Help Me to Understand Comparisons

[Steve Dyke]

The following code is always false even though it should be true. The System.out.println displays O Adinistrator

LastName lastName = LastName.getLname(conn.lname);
UserType userText = UserType.getUType(MainMenuServlet.userTypeString);

if(userText.getUserType().equals("Administrator")){
UserType userText2 = UserType.getUType("true");
session.setAttribute("userTypeBean", userText2);
}
else{
System.out.println(userText.getUserType());
UserType userText2 = UserType.getUType("false");
session.setAttribute("userTypeBean", userText2);
}
    

[Steve Dyke]

Mis-spelled should be:

The following code is always false even though it should be true. The System.out.println displays O Administrator

[Bear Bibeault]

You can edit your posts by clicking the icon.

[Bear Bibeault]

Not sure what this has to do with JSP, but where's the extra "O" coming from?

[Steve Dyke]

Thats part of the string that displays in the console:

[5/10/07 10:25:33:543 CDT] 00000033 SystemOut O Administrator

You are right, this is actually servlet code. Sorry.
[ May 10, 2007: Message edited by: Steve Dyke ]

[Ben Souther]

This has nothing to do with servlets either.

You're comparing a type UserType with a String literal using the 'equals' method of UserType:

 
UserType userText = UserType.getUType(MainMenuServlet.userTypeString);
 
if(userText.getUserType().equals("Administrator")){
 

Without seeing the implementation for your UserType.equals method nobody on this site will be able to tell you why it's not behaving the way you expect it to.

Moving to Java In General (Beginner) since this is not a Servlet issue.

[Steve Dyke]

the .equals is mot a method in the UserType class. I just thought this was a built in Java method to use for comparisons.

[Ernest Friedman-Hill]

Originally posted by Steve Dyke:
. I just thought this was a built in Java method to use for comparisons.

It is -- every class has an equals() -- but you have to be using it correctly. Here, if "getUserType()" returns String, then this is OK code, but if it returns any other class, then that's the problem right there. What class does it return?

[Steve Dyke]

This is my class:

 
package com.ibm.drawinginquiry.rational;
 

public class UserType {
private String userType;

public UserType(String userType) {
this.userType = userType;

}

public String getUserType() {
return this.userType;

}

public static UserType getUType(String userType) {
return new UserType(userType);

}
}
   

[Herman Schelti]

hi Steve,

It's possible that userText.getUserType() contains spaces at the end, and then your comparison will fail.

To fix that: just add the trim() function
if( userText.getUserType().trim().equals("Administrator")){

Herman

[Steve Dyke]

Herman you have saved the day. The trim() made all the difference.

[Ben Souther]

This might be a good time for you to read up on the new Enum type provided by Java 5. This is exactly the type of confusion and ambiguity that they were designed to eliminate.