PHP with CMD display data when add variable

Question

I have test.php like below:

<?
    class Track{

        public function displaySong($title, $duration){

            $data = $title + $duration

            return $data;
        }

    }
?>

how to run it on CMD?... like if i do like this with cmd

song1 = Track("Britney - Toxic", "0:3:51")
song2->displaySong()

It returns

Song :  Britney - Toxic, Duration:  0:3:51

on cmd display

[Are you asking how to execute a PHP file via CMD?](https://stackoverflow.com/questions/10262532/running-php-script-from-the-command-line) — Script47, Jun 11 '18 at 03:07
Possible duplicate of [Running PHP script from the command line](https://stackoverflow.com/questions/10262532/running-php-script-from-the-command-line) — Script47, Jun 11 '18 at 03:07
when I execute `song1 = Track("Britney - Toxic", "0:3:51")` its return `syntax error near unexpected token `('` — Dicky Raambo, Jun 11 '18 at 03:14
Secondly, where is your constructor? Why are you passing values to `Track` without a constructor to mange them? — Script47, Jun 11 '18 at 03:18

Devin.Jin · Accepted Answer · 2018-06-11T03:27:53.080

0

you should use "",this is a suggestion php.net give (this is been blocked by stackoverflow)
1. you have an error at this : "$data = $title + $duration" => "$data = $title + $duration;"
2. if you want run a php file in CMD, you need use as : php test.php argv1 argv[2] ... , argv is the data you want to give the test.php
3. and the last , your test.php should add some code:
```
  public function displaySong($title, $duration){

     $data = $title + $duration;

     echo $data;
  }
```
  } $song = new Track(); $song->displaySong($argv1, $argv[2]);

edited Jun 11 '18 at 03:27

answered Jun 11 '18 at 03:21

Devin.Jin

32
3

I run this php file and I get a result ➜ ~ php c.php 1 2 3 – Devin.Jin Jun 11 '18 at 03:22

PHP with CMD display data when add variable

1 Answers1