Clarissa M.

asked • 02/21/21

NEED HELP JAVA CODE ASAP

How do I find the average of the marks (may use parseInt() and print. There are 3 marks.


if average>= 90 then print "Awesome";

if average>= 80 then print "Very Good";

if average>= 50 then print "Ok";

if average less than 50 then print "Fail";


I use document.getElementById ("result)".innerHTML=;

Clarissa M.

Both HTML and JAVA
Report

02/21/21

1 Expert Answer

By:

Patrick B. answered • 02/21/21

Tutor
4.7 (31)

Math and computer tutor/teacher

See tutors like this
See tutors like this

<!DOCTYPE HTML>


<HTML>



<label> 1st Mark </label> <input type="text" id="txtNum1"/><br>


<label> 2nd Mark </label> <input type="text" id="txtNum2"/><br>


<label> 3rd Mark </label> <input type="text" id="txtNum3"/><br>


<button type="button" onclick="CalcAvg()"> Calculate Average </button>




<script>


function CalcAvg()

{

alert("CalcAvg");


var iNumbuff1 = document.getElementById("txtNum1").value;

var iNumbuff2 = document.getElementById("txtNum2").value;

var iNumbuff3 = document.getElementById("txtNum3").value;



var regex=/^[0-9]+$/;

if (!iNumbuff1.match(regex)) { alert("Invalid input for num1"); return; }

if (!iNumbuff2.match(regex)) { alert("Invalid input for num2"); return; }

if (!iNumbuff3.match(regex)) { alert("Invalid input for num3"); return; }




var iNum1 = parseInt(iNumbuff1)

var iNum2 = parseInt(iNumbuff2)

var iNum3 = parseInt(iNumbuff3)


if (isNaN(iNum1))

{

alert('invalid data for num1 ');

}

else

{

alert('1st # is valid');

if (isNaN(iNum2))

{

alert('invalid data for num2');

}

else

{

alert(' 2nd # is valid');

if (isNaN(iNum3))

{

alert('invalid data for num3');

}

else

{

var avg = (iNum1+iNum2+iNum3)*1.000/3;

alert(" average is " + avg );



if (avg>=90) { alert("AWESOME!"); }

else if (avg>=80) { alert(" Very Good"); }

else if (avg>=50) { alert(" ok "); }

else { alert("FAIL"); }

}


}

}


}

</script>


</HTML>

Upvote • 0 Downvote
More
Reported

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.