我需要在类的构造函数中增加一个数字，而不调用静态函数。我的代码如下:
class Addup {
static public $num;
function __construct($num) {
echo 'My number is==>' . $num;
static::$num++;
}
}
$inc = new Addup();//instantiated an object
echo '<br>';
echo $inc::$num;//should be 2, but still 1, should it be $inc = new Addup() or $inc->addup() implemented in my class?
echo '<br>';
echo $inc::$num;//should be 3, but still 1
echo '<br>';
echo $inc::$num;//should be 4, but still 1
欢迎大家多多指教，谢谢
UPD做了如下重构:
$inc = new Increment();//My number is==>2
echo '<br>';
$inc = new Increment();//My number is==>3
echo '<br>';
$inc = new Increment();//My number is==>4
echo '<br>';
$inc = new Increment();//My number is==>5
这是唯一的方法，做到没有调用函数在一个类吗?
静态属性意味着静态属性的值在类的多个实例化中是相同的。
这可能比您的示例
更好地演示代码发生了什么。旁注:你传递的参数对任何东西都没有影响，也没有任何作用
<?php
class Addup {
static public $num;
function __construct() {
static::$num++;
}
}
$inc1 = new Addup(); // calls the constructor therefore +1
echo PHP_EOL;
echo $inc1::$num; // 1
$inc2 = new Addup(); // calls the constructor therefore +1
echo PHP_EOL;
echo $inc2::$num; // 2
$inc3 = new Addup(); // calls the constructor therefore +1
echo PHP_EOL;
echo $inc3::$num; // 3
// now if you look at $inc1::$num; after instantiating inc2 and inc3
echo PHP_EOL;
echo $inc1::$num; // NOT 1 but whatever it was last incremented to i.e. 3