Convert String to Int with Integer.parseInt don't works

Question

I'm working with JavaEE i need to convert this: request.getParameter("id") to int. The value of request.getParameter("id") is "9" (String). When I'm trying to convert to int I have

java.lang.NumberFormatException

I've tried java.lang.Integer.parseInt(request.getParameter("id")) and request.getParameter("id",10) but it donsen't works...

Any solutions? Thank you.

are you sure id parameter is a number?try trimming the string — singhakash
– singhakash, Commented Apr 9, 2015 at 17:57
Is there any leading or trailing spaces? If there is, you would need to trim() it before calling parseInt() — Victor Stafusa
– Victor Stafusa, Commented Apr 9, 2015 at 17:57
Hidden characters? Try trimming the string: java.lang.Integer.parseInt(request.getParameter("id").trim()). If that doesnt work maybe try printing out the character codes in the string. — ug_
– ug_, Commented Apr 9, 2015 at 17:57
Incredible - At least three persons had the same idea at the same time - trim the string. — Victor Stafusa
– Victor Stafusa, Commented Apr 9, 2015 at 17:58
@VictorStafusa I'm wondering if OP failed to trim() the string. What do you think? — JNYRanger
– JNYRanger, Commented Apr 9, 2015 at 18:09

Aniket Thakur · Accepted Answer · 2015-04-09 18:00:56Z

2

A complete full proof code would be

    String idString = request.getParameter("id");
    if(idString != null) {
        try {
            System.out.println(idString.trim()); // print to verify
            int idInt = Integer.parseInt(idString.trim());
        }
        catch(NumberFormatException nbe) {
            nbe.printStackTrace();
        }
    }

answered Apr 9, 2015 at 18:00

Aniket Thakur

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Comments

Prashant · Accepted Answer · 2015-04-09 18:10:31Z

0

First you need to check whether String returned by getParameter() is null or not then check whether it is empty ("") String or not then use Integer.parseInt().

String id = request.getParameter("id");
    if(null != id && !("".equals(id))) {
        try {
            int number = Integer.parseInt(id.trim());
        }
        catch(NumberFormatException e) {
            e.printStackTrace();
        }
    }

answered Apr 9, 2015 at 18:10

Prashant

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