Sdaram — Mon, 30 Aug 1999 21:20:00 +0000

Well, thank you very much for the help Mr. John.

However, parseInt() does return the integer part of a string. Unfortunately it can't detect an invalid string consisting of digits and other chars. For example, parseInt() returns "1" for the string "1as#" which is invalid input for a specified number field. I have modified the code. Below is the code:-

function chkyear(theyr) {
y=new Date();
limit=1900 + y.getYear()

if(theyr.length==0)
{alert("Input valid year")}
for (var i in theyr)
{if(isNaN(theyr[i]))
{found=1
if (found==1)
{alert("Input valid year")}
break}

else {
if(theyr.length < 4)
{theyr=19 + theyr
if ((theyr > limit)¦¦(theyr < 1960))
{alert("You are not qualified!")}}
else {
if ((theyr >= limit)¦¦(theyr < 1960))
{alert("You are not qualified!")}}
}}
}

It seem to work. Thanks again for the help.
Sundaram.

John Pollock — Mon, 30 Aug 1999 01:06:00 +0000

Hmm...try this instead:

function chkyear(inyear) {

theyr=parseInt(inyear);

y=new Date();
limit=1900 + y.getYear()

etc..

}

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Sdaram — Sat, 28 Aug 1999 18:35:00 +0000

I get an error!
missing ) formal parameter

function chkyear(parseInt(theyr))
------------------------^

John Pollock — Thu, 26 Aug 1999 20:03:00 +0000

Try using the parseInt() function to get just the number from the variable once you have it, try placing it at the beginning of the function you have:

function chkyear(parseInt(theyr)) {

y=new Date();
limit=1900 + y.getYear()

etc..

}

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