Eval method returns wrong output

Asked May 24 '18 at 11:10

Active May 24 '18 at 13:04

Viewed 32 times

eval method of JavaScript return wrong output

eval(19 > 0022)

the above code returns true. Is there any way to fix this.

Thanks for all your comments its fixed now. with these comments

const b = '0022'.replace(/^[0]+/g,"");

and

eval(Number('19') > Number('0022'))

edited May 24 '18 at 13:04

asked May 24 '18 at 11:10

Sandeep Bhardwaj

1

leading zero numbers could be read as octal numbers. – Nina Scholz May 24 '18 at 11:11
Any way to fix it , i have a dynamic string that need to be evaluated. – Sandeep Bhardwaj May 24 '18 at 11:12
And the function expect a sting BTW – Alon Eitan May 24 '18 at 11:13
even if condition also allowing true.. it's not a eval problem – Ramesh Rajendran May 24 '18 at 11:14
Thanks for lots of comments , but how to deal with this condition. Should i use parseInt – Sandeep Bhardwaj May 24 '18 at 11:16
1

You can remove the leading zero's using regex. E.g. const b = '0022'.replace(/^[0]+/g,""); – Gary Thomas May 24 '18 at 11:16
1

take `parseInt(value, 10)` with a base of `10`. this ignores the octal default. – Nina Scholz May 24 '18 at 11:18
1

eval(Number('19') > Number('0022')) === false – RobC May 24 '18 at 11:18
Thanks a lot @RobC – Sandeep Bhardwaj May 24 '18 at 11:19
Why are you using `eval` at all? It's needed only when you have a *string* of code - `19 > 022` is an expression that evaluates to a boolean – Bergi May 24 '18 at 12:54
Yes @Bergi, i have a string as an expression – Sandeep Bhardwaj May 24 '18 at 13:00
@SandeepBhardwaj Then you should have used that string in your question – Bergi May 24 '18 at 13:31

0 Answers0