如何使方法和功能工作 - Javascript

时间:2016-09-28 15:56:02

标签: javascript arrays function object methods

以下方法和功能不起作用,有人可以帮助我吗?

hasMoreOscarsThan - this method accepts one actor object as a parameter and
    returns true if the actor has more Oscars than the one that is passed as
    a parameter and false otherwise.

现在编写以下函数:

getActorByName - this function expects a string as a parameter and returns
    the object in the actors array whose name property is equal to the
    string that is passed in (if there is one).

我的代码:

function Person(firstName, lastName, age, numOscars) {

    this.firstName = firstName;
    this.lastName = lastName;
    this.age = age;
    this.numOscars = numOscars;
    this.hello = function() { console.log("Hello, my name is " + this.firstName); }
    this.hasMoreOscarsThan = function(x, y) {
        if (this.numOscars > this.numOscars) {
            return this.firstName;
        } else {
            return "False!";
        }
    }

};


var actors = new Array();
actors[0] = new Person("Leonardo", "DiCaprio", 41, 1);
actors[1] = new Person("Jennifer", "Lawrence", 25, 1);
actors[2] = new Person("Samuel L.", " Jackson", 67, 0);
actors[3] = new Person("Meryl", "Streep", 66, 3);
actors[4] = new Person("John", "Cho", 43, 0);

actors.forEach(function(item) {
    item.hello();
})

actors.forEach(function(item) {
    item.hasMoreOscarsThan();
})


function getActorByName(person) {
    console.log(actors.firstName + " " + actors.lastName);
}


function list() {
    var actorsLength = actors.length;
    for (var i = 0; i < actorsLength; i++) {
        getActorByName(actors[i]);
    }
}


var search = function(lastName) {
    var actorsLength = actors.length;
    for (var i = 0; i < actorsLength; i++) {
        if (lastName == actors[i].lastName) {
            getActorByName(actors[i]);
        }
    }
}


search("DiCaprio");

var getAverageAge;

getAverageAge = (actors[0].age + actors[1].age + actors[2].age + actors[3].age + actors[4].age) / actors.length;
console.log(getAverageAge);

提前非常感谢!!!

2 个答案:

答案 0 :(得分:0)

参数是person对象,但你指的是一些actors,它没有在函数内的任何地方定义,因此改为person.firstName&amp; person.lastName

function getActorByName(person) {
    console.log(person.firstName + " " + person.lastName);
}

JSFIDDLE

答案 1 :(得分:0)

我编辑了你的代码:

function Person(firstName, lastName, age, numOscars) {

    this.firstName = firstName;
    this.lastName = lastName;
    this.age = age;
    this.numOscars = numOscars;
    this.hello = function() { console.log("Hello, my name is " + this.firstName); }
    this.hasMoreOscarsThan = function(x) { // x is a compare parameter
        if (this.numOscars > x) { // comparing numOscars with argument x
            return this.firstName;
        } else {
            return "False!";
        }
    }

};


var actors = new Array();
actors[0] = new Person("Leonardo", "DiCaprio", 41, 1);
actors[1] = new Person("Jennifer", "Lawrence", 25, 1);
actors[2] = new Person("Samuel L.", " Jackson", 67, 0);
actors[3] = new Person("Meryl", "Streep", 66, 3);
actors[4] = new Person("John", "Cho", 43, 0);

actors.forEach(function(item) {
    item.hello();
})

actors.forEach(function(item) {
    console.log(item.hasMoreOscarsThan(2)); // I put compare argument 2 and print result to console
})


function getActorByName(person) {
    console.log(person.firstName + " " + person.lastName); // changed actors to person
}


function list() {
    var actorsLength = actors.length;
    for (var i = 0; i < actorsLength; i++) {
        getActorByName(actors[i]);
    }
}


var search = function(lastName) {
    var actorsLength = actors.length;
    for (var i = 0; i < actorsLength; i++) {
        if (lastName == actors[i].lastName) {
            getActorByName(actors[i]);
        }
    }
}


search("DiCaprio");

var getAverageAge;

getAverageAge = (actors[0].age + actors[1].age + actors[2].age + actors[3].age + actors[4].age) / actors.length;
console.log(getAverageAge);

结果:

  

您好,我的名字是Leonardo

     

您好,我的名字是詹妮弗

     

您好,我的名字是Samuel L.

     

您好,我的名字是Meryl

     

您好,我的名字是John

     

假的!   假的!
  假的!
  梅丽尔
  假的!
  莱昂纳多·迪卡普里奥   48.4