PHP exception Parse error: syntax error... error solution

There is no need to use var declaration in PHP, but when a variable is used as a member variable of a class, there is no problem in using var

In fact, this is a very easy problem to solve . In my opinion, it seems familiar, haha, I recently learned JavaScript and learned to use var to declare variables.

In fact, there is no need to use var declaration in PHP, but when a variable is used as a member variable of a class , there is still no problem in using var.

When using var externally, an error occurs. Parse error: syntax error, unexpected T_VAR in..., for example, my error message:

Parse error: syntax error, unexpected T_VAR in D:Apache2. 2htdocsshirdrnpagep2pageUtil.inc on line 34

I was testing: When using a self-defined class object as a member of this class inside a class, something went wrong.

The address.inc code corresponding to the Address class:

The code is as follows:

<?php
class Address {
   var $road;
   function Address(){}
   function setRoad($road){
    $this->road = $road;
   }
}
?>

Person class and its test code are person .php is as follows:

The code is as follows:

<?php
require("address.inc");
class Person {
   var $name;
   var $address;
   function Person(){
   }
   function display(){
    echo "Name : ".$this->name."<BR>";
    echo "Road : ".$this->address->road."<BR>";
   }
}
var $p = new Person();
$p->address = new Address();
$p->address->setRoad("Chagnchun Road");
$p->name = "Shirdrn";
$p->display();
?>

Exception occurs in test output:

Parse error: syntax error , unexpected T_VAR in D:Apache2.2htdocsshirdrnpagep2pageUtil.inc on line 34

It’s because var is used to declare variables in the person.php code. This cannot be done in PHP. Just use the "$" symbol. It means that what follows this character is a PHP variable.