Problem sorting data in a file?

I have a problem sorting this file alphabetically by second name. Basically, my method sorts each column alphabetically but i would like to sort the file according to the second name. I really really need help. Thanks
File:
Moe     Carl
Saul     Sergio
Rocky     Louis
Ike     Ziken     
This is how my method sorts the file:
Ike          Carl
Moe          Louis
Rocky          Sergio
Saul          ziken
Instead
Moe     Carl
Rocky      Louis
Saul     Sergio
Ike     Ziken
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
import java.io.*;
import java.util.*;
class thedata implements Comparable
          private String firstname;
          private String secondname;
    public void setFirstname(String firstname)
             this.firstname = firstname;
    public String getFirstname()
              return firstname;
    public void setSecondname(String secondname)
            this.secondname = secondname;
    public String getSecondname()
            return secondname;
    public int compareTo(Object Student) throws ClassCastException
            if (!(Student instanceof ShowData))
            throw new ClassCastException("Error");
            String sn = ((ShowData) Student).getSecondname();
            return this.secondname.compareTo(sn);
public class sort {
        public static void main(String[] args) {
        sortmethod();
        public static void sortmethod(){
          int j = 0;
          thedata data[] = new thedata[4];
         try
                  FileInputStream fstream = new FileInputStream("datafile.txt");
                  DataInputStream in = new DataInputStream(fstream);
                  BufferedReader br = new BufferedReader(new InputStreamReader(in));
                  String line;
                  String[] firstname = new String[4];
                  String[] secondname = new String[4];
                  ArrayList<String> list = new ArrayList<String>();
                  while ((line = br.readLine()) != null)
                          list.add(line);
                  Iterator<String> itr = list.iterator();
                  int k = 0;
                  for (itr = list.iterator(); itr.hasNext();)
                          String str = itr.next().toString();
                          String[] splitSt = str.split("\t");
                 for (int i = 0; i < splitSt.length; i++)
                             firstname[k] = splitSt[0];
                             secondname[k] = splitSt[1];
                  k++;
                  arraysort(firstname);
                  arraysort(secondname);
                     for(j = 0;j < 4;j++)
                             data[j] = new thedata();
                             data[j].setFirstname(firstname[j]);
                             data[j].setSecondname(secondname[j]);
                  for (int i = 0; i < 4; i++)
                             thedata show = data;
                         String firstname1 = show.getFirstname();
                         String secondname1 = show.getSecondname();
                              System.out.println(firstname1 + "\t\t" + secondname1);
                    catch (Exception e)
     private static void arraysort(String[] array)
          for(int i=0; i<array.length; i++)
               for(int j=0; j<array.length-1-i; j++)
                    if(array[j].compareTo(array[j+1])>0)
                              String temp1 ="";
temp1= array[j];
array[j] = array[j+1];
array[j+1] = temp1;
Edited by: 999363 on Apr 11, 2013 3:41 AM

You're not sorting your objects, you're only sorting the firstname and secondname arrays separately, so they get scrambled. Sort the array of `thedata` objects.
BTW this:
    public int compareTo(Object Student) throws ClassCastException
            if (!(Student instanceof ShowData))
            throw new ClassCastException("Error");
            String sn = ((ShowData) Student).getSecondname();
    }contains a complete waste of time. You should just remove 'throws ClassCastException' and the first two lines of the method. The test and throw happens anyway at the cast. If Student isn't an instance of ShowData the cast will fail with the same exception (and a much better error message).

