- 1、php构造方法对成员变量赋值 function__construct
- 2、php变量赋值的方法
- 3、php中如何给成员变量，赋值？
function __construct
function 是系统关键词，表示定义一个方法，后面加空格然后根方法名（ __construct
是方法名）。你的未加空格
__construct 是系统内置的，叫魔术方法，每次实例化类是会自动执行此方法。
还有就是调用类的成员变量不需要在变量前面加 $ 比如：
$this-school_name = $name; 正确
$this-$school_name = $name; 错误
?php
class School {
public $school_name;
public $school_student;
public $school_room;
public $school_teacher;
function __construct($name, $student, $room, $teacher) {
$this-school_name = $name;
$this-school_student = $student;
$this-school_room = $room;
$this-school_teacher = $teacher;
}
function show() {
echo “!@#$%^*”;
}
}
class People extends School {
public $teachername;
function __construct($tname, $studentconsts) {
$this-teachername = $tname;
$this-school_student = $studentconsts;
}
function show() {
return “今天上课” . $this-teachername . “讲课,学生” . $this-school_student . “人”;
}
}
class Tongji extends School {
function show() {
return “学校名：” . $this-school_name . “学生数:” . $this-school_student . “教室数:” . $this-school_room . “教师数:” . $this-school_teacher;
}
}
$a = new People ( “周周周”, 20 );
$b = new School ( “DZ”, 20, 50, 2 );
echo $a-show () . “br”;
echo $b-show ();
?
不是变量问题，是if的语法问题，应该是
if($view[$typeid]==0) {
….
}
当然，最好增加一个判断，以免发生$view[$typeid]未定义的错误，如
if(! isset($view[$typeid])) {
die(‘$view中并没有定义下标’ . $typeid . ‘哦’);//当然你可以修改为其他处理
}
if($view[$typeid] ==0) {
…
}
如果，你的意思是你有$view1,$view2,$view3，然后想根据$typeid动态调用变量的话，这样写
$str = ‘view’ . $typeid; //得到类似view1,view2的字符
if ($$str == 0) { //连续两个$$表示变量的变量，即已$str的值为变量名的变量的值
…
}
class ren_min
{
private $aaa;
function _loveyou($inp)
{
$this-aaa = $inp + 1;
return $this-aaa;
}
}
$ceshi = new ren_min;
echo $ceshi-_loveyou(800);
原创文章，作者：CBZ9S，如若转载，请注明出处：https://www.506064.com/n/127240.html
微信扫一扫 
支付宝扫一扫 