My simple Javascript exercises colection starts from page: Some simple Javascript exercises. In this article I propose more.
Exercise 4
Solve a second order equation
<!DOCTYPE html>
<html><head>
<script>
function equation2degree() {
var a = document.getElementById("number1").value;
a=parseFloat(a);
var b = document.getElementById("number2").value;
b=parseFloat(b);
var c = document.getElementById("number3").value;
c=parseFloat(c);
var delta = b*b-(4*a*c);
if (delta < 0)
{
document.getElementById("solution").innerHTML = "Impossible";
}
else
{
var x1=(-b+Math.sqrt(delta))/(2*a);
var x2=(-b-Math.sqrt(delta))/(2*a);
document.getElementById("solution").innerHTML = "x<sub>1</sub>=" + x1 + " x<sub>2</sub>=" + x2;
}
}
</script>
</head><body>
<h3>Solve a second order equation in the form ax<sup>2</sup>+bx+c=0</h3>
<p>Insert the coefficients (integer or real numbers, with negative sign if present)<BR><BR>
a: <input type="text" id="number1"><BR><BR>
b: <input type="text" id="number2"><BR><BR>
c: <input type="text" id="number3"><BR><BR>
<button onclick="equation2degree()">Solve!</button></p>
<p>Solution: <span id="solution"></span></p>
</body></html>
Exercise 5
Determine if a year is a leap year.
All years whose number represents them is a multiple of 4, are leap years, except for secular years (ie multiples of 100) if their number is not a multiple of 400.
<!DOCTYPE html>
<html><head>
<script>
function leapYear() {
var year = parseInt(document.getElementById("year").value);
var leap = false;
if (year%100==0) // is it a secular year?
{
if (year%400==0)
{
leap = true;
}
}
else // all other years!
{
if (year%4==0)
{
leap = true;
}
}
document.getElementById("print").innerHTML = leap;
}
</script>
</head><body>
<h3>Tell me a year and I determine if it is a leap year!</h3>
Gimme the year: <input type="number" id="year">
<button onclick="leapYear()">Submit</button>
<BR><p>Answer: <span id="print"></span></p><BR>
</body></html>