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    Pavel Vera,
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                      <TrackingNumber>1ZR3W891PG47477811</TrackingNumber>
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          gs_shipmentdetail         TYPE ts_shipmentdetail.
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      WRITE: / 'Date  :',      gs_shipment-Date.
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      WRITE: / 'Freight :',           gs_shipmentdetail-Freight.
      WRITE: / 'Weight  :',           gs_shipmentdetail-Weight.
      WRITE: / 'ShipmentDate  :',     gs_shipmentdetail-ShipmentDate.
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      WRITE: / 'OrderedQuantity  :',  gs_productinformation-OrderedQuantity.
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            <xsl:value-of select="VendorNumber"/>
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            <xsl:value-of select="OrderNumber"/>
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          <DATE>
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      <xsl:template match="/">
        <asx:abap xmlns:asx="http://www.sap.com/abapxml" version="1.0">
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            <ISHIPDET>
              <xsl:apply-templates select="//OrderShipment/ShipmentDetail"/>
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          </asx:values>
        </asx:abap>
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      <xsl:template match="ShipmentDetail">
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          <TRACKINGNUMBER>
            <xsl:value-of select="TrackingNumber"/>
          </TRACKINGNUMBER>
          <FREIGHT>
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          </FREIGHT>
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            <xsl:value-of select="ShipmentDate"/>
          </SHIPMENTDATE>
          <SHIPMENTMETHOD>
            <xsl:value-of select="ShipmentMethod"/>
          </SHIPMENTMETHOD>
        </item>
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    </xsl:transform> .
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    Any Help will be highly apppreciated. Thanks in advance
    Regards,
    Jessica   Sam

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    I want to display the data in alphabetical order by last name and by first name. what is the best way to do this? I found java.util.Arrays
    but i thought that it might be best if i read the data into a linked list. I want to display it on a form, so that I can press a next and previous button. Next will show the next record in alphabetical order after the previous one(B->C) and previous would go (C->B)
    thanks.

    How about this:
    Create a PersonRecord object, and have it implement Comparable.
    import java.util.Comparator;
    public class PersonRecord implements Comparable < PersonRecord >
        public static Comparator < PersonRecord > surnameComparator ()
            public int compare (PersonRecord pr1, PersonRecord pr2)
                return pr1.getSurname ().compareTo (pr2.getSurname ());
        public static Comparator < PersonRecord > nameComparator ()
            public int compare (PersonRecord pr1, PersonRecord pr2)
                return pr1.getName ().compareTo (pr2.getName ());
    }See java.util.Comparator, java.lang.Comparable.
    Then use Arrays.sort(List, Comparable) at your own discretion...
    I've said enough; work that part out yourself.
    lutha

  • Problem : Inserting Data from flat file  into  databse

    Hi,
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  • Problem with Data load from file

    Hi,
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    A known bug. See below for a solution to set the timeout. Remember to reboot the PC for the changes take effect.
    See Re: Problem with importing HTML DB applications

  • Problem grouping data in xml file

    Hello All,
    Following is the sample xml file used as input source into my BIP stand alone report:
    <ROWSET>
    <SAMPLE_ROW>
    <ID>1234</ID>
    <CODE>A1</CODE>
    <TEXT>FIRST GRADE</TEXT>
    <CODE>B1</CODE>
    <TEXT>SECOND GRADE</TEXT>
    </SAMPLE_ROW>
    <SAMPLE_ROW>
    <ID>6789</ID>
    <CODE>C1</CODE>
    <TEXT>THIRD GRADE</TEXT>
    <CODE>B1</CODE>
    <TEXT>SECOND GRADE</TEXT>
    </SAMPLE_ROW>
    </ROWSET>
    The expected output should be:
    ID# 1234
    Grades: A1
    FIRST GRADE
    B1
    SECOND GRADE
    ID# 6789
    Grades: C1
    THIRD GRADE
    B1
    SECOND GRADE
    And the design i've created is:
    <?for-each-group:SAMPLE_ROW;./ID?>
    ID#<?ID?>
    Grades:<?CODE?>
    <?TEXT?>
    <?end for-each-group?>
    I could only display partial result but not able to achieve my desired output. Please guide.
    Thanks.

