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。
以上就是PHP异常Parse error: syntax error错误解决方法的内容,更多相关内容请关注PHP中文网(www.php.cn)!

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