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View Full Version : I need some help please trying to get this converted rrom metric to imperial Please



Alimcvev
07-19-2006, 10:27 AM
Urgent please. Bimbo on code. Can any one help me please my brain is shutting down now after working on this all night. Im a lil bit of a newbie here so i was hoping if you guys could help me .Is there any way to do this script in both metric and imperial as it wouuld prove usefull to my draugthsmen at my work environment any help in this matter would be fantastic

My e-mail is alimcvev@hotmail.com

HTML>
<HEAD>
<TITLE>example </TITLE>
<SCRIPT type="text/javascript">
var shapeArray = new Array (5);
var RectangleLength;
var RectangleWidth;
var sum1;
var sum2;
var sum3;
var circle;
function test(){
for (var i= 0; i< shapeArray.length; i++){
Mesure=window.prompt('Please enter the format of the size ("M" for metric or "I" for imperial)');

shape=window.prompt('Please enter shape "r" for rectangle or "c" for Circle');
if(shape=='r'){
RectangleLength = window.prompt('Please enter the length of the rectangle in centimetres','');
while(mesure== I * 2.54)
RectangleLength=parseFloat(RectangleLength);
RectangleWidth = window.prompt('Please enter the width of the rectangle in centimetres','');
while(mesure== I * 2.54)
RectangleWidth=parseFloat(RectangleWidth);
shapeArray[i]='The area of the '+shape+' is ' +RectangleLength * RectangleWidth+ ' square centimetres <br>'
}
while(mesure== I * 2.54)
if (shape=='c'){
circle = window.prompt('Please enter the radius','');
circle=parseFloat(circle);
shapeArray[i]='The radius of the circle is '+circle * circle * 3.142+'<br>'
}
document.getElementById("display").innerHTML+=shapeArray[i]
}
}
document.write('<BR>' + 'Thank you for using this program')
</SCRIPT>
</HEAD>
<BODY onload="test()">
<div id="display"></div>
</BODY>
</HTML>

vwphillips
07-19-2006, 05:39 PM
not understaning this as if you leave out the units promps

the calculation will be correct for whatever units are used

ie
2 inch *3 inch = 6 inch

2 cm * 3 cm = 6 cm

Alimcvev
07-19-2006, 05:43 PM
I realy need to get a metric result from the imperial value. Therefore multiplying the end result of an imperial mesurement. Basicaly in the case of an imperial number added to the program multiplying this by 2.54



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