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PHP exception Parse error: syntax error, unexpected error solution
PHP exception Parse error: syntax error, unexpected 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";
$p->display();
?>The test output is abnormal:
Parse error: syntax error, unexpected T_VAR in D:Apache2.2htdocsshirdrnpagep2pageUtil.inc on line 34
It is because var is used to declare variables in the person.php code. This cannot be done in PHP. Just use the "$" symbol to start. What follows this character is a PHP variable.
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