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Homephp教程php手册PHP异常Parse error: syntax error, unexpected T_VAR错误

在PHP中根本不需要使用var声明的,但是当一个变量作为一个类的成员变量的时候,使用var还是没有问题的

其实,这是一个非常容易解决掉的问题。在我看来,似曾相识,呵呵,最近学JavaScript可是学会了使用var声明变量。

其实,在PHP中根本不需要使用var声明的,但是当一个变量作为一个类的成员变量的时候,使用var还是没有问题的。

在外部使用var就报错Parse error: syntax error, unexpected T_VAR in...,例如我的出错信息:

Parse error: syntax error, unexpected T_VAR in D:\Apache2.2\htdocs\shirdrn\page\p2\pageUtil.inc on line 34

我在测试:在一个类的内部,使用一个自己定义的类对象作为这个类的成员时,出错了。

Address类对应的address.inc代码:

代码如下:

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

Person类及其测试代码为person.php如下:

代码如下:

<?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();
?>

测试输出现异常:

Parse error: syntax error, unexpected T_VAR in D:\Apache2.2\htdocs\shirdrn\page\p2\pageUtil.inc on line 34

就是因为在person.php代码中使用var声明变量,PHP中是不能这样的,只要使用“$”符号起始就表示这个字符后面的是一个PHP变量。

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