Error while handling constructors in php

Question

I am getting the following error while running the following PHP code

class tableData
{
    private $row1 = array("Kalle1", "address1", "postal code1", "[email protected]", "070111001", "08111001");
    private $row2 = array("kalle2", "address2", "postcode2", "[email protected]", "070111002", "08111002");
    private $row3 = array("kalle3", "address3", "postcode3", "[email protected]", "070111003", "08111003");
    private $row4 = array("kalle4", "address4", "postcode4", "[email protected]", "070111004", "08111004");
    private $rader = array(array($row1),array($row2),array($row3),array($row4));
}

The error is corresponding to the variable $rader:

Parse error: syntax error, unexpected '$row1' (T_VARIABLE), expecting ')' in C:\xampp\htdocs\test_k.php on line 15

Can anyone help me correct this error?

Line 15 is the variable $rader in the above code.

Chalasani Arjith
– Chalasani Arjith

2018-04-09 01:22:37 +00:00
Commented Apr 9, 2018 at 1:22 — Chalasani Arjith
– Chalasani Arjith, Commented Apr 9, 2018 at 1:22

James · Accepted Answer · 2018-04-09 01:41:33Z

2

The PHP manual states:

This declaration [of class properties] may include an initialization, but this initialization must be a constant value--that is, it must be able to be evaluated at compile time and must not depend on run-time information in order to be evaluated.

private $rader = array(array($row1),array($row2),array($row3),array($row4));

So your $row1 is not a constant value, it requires run time information, ie the value that the variable holds.

When PHP is compiling your class, at the point of property initialisation, it has no idea what value your $row1 variable holds.

You'll have to set that variable to empty or null (whatever suits), and either have a setter method which sets it and call that first, OR use a constructor if this is always going to be the same data.

Basically, you cannot set properties from values that would come later from the object at run time.

edited Apr 9, 2018 at 1:41

answered Apr 9, 2018 at 1:28

James

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Comments

csb · Accepted Answer · 2018-04-09 02:17:31Z

1

Use a constructor to declare class variables that are populated with other variables.

class tableData
{
    private $row1 = array("Kalle1", "address1", "postal code1", "[email protected]", "070111001", "08111001");
    private $row2 = array("kalle2", "address2", "postcode2", "[email protected]", "070111002", "08111002");
    private $row3 = array("kalle3", "address3", "postcode3", "[email protected]", "070111003", "08111003");
    private $row4 = array("kalle4", "address4", "postcode4", "[email protected]", "070111004", "08111004");
    private $rader;

    public function __construct() {
        $this->rader = array(
            array($this->row1),
            array($this->row2),
            array($this->row3),
            array($this->row4)
        );
    }

}

This assumes you do intend to create a 3-level nested array.

edited Apr 9, 2018 at 2:17

answered Apr 9, 2018 at 1:29

csb

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1 Comment

Karlo Kokkak Over a year ago

You're missing $this-> in $row1 ...`

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