在函数Javascript中调用函数

Question

Hi everyone, I'm learning JS and trying to practice using a function inside a function. So, I've come up with my own problem to code. I understand that there's an easier way to do this and the problem could've been written better, but for the purpose of practicing, this is what I've come up with:

John is 13 years old. Write a function to find out what year John was born, and what school type he is attending BASED ON his birth year -- high school or middle school.

Middle school = 2003 - 2004
High school = 1999 - 2002

function calculateBirthYearJohn(ageJohn) {
    var currentYear = 2017;
    var birthYearJohn = currentYear - ageJohn;
    return birthYearJohn;
}

function defineSchoolTypeJohn(birthYear) {
    var birthYear = calculateBirthYearJohn(ageJohn);
    if (birthYear >= 2003 & <= 2004) {
        console.log('John goes to Middle School.');
    } else if (birthYear >= 1999 & <= 2002) {
        console.log('John goes to High School.');
    }
}

Console will say "John goes to Middle School" if he's born between 2003 and 2004; "John goes to High School" if he's born between 1999 and 2002.

Blank result -- no error.

I've made a newbie mistake somewhere; can someone please point me in the right direction? Much appreciated!

Answer 1

Couple problems.

line 8: ageJohn == undefined

lines 9 and 11: In JavaScript, && is the Logical AND operator (two ampersands). One singular ampersand is the Bitwise AND operator. Also, comparison only works between two variables at a time, hence, your attempt to chain is invalid.

See below for a working example.

 function calculateBirthYear(ageJohn) { const currentYear = 2017 const birthYear = currentYear - ageJohn return birthYear } function defineSchoolType(ageJohn) { const birthYear = calculateBirthYear(ageJohn) if (birthYear >= 2003 && birthYear <= 2004) console.log('John goes to Middle School.') else if (birthYear >= 1999 && birthYear <= 2002) console.log('John goes to High School.') else console.log('John doesn\\'t go to Middle School or High School') } defineSchoolType(13) // Middle School. defineSchoolType(15) // High School. defineSchoolType(20) // Neither. 

Answer 2

You have:

var birthyear = calculateBirthYear(ageJohn);

but you have not defined ageJohn anywhere. Up in calculateBirthYear , you gave it as the argument, but that is not the same as a variable that can be accessed anywhere. I think you meant to write:

function defineSchoolTypeJohn(ageJohn) {
    var birthYear = calculateBirthYear(ageJohn);
    ...

Now, you have defined ageJohn as the argument to the function.