    Thanks Jorge,
    This structure can be arranged by my third party and the output came out as required.
    But we also have another report which would use the same xml file as one data source and a sql query as another data source. The total xml generated would look like:
    <DATA>
    <DT>
    <CLASS>
    <ID>1234</ID>
    <NAME>Rich Walker</NAME>
    <Dept>IT</Dept>
    </CLASS>
    <CLASS>
    <ID>6789</ID>
    <NAME>Steve Round</NAME>
    <Dept>Arts</Dept>
    </CLASS>
    <CLASS>
    <ID>1873</ID>
    <NAME>Mary Joyce</NAME>
    <Dept>Arts</Dept>
    </CLASS>
    </DT>
    <Main_XML>
    <ROWSET>
    <SAMPLE_ROW>
    <ID>1234</ID>
    <CODEROW>
    <CODE>A1</CODE>
    <TEXT>FIRST GRADE</TEXT>
    </CODEROW>
    <CODEROW>
    <CODE>B1</CODE>
    <TEXT>SECOND GRADE</TEXT>
    </CODEROW>
    </SAMPLE_ROW>
    <SAMPLE_ROW>
    <ID>6789</ID>
    <CODEROW>
    <CODE>C1</CODE>
    <TEXT>THIRD GRADE</TEXT>
    </CODEROW>
    <CODEROW>
    <CODE>B1</CODE>
    <TEXT>SECOND GRADE</TEXT>
    </CODEROW>
    </SAMPLE_ROW>
    </ROWSET>
    </Main_XML>
    </DATA>
    And the rtf template i've designed(which is not working):
    <?for-each-group:CLASS;./ID?>
    ID# <?ID?>
    <?variable@incontext:v_id;ID?>
    Name: <?NAME?>
    Dept: <?Dept?>
    <?for-each-group:current-group();./ID[../ID=$v_id]?>
    Grades: <?/DATA/Main_XML/ROWSET/SAMPLE_ROW/CODEROW/CODE?>
    <?/DATA/Main_XML/ROWSET/SAMPLE_ROW/CODEROW/TEXT?>
    <?end for-each-group?>
    <?end for-each-group?>
    The ideal output should be:
    ID# 1234
    Name: Rich Walker
    Dept: IT
    Grades: A1
    FIRST GRADE
    B1
    SECOND GRADE
    ID# 6789
    Name: Steve Round
    Dept: Arts
    Grades: C1
    THIRD GRADE
    B1
    SECOND GRADE
    Please assist me in getting the desired output.

  • Problem Loading Data Using UTL File Package

    Hi Friends..
    My Database is Oracle 10gR2 and OS is WINDOWS.
    I have one excel file which contains 10 fields.
    My requirement is to Load data from 1 excel file into two tables..
    tables are master and detail
    Below are sample data and structure for it..
    Excel file format
    TEST.CSV
    Srno Empno Empname City Challanno challandate Materialno materialname Materialqty Materialcost
    1     232 raj      Hyderabad      533     20/04/2010     11     abc      34     10
    1     232 raj      Hyderabad      533     20/04/2010     12      aa      4     110
    1     231 ram Baroda      533     20/04/2010     14      abcd      33     210
    Master table
    empno
    Challanno
    challandate
    Detail table
    empno
    materialno
    materialname
    materialqty
    materialcost
    My question is ..While reading 1st line if its empno is new then record is entered in master table for the first time and remaining records of same empno is entered in detail table.. now when empno no changes then new entry is done in master table and associated records are entered in detail table..
    So in this case for empno 232 master table would have
    232,533,20/4/2010
    Detail table for empno 232 would now have 2 records..
    232,11,abc,34,10
    232,12,aa,4,110
    I am using UTL_FILE package to achieve this as the file is on server...
    Kindly please help me to proceed in this..
    Really appreciate your help...

    sai121 wrote:
    Its ok..if u dont wanna reply sir... but thats what i m told to do..and i have 4yrs of industry experience too...but cant argue with boss..u know that right.It's not that people don't want to reply, it's that what you're asking for is something achieved very simply in a few lines of code using External tables but is a convoluted and complex thing to do using UTL_FILE, so why would anybody want to waste their time giving you a load of code to achieve what you want when they know it's the wrong approach anyway.
    I've been computer programming for 28 years (jeez has it really been that long :D ), so I wouldn't be telling you that you're doing it the wrong way without knowing that there are better ways to do it and you're asking for the wrong way. Speak with your boss, tell him that you've been recommended to use External Tables instead because they're the right way to read such data and UTL_FILE is not the right approach.

