Question

我创建了Person类，然后将其扩展到Student类。当我在Student对象上调用方法时，它不能按预期工作。

这是我的代码

class Person{
    static public $IsAlive;
    public $FirstName;
    public $LastName;
    public $Gender;
    public $Age;
    static public $total_person = 1;
    static public $type;

    function __construct($FirstName,$LastName,$Gender,$Age,$IsAlive,$type=""){
        self::$IsAlive = $IsAlive;
        if(self::$IsAlive === TRUE){
            echo "<strong>Person Number#" . self::$total_person++ . "</strong> details are given below: <br />";
            $this->FirstName = $FirstName;
            $this->LastName = $LastName;
            $this->Gender = $Gender;
            $this->Age = $Age;
            self::$type = $type;
        }else{
            echo "This Person is not longer available <br />";
        }
    }
    function PersonDetail(){
        if(self::$IsAlive === TRUE){
            echo "Person Name: " . $this->FirstName . " " . $this->LastName . "<br />";
            echo "Person Gender: " . $this->Gender . "<br />";
            echo "Age: " . $this->Age;
        }
    }

}
class Student extends Person{
    public $standard = "ABC";
    public $subjects = "ABC";
    public $fee = "123";
    function StudentDetail(){
        if(parent::$type === "Student"){
            return parent::PersonDetail();
            echo "Education Standard: " . $this->standard;
            echo "Subjects: " . $this->subjects;
            echo "Fee: " . $this->fee;
            echo parent::$type;
        }
    }
}

$Hamza = new Student("Muhammad Hamza","Nisar","Male","22 years old",TRUE,"Student");
$Hamza->StudentDetail();

仅打印PersonDetail()，但不打印教育标准，科目和费用等完整信息。

这是我的输出

Person Number#1 details are given below: 
Person Name: Muhammad Hamza Nisar
Person Gender: Male
Age: 22 years old

我还需要在其下方看到StudentDetail()输出。怎么了？怎么解决？

Answer 1

如果在函数（或方法）中放置return语句，PHP将不会在此点之后继续执行（大部分时间）*。因此，当执行以下代码（取自您的问题）时，它会在您致电return parent::PersonDetail();后停止。

function StudentDetail() {
    if(parent::$type === "Student") {
        return parent::PersonDetail();
        echo "Education Standard: " . $this->standard;
        echo "Subjects: " . $this->subjects;
        echo "Fee: " . $this->fee;
        echo parent::$type;
    }
}

由于PersonDetail()方法本身会显示信息，因此您无需返回其值。在这种情况下只需调用它就足够了：

function StudentDetail() {
    if(parent::$type === "Student") {
        parent::PersonDetail(); // this will echo parent output

        // and this code will still be executed
        echo "Education Standard: " . $this->standard;
        echo "Subjects: " . $this->subjects;
        echo "Fee: " . $this->fee;
        echo parent::$type;
    }
}

在这个问题的情况下，这是无关紧要的，但有时即使在函数返回后执行仍在进行，例如finally子句

继承类在PHP

1 个答案: