如何使用箭头指针运算符将PHP对象引用到动态生成的值

Question

I want to be able to do an object reference with a pointer operator to a dynamically generated value

//Jobs

class Jobs 
{
    private $orderid;
    private $quantity;
    private $contactno;
    private $contactname;
}

Hence I will create object keys

$keys = array(
    'orderid',
    'quantity',
    'contactno',
    'contactname'
);

Then do

$size = 4;
$i = 0;

//object instantiation

$jobObject = new Jobs();

$row = array('894949','45','08097577580','Emi');

for($i=0;$i<$size;$i++){

    //This is my challenge

    $jobObject->$keys[$i] = $row[$i];
}

I am very sure there should be a way to get the value of $keys[$i] that can be referenced by "$jobObject->" to do object property initialization.

I have tried to enclose $keys[$i] with braces by doing

$jobObject->{$keys[$i]} = $row[$i];

Yet it throws error. Please I need help on how to get this working or a way out.

Answer 1

You cannot assign a value directly to a protected property. Take a look to visibility reference .

The correct way to achieve what you want is the following.

<?php
class Jobs 
{
    // Change access to public
    public $orderid;
    public $quantity;
    public $contactno;
    public $contactname;
}

$keys = array(
    'orderid',
    'quantity',
    'contactno',
    'contactname'
);

$size = 4;
$i = 0;

//object instantiation

$jobObject = new Jobs();

$row = array('894949','45','08097577580','Emi');

for($i=0;$i<$size;$i++){

    // This is the way

    $jobObject->{$keys[$i]} = $row[$i];
}

var_dump($jobObject);

Answer 2

You have to use brackets for variable variables. You'll then be able to dynamically create your object:

$jobObject->{$keys[$i]} = $row[$i];

Take a look at the php documentation this subject: http://php.net/manual/en/language.variables.variable.php