Contact sorting and nicknames

I am one of those people who has a well organized Address Book so if need be I can print out nice formal address slips or when needed I have all my friends listed by their nicknames.
Am I not finding the right settings page or isn't it possible at all to organize people by nicknames on the iPhone? I have a lot of business contacts for whom I don't mind having last and first name, but I'd still want my best friend to display by his nickname.
Thanks in advance,
eero

hi..jain...
ROLE MERGING
If you want to assign two roles to a user but dont want them to appear as two roles,
you can use the concept of Merging of Roles.
Here,one can allow the user to see only that role in the top level navigation,whose
priority has been set as low.Although the contents of the other role are visible too,
but only n the second level navigation,with the contents of the higher priority role.
This is done by setting the merge-id of the two roles as one string(e.g. merge), and
then setting the merge-priority of the role you want to be visible in Top level navigation
as 99,while setting the other role's priority as 100.
This ensures that only one of the two (or many roles are visible).
SORTING is...
You can define the display sequence of the nodes comprising the various levels of the navigation hierarchy by prioritizing the items on a per level basis.
You define priorities by assigning a value between 0 and 100 to each item. Values may include decimal numbers.
The smaller the value of the navigation node, the further left it displays on the navigation bar, or the higher up it displays in detailed navigation. The item with the lowest value appears to the far left or the top, respectively, and the others follow in the order of their priorities.
thanx..
rewards points if useful..

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Hi, Could somn please be of any help here. I am trying to sort this string in alphabetical order and then search for data available but I am not makingprogress with the code. Does anyone have any advise or better still the code to solve this problem... This is hat I have at the moment
import java.util.*;
import java.io.PrintStream;
import java.io.File;
import java.io.FileNotFoundException;
public class coursework2
     public static final int MAX_RECORDS = 20;
     public static String lastName[] = new String[MAX_RECORDS];
     public static String firstName[] = new String[MAX_RECORDS];
     public static String telNumber[] = new String[MAX_RECORDS];
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                    telNumber[record_count] = line.next();
                    emailAddress[record_count] = line.next();
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               e.printStackTrace();
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               write_out.close();
          catch (FileNotFoundException e)
               e.printStackTrace();
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Am very sorry this is long, the file should be sorted after it is loaded and I have that, You can make up some data with last ane, first name, tel and email to show me an example. Thanks alot
Joseph

Hi Monica, Thanks for you patience with me.
I have tried writing a selection sort method but am having some errors. Could you kindly have a lookand correct me please.
This is the code i wrote
public static void selectionSort(Comparable [] data)
          String strFirstName;
          String strLastName;
          Comparable temp;
          for(int i = 0; i < lastName.length; i++)
               lastName = index;
               for (int j = i; j < lastName.length; j++)
                    if (lastName[j].compareTo(lastName) < 0)
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               temp = lastName[j];
               lastName[j] = lastName[i];
               lastName[i] = temp;
As you requested, below are 4 error messages I had
E:\coursework2.java:101: cannot find symbol
symbol : variable index
location: class coursework2
               lastName = index;
^
E:\coursework2.java:107: cannot find symbol
symbol : variable j
location: class coursework2
               temp = lastName[j];
^
E:\coursework2.java:108: cannot find symbol
symbol : variable j
location: class coursework2
               lastName[j] = lastName[i];
^
E:\coursework2.java:109: incompatible types
found : java.lang.Comparable
required: java.lang.String
               lastName[i] = temp;
^
4 errors
JCompiler done.
JCompiler ready.
Joseph

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