  • Problem reading data from a file

    this is inside a servlet.
    Re: servlet question
    Author: gramberto
    In Reply To: servlet question Oct 1, 2004 10:24 AM
    Reply 6 of 6
    I'm getting the following errors:
    java.util.NoSuchElementException     at java.util.StringTokenizer.nextToken(StringTokenizer.java:259)     at myClass.readFile(myClass.java:387)     at myClass.doPost(myClass.java:421)I am reading a file of the form:
    Blah|blah2||
    '|' is the delimiter. A null would just have another |
    Here is my code:
    public void readFile(HttpServletRequest request, HttpServletResponse response)
        throws IOException, ServletException
              PrintWriter out = response.getWriter();
             String row;
             BufferedReader infile1 = new BufferedReader(new FileReader("c:\\file1.txt"));
             response.setContentType("text/html");
             String FName="",MName="",LName="",Sex="",Street1="",Street2="",City="",State="",Zip="",PhoneLocalCode="",PhoneAreaCode="",PhoneLastCode="",Email="",OtherInfo="";
            try {
                    while ((row = infile1.readLine()) != null)
                      StringTokenizer in_str = new StringTokenizer(row, "|");
                      FName = in_str.nextToken();
                      MName = in_str.nextToken();
                      LName = in_str.nextToken();
                      Sex = in_str.nextToken();
                      Street1 = in_str.nextToken();
                      Street2 = in_str.nextToken();
                      City = in_str.nextToken();
                      State = in_str.nextToken();
                      Zip = in_str.nextToken();
                      PhoneLocalCode = in_str.nextToken();
                      PhoneAreaCode = in_str.nextToken();
                      PhoneLastCode = in_str.nextToken();
                      Email = in_str.nextToken();
                      OtherInfo = in_str.nextToken();
                 } catch (IOException ioException) {
                   out.println("Error: "+ioException);
                 out.println(FName);
                 out.println(MName);

    Well, you are probably passing in a line that is tokinized into less than 14 tokens.
    Add some debug output that prints the line that you are tokenizing.
    Use hasMoreTokens and countTokens to determine if it is ok to call nextToken or if you are already out of tokens.
    E.g. you could do something along these lines:
    fixed token count formatprivate void tokenize(final String line)
      final StringTokenizer st = new StringTokenizer( line, "|" );
      if ( st.countTokens() != 14 )
         throw new RuntimeException("Invalid token count: " + st.countTokens());
      FName = st.nextToken();
    }variable token count formatprivate void tokenize(final String line)
      final StringTokenizer st = new StringTokenizer( line, "|" );
      if (st.hasMoreTokens())
        FName = st.nextToken();
      if (st.hasMoreTokens())
        MName = st.nextToken();
    }

  • Sort data does not match song info

    I've just discovered that some of my songs have sort data from different files. Short of going though all my music song by song is there a way to clear all sort data?

    Generally setting a common Album Artist will fix split albums.
    For deeper problems see Grouping tracks into albums.
    tt2

  • In my opinion, the real solution is for Apple to offer us a choice of the photo sort order in ITunes. My preference would be filename, perhaps with options to choose the Date Taken attribute, file timestamp or date modified filestamp (EXIF date fields as

    I tryed to sort my pictures with buying Apps and following suggestion from apple, but without any success. In my opinion, the real solution is for Apple to offer us a choice of the photo sort order in ITunes. My preference would be filename, perhaps with options to choose the Date Taken attribute, file timestamp or date modified filestamp.

    Not a problem when using iPhoto on a Mac, which the transfer of photos is primarily based on - not manually managing photo storage as seems to be common with Windoze.
    The same should be available with a supported photo management app on a PC.
    http://support.apple.com/kb/HT4221

  • Efficent method to sort data from tab delimited text file

    I am currently writing a program to sort through data that was acquired and display it on a graph and some other indicators.  The file is a tab delimited text file with possibly 100,000s of data points.  the current method that I have tried using was that if I wanted all of the data from Oct, I would parse out the month from the timestamp, compare that to the desired month, and add it to the array if it is the same.  Other possible options of sorting are yearly and daily, possibly even hourly.
    The method does work, however it does take some time (up to a minute on a P4 3.6 GHz with 2 gb ram), and most of the other computers are not nearly as fast or with as much memory.  Is there a more efficent method to sorting the data??
    I attached my sorting vi as well as a sample data file.
    thanks for the advice.  It is saved in LV8.0.1
    Kenny
    Kenny
    Attachments:
    data sort.zip ‏84 KB
    oven1.txt ‏21 KB

