Create a PHP Object using an Array List dynamically

Question

I want to create an Object in PHP using, as element, a list of fields taken from an array list obtained from a function outside the class. I need to do that dinamically, cause I have to compile the array from a DB and this list of data could change. Is this possible? I've a function called:

public $classKeyword = "contratti";
public $arrayList = getDbTableStructure($classKeyword);

And this function return me the array list and I'll cicle it to build the object. This function work in all other the project (not using class), but seems not working if I call it from a class... I'm losing something?

Don't know if you lost something, but you might want to give us a little more code so we have a better understanding of the problem you are having. Sorry for posting it as an answer, hit the wrong button, didn't really think about it. — user936965
– user936965, Commented Oct 24, 2014 at 13:40

jewelnguyen8 · Accepted Answer · 2014-10-24 13:30:42Z

1

just do like this: (object) $array

answered Oct 24, 2014 at 13:30

jewelnguyen8

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1 Comment

Alexander Guz Over a year ago

Right, I haven't thought about that.

Logan Murphy · Accepted Answer · 2014-10-24 13:22:59Z

0

You cannot call the function where you are calling it. It is a syntax error.

Instead you must call it in the constructor.

<?php
    class MyClass {
        public $classKeyword;
        public $arrayList;
        public function __construct() {
            $this->classKeyword = "contratti";
            $this->arrayList = getDbTableStructure($this->classKeyword);
        }
    }
?>

answered Oct 24, 2014 at 13:22

Logan Murphy

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Comments

Alexander Guz · Accepted Answer · 2014-10-24 13:27:22Z

0

For creating dynamic objects you might consider using something like this:

<?php

function getList() {
    return [
        'fieldName' => 'fieldValue',
        // other columns
    ];
}

$object = new stdClass();
foreach (getList() as $fieldName => $fieldValue) {
    $object->{$fieldName} = $fieldValue;
}

echo $object->fieldName . PHP_EOL;

answered Oct 24, 2014 at 13:27

Alexander Guz

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