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PHP在调用带有参数的函数时初始化类_Php_Class_Object_Arguments - Fatal编程技术网

PHP在调用带有参数的函数时初始化类

PHP在调用带有参数的函数时初始化类,php,class,object,arguments,Php,Class,Object,Arguments,在php中,我想在调用带有参数的函数时调用类的初始化属性,但似乎无法理解。下面是我的源代码 任务2015年11月25日 您正在将代码中的初始值替换为空字符串: echo $object1->method1("","",""); 尝试: <?php class Task { public $name = "Naveed"; public $fname = "Zahid"; public $dob = "10-04-1991"; public $co

在php中,我想在调用带有参数的函数时调用类的初始化属性,但似乎无法理解。下面是我的源代码


任务2015年11月25日

您正在将代码中的初始值替换为空字符串:

echo $object1->method1("","","");
尝试:

<?php
class Task
{
    public $name = "Naveed";
    public $fname = "Zahid";
    public $dob = "10-04-1991";
    public $country = "Pakistan";
    public $ddress = "shergarh mardan kpk";
    public $cgpa = 3.83;
    public $degree = "MCS";
    public $year = "2014";

    public function method1($arg1, $arg2, $arg3) {
        $this->name = $arg1;
        $this->fname = $arg2;
        $this->dob = $arg3;

        return "My name is " . $this->name . "  " . ", and my father name is " . $this->fname;

        //i want to print the above values ... without giving in the function args


    }

    public function method2($arg4, $arg5) {
    }

    public function method3() {
    }

    public function method4() {
    }

    public function method5() {
    }
}

$object1 = new Task();

echo $object1->method1("Naveed", "Zahid", "10-04-1991");
?>

或者,如果确实需要使用初始化值,请不要放置参数,也不要覆盖初始值:

<?php
class Task
{
    public $name = "Naveed";
    public $fname = "Zahid";
    public $dob = "10-04-1991";
    public $country = "Pakistan";
    public $ddress = "shergarh mardan kpk";
    public $cgpa = 3.83;
    public $degree = "MCS";
    public $year = "2014";

    public function method1() {
        return "My name is " . $this->name . "  " . ", and my father name is " . $this->fname;
        //i want to print the above values ... without giving in the function args
    }

    public function method2($arg4, $arg5) {
    }

    public function method3() {
    }

    public function method4() {
    }

    public function method5() {
    }
}

$object1 = new Task();

echo $object1->method1();
?>


另外,尝试了解更多关于PHP的基本/高级功能。您可以开始使用此链接:tutorialspoint在学习新语言时帮助我很多

当您做一些非常简单的事情时,不使用类会更容易。这与java不同,java必须使用类。在php中,您不必这样做。感谢您的回复,,,但这是我们软件公司的任务,,,我必须这样做…非常感谢。。这是我们的任务,使一个类具有8个属性和5个函数,这些函数在方法中使用“&”。。。我需要这个solution@user3719190这能解决你的问题吗?如果是,则将其标记为已接受。如果没有,您能否加强上述问题,以便我们更好地帮助您