PHP - using methods to input data into Array

Question

This is my first post here, so forgive me if I do a mistake or two :)

Few days ago I had an interview (the job offer is even not related to PHP), the employer gave me few excercises to show him my skill. Currently I'm stuck at PHP, this is my 1st time doing anything in PHP, I'm totally new to this one. I really want this job, so I hope you can help me out :)

The task: Create class "Person", inside Person create an Array variable named "data" with keys: "name", "surname" and "age", then create method addPerson(), which will fill the array with data. Create another method - showPerson() to show the gathered data (you can use print_r();).

So far, I've made this, but this is totally mess, it doesn't work. I have no knowledge about PHP, the code comes strictly from here and other websites.

<?php

class Person
{

    private $data = Array(
        'name'    => array(),
        'surname' => array(),
        'age'     => array()
    );

    function addPerson($name, $surname, $age)
    {
        array_push($this->$data['name']    = $name);
        array_push($this->$data['surname'] = $surname);
        array_push($this->$data['age']     = $age);
    }

    function showPerson()
    {
        echo json_encode($data);
        //print_r($data);
    }

}

$human = new Person;
$man->addPerson("John", "Johnes", 50);
$woman->addPerson("Maria", "Johnes", 45);
$human->showPerson();

Notice: Undefined variable: data; Fatal error: Cannot access empty property

The error marks first array_push in addPerson method. I don't know how to repair the code to make it work. Thank you in advance.

Im not a PHP person either but here are my thoughts: $this->$data['name'] = $name is assigning a string to $data['name']. however you declared $data['name'] to point to an array, not a string — Rana
– Rana, Commented Feb 2, 2016 at 20:12

cgalev · Accepted Answer · 2016-02-02 20:19:20Z

1

Use: $this->data['name'] instead of $this->$data['name'] (no dollar sign in front of data).

Also, No need to initialize name, surname and age as arrays. You could try something like :

private $data = Array(
'name' => '',
'surname' => '',
'age' => 0
);

Then you could try the addPerson method suggested by Nechemya-Kanelsky

answered Feb 2, 2016 at 20:19

cgalev

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Comments

RomanPerekhrest · Accepted Answer · 2016-02-02 20:18:30Z

0

Change your code as shown below:

class Person {

    private $data = Array(
        'name' => array(),
        'surname' => array(),
        'age' => array()
    );

    function addPerson($name, $surname, $age) {
        array_push($this->data['name'], $name);
        array_push($this->data['surname'], $surname);
        array_push($this->data['age'], $age);
    }

    function showPerson() {
        echo json_encode($this->data);
    }

}

$human = new Person;
$human->addPerson("John", "Johnes", 50);
$human->showPerson();

http://php.net/manual/en/function.array-push.php
array_push accepts two parameters: the array and the first value to push onto the end of the array

answered Feb 2, 2016 at 20:18

RomanPerekhrest

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5 Comments

Haseoh Over a year ago

@RomanPerekherst Thanks man! It works, but still, before the result there's a warning: "array_push() expects at least 2 parameters, 1 given" and points at those array_push lines. How do I get rid of it?

RomanPerekhrest Over a year ago

welcome! are you sure you have passed two parameters into each array_push call ? can you show me one of those array_push lines ?

Haseoh Over a year ago

@RomanPerekherst Here it is: array_push($this->data['name'] = $name); - I have to use "=" instead of "," because it doesn't show the result properly. With "," there's "array_push() expects parameter 1 to be array, string given" warning.

RomanPerekhrest Over a year ago

I suppose that you have changed $data property content. If you are going to use array_push then your $data property should have the following structure: private $data = Array( 'name' => array(), 'surname' => array(), 'age' => array() );. If you have changed nested arrays with strings, such as 'name' => '' then change your addPerson method content to this: $this->data['name'] = $name; $this->data['surname'] = $surname; $this->data['age'] = $age;

Haseoh Over a year ago

Yes, you're right. I'm sorry I caused some problems here. Thanks to you, I may be one step closer to finish my tasks :)

Nazmul Alam · Accepted Answer · 2016-02-02 20:21:57Z

0

Try this.

    <?php
    class Person {
    private $data = Array(
    'name' => array(),
    'surname' => array(),
    'age' => array()
    );

    function addPerson($name, $surname, $age){
        $this->data['name'] = $name;
        $this->data['surname'] = $surname;
        $this->data['age'] = $age;
    }
    function showPerson(){
        echo json_encode($this->data);
        //print_r($this->data);
    }   
    } 
    $man = new Person;
    $man -> addPerson("John", "Johnes", 50);
    $man -> showPerson();
    $woman = new Person;
    $woman -> addPerson("Maria", "Johnes", 45);
    $woman -> showPerson();
    ?>

answered Feb 2, 2016 at 20:21

Nazmul Alam

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Comments

Mdalz · Accepted Answer · 2016-02-02 20:25:12Z

0

Not quite sure why you have to predefine the array. This example works, similar to the answer by @RomanPerekhrest except without all the pushing business.

class Person {

    private $data = array();

    function addPerson($name, $surname, $age){
        $this->data['name'] = $name;
        $this->data['surname'] = $surname;
        $this->data['age'] = $age;
    }

    function showPerson(){
        echo json_encode($this->data);
        //print_r($data);
    }

}
$human = new Person;
$human -> addPerson("John", "Johnes", 50);
$human -> showPerson();

answered Feb 2, 2016 at 20:25

Mdalz

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Comments

David D · Accepted Answer · 2016-02-02 20:30:22Z

0

for multiple people:

class Person {
    public $data = array();

    public function addPerson($name, $surname, $age){
        $this->data[] = array(
            'name'=>$name,
            'surname'=>$surname,
            'age'=>$age
        );
    }

    public function showPeople(){
        // echo json_encode($data);
        print_r($this->data);
    }   

} 

$Person = new Person;
$Person->addPerson("John", "Johnes", 50);
$Person ->addPerson("Maria", "Johnes", 45);
$Person->showPeople();

for single person:

class Person {
    public $data = array();

    public function addPerson($name, $surname, $age){
        $this->data = array(
            'name'=>$name,
            'surname'=>$surname,
            'age'=>$age
        );
    }

    public function showPerson(){
        // echo json_encode($data);
        print_r($this->data);
    }   

} 

$Man = new Person;
$Man->addPerson("John", "Johnes", 50);
$Man->showPerson();

$Woman = new Person;
$Woman->addPerson("Maria", "Johnes", 45);
$Woman->showPerson();

It's clear you should read some PHP OOP beginner guides to understand better.

edited Feb 2, 2016 at 20:30

answered Feb 2, 2016 at 20:24

David D

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1 Comment

u_mulder Over a year ago

You needed to change "private" to "public" Serioulsy? Why?

LuanComputacao · Accepted Answer · 2016-02-02 21:03:13Z

0

According to the task you need create an array called 'data' in your Person object using the below format

Array
(
    [name] => Maria
    [surname] => Johnes
    [age] => 45
)

To this you can use this class

<?php

class Person
{
    private $data = Array();

    function addPerson($name, $surname, $age)
    {
        $this->data = ['name' => $name, 'surname' => $surname, 'age' => $age];
    }

    function showPerson()
    {
        print_r($this->data);
    }

}

//Using to create a woman
$woman = new Person;
$woman->addPerson("Maria", "Johnes", 45);
$woman->showPerson();


//Using to create a man
$man = new Person;
$man->addPerson("John", "Johnes", 50);
$man->showPerson();

Use this link to view this running https://eval.in/512101

edited Feb 2, 2016 at 21:03

answered Feb 2, 2016 at 20:54

LuanComputacao

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