    First of all, "sorting" has usually a different meaning (Sorting and numeric array ascending or descending, a string array aphabetically, etc.). Your data already seems sorted by date and time, you just want to pick a subset having certain characteristics.
    The main problem that is slowing you down is your constant growing of large arrays. This causes constant memory reallocations.
    Since your data is already sorted by date and time, all you need is to place your data in a sutable data structure, find the start and end point of your selection, then use "array subset" for example.
    Your code also seems to have a lot of unecessary complexity. See for example your "test for sort data" (see image below).
    the four cases only differ by filename --> only the file name belongs into the case and the file operation outside the inner case. Even better, just use autoindexing.
    that shift register does not do anything, because it always contains the same data. Using "index array" with index wired to [i] is equivalent to an autoindexing tunnel.
    You have a case structure to select which files to read, skipped files give you an empty array. Do you really need to do all these operations on an empty array. Why not place all code inside the TRUE case??
    Below is an image of one possible code alternative that addresses some of these points.
    Message Edited by altenbach on 10-26-2006 09:32 AM
    LabVIEW Champion . Do more with less code and in less time .
    Attachments:
    testForSortData.png ‏33 KB

  • Can no longer sort RAW with JPG files together by capture date.

    Suddenly I can no longer sort  JPG and RAW files together by capture date. I have two Canon SLR's shooting in RAW and one Canon point an shoot shooting in JPG and want to collate the photos in LR4, which I was able to do up to now. I cannot find out what I did to make this happen or correct it in LR preferences. Any solutions?
    JimB.

    View->Sort->Capture Time

  • I have one problem with Data Guard. My archive log files are not applied.

