16 '18 at 23:32
I defined it as an empty array in the beginning, but the test doesn't pass :/ – Bab Oct 16 '18 at 23:44

score 0 · Answer 2 · answered Oct 16 '18 at 23:54

0

Try calling with test(array('20' => 0)); and you will not get the error. You are probably calling it with something that is not an array?

answered Oct 16 '18 at 23:54

Tommy Pedersen

33
7

I am not sure what you actually are trying to achieve. Can you please explain further? – Tommy Pedersen Oct 17 '18 at 00:03
In your revised question you don't use the input. Please explain. – Tommy Pedersen Oct 17 '18 at 00:03

k.mak · Answer 3 · 2018-10-17T00:25:45.757

$cash[$key] += 1;

should be

$cash[$k] += 1;

I have included the code that I used for testing here:-

<?php
function test($a){
  $cash = array(
    '20' => 0,
    '30' => 0,
    '40' => 0,
    '50' => 0,
    '60' => 0,
    '70' => 0,
    '80' => 0,
  );

  foreach($a as $k => $v) {
      if (array_key_exists("$k",$cash) && $v != NULL ) {
        $cash[$k] += 1;
      }
    }
  return  $cash;
}

// code that I have added to test the function from the post question
$testArray = array(
  '20' => 1,
  '30' => 2,
  '40' => 3,
  '50' => 4,
  '60' => 5,
  '70' => 6,
  '80' => 7,
);

$cash = test($testArray);

foreach($cash as $k => $v) {
  echo("\n$k : $v");
}

?>

Results:-

score -2 · Answer 4 · edited Oct 17 '18 at 04:56

-2

Maybe you can try removing the comma at the end of your array.

change: '80' => 0,

to: '80' => 0

edited Oct 17 '18 at 04:56

Rafael

7,605
13
31
46

answered Oct 16 '18 at 23:25

Kenny's Labotatory

11
2

Sorry, it's a typo here on Stackoverflow. It's not there in my code – Bab Oct 16 '18 at 23:26
3

that's actually common practice, to leave the trailing comma. See https://stackoverflow.com/questions/2829581/why-do-php-array-examples-leave-a-trailing-comma – Yoshimitsu Oct 16 '18 at 23:27

How to use foreach() properly in PHP?

4 Answers4