# Could someone please tell me how I could simplify this code?

In practice I have this code and it seems very tiring to do all those if there is a way to make it easier?
But that it works correctly? Because this code not only has * (multiplications) it also has 3 / (divisions).
If it only had multiplications it was easy to do it but it also has those divisons so it becomes more difficult to do so.
If anyone could help me thanks.

``````        function calculateLength() {

var lengthinput = parseFloat(document.getElementById('lengthinput').value);

var oper = document.getElementById('lengthselector1').value;
var oper2 = document.getElementById('lengthselector2').value;

if(oper === 'k' && oper2 === 'm')
{
document.getElementById('lengthresult').value = lengthinput*1000 || 0;
}

if(oper === 'k' && oper2 === 'd')
{
document.getElementById('lengthresult').value = lengthinput*10000 || 0;
}

if(oper === 'k' && oper2 === 'c')
{
document.getElementById('lengthresult').value = lengthinput*100000 || 0;
}

if(oper === 'k' && oper2 === 'mi')
{
document.getElementById('lengthresult').value = lengthinput*1000000 || 0;
}

if(oper === 'k' && oper2 === 'mic')
{
document.getElementById('lengthresult').value = lengthinput*1.0000E+9 || 0;
}

if(oper === 'k' && oper2 === 'de')
{
document.getElementById('lengthresult').value = lengthinput*100 || 0;
}

if(oper === 'k' && oper2 === 'h')
{
document.getElementById('lengthresult').value = lengthinput*10 || 0;
}

if(oper === 'k' && oper2 === 'me')
{
document.getElementById('lengthresult').value = lengthinput/1000 || 0;
}

if(oper === 'k' && oper2 === 'g')
{
document.getElementById('lengthresult').value = lengthinput/1000000 || 0;
}

if(oper === 'k' && oper2 === 'z')
{
document.getElementById('lengthresult').value = lengthinput/1.0000E+18 || 0;
}

if(oper === 'k' && oper2 === 'i')
{
document.getElementById('lengthresult').value = lengthinput*39370.0787 || 0;
}

if(oper === 'k' && oper2 === 'a')
{
document.getElementById('lengthresult').value = lengthinput*1.0000E+13 || 0;
}

}
``````

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