Help! problems with functions and forms

Toka1 · December 12th, 2003, 06:14 AM

I am very new to javascript, I am trying to write a piece of code that will take input from a textbox and do certain calculations depending on the input.
my confusion is with the function, I don't quite get the parameters bit and I think I have missed something relating to this on the form in the function()
Please could someone point me in the right direction, here is my code so far.

<html>
<head>
<title>Calculator</title>

<script language="JavaScript">

function butCheckForm_onclick()

{

var myForm = document.form1;

var Result;

var SUS=myForm.txtSaybolt.value

SuS=parseInt(SUS);

var Result= return Result;

if (SUS >= 45 && SUS<= "100")
{
Result=SUS*0.2253;
}
else if (SUS>=100 && SUS<=240)
{
Result=SUS*.02193;
}

return Result;

}
alert(Result);

</script>

</head>

<body>

<form name=form1>

<p><pre>
Please Enter SUS
<input type="text" name="txtSaybolt" value=""></input>
<input type="button" value="Check Details"
Name=butCheckForm onclick="butCheckForm_onclick()"></input>
</pre></p>

</form>
</body>
</html>

Thank you in advance

Toka

pgtips · December 12th, 2003, 06:57 AM

Just a bit of syntax problem:

var Result= return Result;
take out this line, its nonsense. You've already declared the result variable, and anyway you can't use the return keyword on the RHS of an assignment.

alert(Result);
take out this line too. the variable Result only exists inside the body of the function, so it doesn't exist when you're trying to alert it. If you want to see the return value use <input type="button" value="Check Details"
Name=butCheckForm onclick="alert(butCheckForm_onclick());">

SuS=parseInt(SUS);
js is case-sensitive so you need to change this to
SUS=parseInt(SUS);

hth
Phil

Toka1 · December 12th, 2003, 07:14 AM

Thanks Phil

That worked a treat, and thank you for explaining what I had done wrong and why it wouldn't work.

Regards
Toka

joefawcett · December 12th, 2003, 10:29 AM

Two more things.

Code:

if (SUS >= 45 && SUS<= "100")

should be

Code:

if (SUS >= 45 && SUS<= 100)

you are comparing a number with a string.

Code:

parseInt(SUS)

should be

Code:

parseInt(SUS, 10)

You must specify the base of the target system or you will fall foul of someone typing in 070 for 70.
Try typing the following into the browser address bar and pressing enter

Code:

javascript:alert(parseInt(070, 10))

compared to

Code:

javascript:alert(parseInt(070))

--

Joe

pgtips · December 12th, 2003, 11:09 AM

Nice one Joe ;)

I suppose one could also argue that
if (SUS >= 45 && SUS<= "100")
{
Result=SUS*0.2253;
}
else if (SUS>=100 && SUS<=240)
{
Result=SUS*.02193;
}
is ambiguous because if SUS = 100 then it satisfies both tests, but that would be too pedantic :D

rgds
Phil

pgtips · December 12th, 2003, 11:12 AM

LOL, Joe you forgot the quotes too. Both your examples will alert 56!
javascript:alert(parseInt("070", 10)) - alerts 70
javascript:alert(parseInt("070")) - alerts 56

Toka1 · December 16th, 2003, 11:15 AM

Thanks All

Toka