Java - Issue with "if" statement

Question

I must have some sort of syntax error somewhere, because both correctip and loginip are equal but my code is still saying they are not equal.

        if(correctip != loginip){
            event.disallow(null, "Access is denied to user " + name + ". If you believe this is in error, contact an admin.");
            System.out.println(loginip);
            System.out.println(correctip);
            System.out.println("[Denied] HIGHRISKUSER " + name + " denied login. IP Att: " + loginip + " | IP Cor: " + correctip);
        }else{
            System.out.println("[Allowed] HIGHRISKUSER " + name + " allowed login. IP Att: " + loginip + " | IP Cor: " + correctip);
        }

Both correctip and loginip are EXACTLY equal. Is there something else that could be causing problems?

Are they strings? If so you have to use .equals() function call — Sagar V, May 07 '12 at 03:53
What type are `correctip` and `loginip`? If they're not ints or some other primitive you should use `loginip.equals(correctip)` — Paul, May 07 '12 at 03:53
Refer [here](http://stackoverflow.com/q/513832/1285418) for comparing String — Pau Kiat Wee, May 07 '12 at 04:02

score 4 · Accepted Answer · answered May 07 '12 at 03:53

4

correctip and loginip might be String. In that case, you should use the equals comparator:

if(correctip != null && correctip.equals(loginip)) {
    //your logic goes here...
}

answered May 07 '12 at 03:53

Luiggi Mendoza

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Thanks, I was not aware of that. – DannyF247 May 07 '12 at 04:22

score 1 · Answer 2 · answered May 07 '12 at 03:54

1

if both variables are strings, you need to use correctip.equals(loginip) or correctip.equalsIgnoreCase(loginip) to compare them.

answered May 07 '12 at 03:54

sgowd

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score 1 · Answer 3 · answered May 07 '12 at 04:09

What is the type of loginip and correctip?

Per your statement, I assume that the type should not be primitive(For example: int, long, short and so on). Instead, it would be a object (probably be string). so loginip or correctip is actually a reference of these two objects.

The equal-sign only assures the equality of its reference address. To compare two objects, you should use method equals or equalsIgnoreCase instead of =.

Java - Issue with "if" statement

3 Answers3