    I have one problem with Data Guard. My archive log files are not applied. However I have received all archive log files to my physical Standby db
    I have created a Physical Standby database on Oracle 10gR2 (Windows XP professional). Primary database is on another computer.
    In Enterprise Manager on Primary database it looks ok. I get the following message “Data Guard status Normal”
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    After I created the Physical Standby database, I have also done:
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    CONNECT SYS/SYS@luda AS SYSDBA
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    STARTUP NOMOUNT PFILE=C:\oracle\product\10.2.0\db_1\database\initluda.ora
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    ALTER DATABASE MOUNT STANDBY DATABASE
    4. I started redo apply on Physical Standby database
    alter database recover managed standby database disconnect from session
    5. I switched the log files on Physical Standby database
    alter system switch logfile
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    select sequence#, first_time, next_time from v$archived_log order by sequence#
    SEQUENCE# FIRST_TIME NEXT_TIME
    3 2006-06-27 2006-06-27
    4 2006-06-27 2006-06-27
    5 2006-06-27 2006-06-27
    6 2006-06-27 2006-06-27
    7 2006-06-27 2006-06-27
    8 2006-06-27 2006-06-27
    7. I verified the archived redo log files were applied on Physical Standby database
    select sequence#,applied from v$archived_log;
    SEQUENCE# APP
    4 NO
    3 NO
    5 NO
    6 NO
    7 NO
    8 NO
    8. on Physical Standby database
    select * from v$archive_gap;
    No rows
    9. on Physical Standby database
    SELECT MESSAGE FROM V$DATAGUARD_STATUS;
    MESSAGE
    ARC0: Archival started
    ARC1: Archival started
    ARC2: Archival started
    ARC3: Archival started
    ARC4: Archival started
    ARC5: Archival started
    ARC6: Archival started
    ARC7: Archival started
    ARC8: Archival started
    ARC9: Archival started
    ARCa: Archival started
    ARCb: Archival started
    ARCc: Archival started
    ARCd: Archival started
    ARCe: Archival started
    ARCf: Archival started
    ARCg: Archival started
    ARCh: Archival started
    ARCi: Archival started
    ARCj: Archival started
    ARCk: Archival started
    ARCl: Archival started
    ARCm: Archival started
    ARCn: Archival started
    ARCo: Archival started
    ARCp: Archival started
    ARCq: Archival started
    ARCr: Archival started
    ARCs: Archival started
    ARCt: Archival started
    ARC0: Becoming the 'no FAL' ARCH
    ARC0: Becoming the 'no SRL' ARCH
    ARC1: Becoming the heartbeat ARCH
    Attempt to start background Managed Standby Recovery process
    MRP0: Background Managed Standby Recovery process started
    Managed Standby Recovery not using Real Time Apply
    MRP0: Background Media Recovery terminated with error 1110
    MRP0: Background Media Recovery process shutdown
    Redo Shipping Client Connected as PUBLIC
    -- Connected User is Valid
    RFS[1]: Assigned to RFS process 2148
    RFS[1]: Identified database type as 'physical standby'
    Redo Shipping Client Connected as PUBLIC
    -- Connected User is Valid
    RFS[2]: Assigned to RFS process 2384
    RFS[2]: Identified database type as 'physical standby'
    Redo Shipping Client Connected as PUBLIC
    -- Connected User is Valid
    RFS[3]: Assigned to RFS process 3188
    RFS[3]: Identified database type as 'physical standby'
    Primary database is in MAXIMUM PERFORMANCE mode
    Primary database is in MAXIMUM PERFORMANCE mode
    RFS[3]: No standby redo logfiles created
    Redo Shipping Client Connected as PUBLIC
    -- Connected User is Valid
    RFS[4]: Assigned to RFS process 3168
    RFS[4]: Identified database type as 'physical standby'
    RFS[4]: No standby redo logfiles created
    Primary database is in MAXIMUM PERFORMANCE mode
    RFS[3]: No standby redo logfiles created
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    SELECT PROCESS, STATUS, THREAD#, SEQUENCE#, BLOCK#, BLOCKS FROM V$MANAGED_STANDBY;
    PROCESS STATUS THREAD# SEQUENCE# BLOCK# BLOCKS
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    ARCH CONNECTED 0 0 0 0
    RFS IDLE 0 0 0 0
    RFS IDLE 0 0 0 0
    RFS IDLE 1 9 13664 2
    RFS IDLE 0 0 0 0
    10) on Primary database:
    select message from v$dataguard_status;
    MESSAGE
    ARC0: Archival started
    ARC1: Archival started
    ARC2: Archival started
    ARC3: Archival started
    ARC4: Archival started
    ARC5: Archival started
    ARC6: Archival started
    ARC7: Archival started
    ARC8: Archival started
    ARC9: Archival started
    ARCa: Archival started
    ARCb: Archival started
    ARCc: Archival started
    ARCd: Archival started
    ARCe: Archival started
    ARCf: Archival started
    ARCg: Archival started
    ARCh: Archival started
    ARCi: Archival started
    ARCj: Archival started
    ARCk: Archival started
    ARCl: Archival started
    ARCm: Archival started
    ARCn: Archival started
    ARCo: Archival started
    ARCp: Archival started
    ARCq: Archival started
    ARCr: Archival started
    ARCs: Archival started
    ARCt: Archival started
    ARCm: Becoming the 'no FAL' ARCH
    ARCm: Becoming the 'no SRL' ARCH
    ARCd: Becoming the heartbeat ARCH
    Error 1034 received logging on to the standby
    Error 1034 received logging on to the standby
    LGWR: Error 1034 creating archivelog file 'luda'
    LNS: Failed to archive log 3 thread 1 sequence 7 (1034)
    FAL[server, ARCh]: Error 1034 creating remote archivelog file 'luda'
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    select name,sequence#,applied from v$archived_log;
    NAME SEQUENCE# APP
    C:\ORACLE\PRODUCT\10.2.0\ORADATA\IRINA\ARC00003_0594204176.001 3 NO
    C:\ORACLE\PRODUCT\10.2.0\ORADATA\IRINA\ARC00004_0594204176.001 4 NO
    Luda 4 NO
    Luda 3 NO
    C:\ORACLE\PRODUCT\10.2.0\ORADATA\IRINA\ARC00005_0594204176.001 5 NO
    Luda 5 NO
    C:\ORACLE\PRODUCT\10.2.0\ORADATA\IRINA\ARC00006_0594204176.001 6 NO
    Luda 6 NO
    C:\ORACLE\PRODUCT\10.2.0\ORADATA\IRINA\ARC00007_0594204176.001 7 NO
    Luda 7 NO
    C:\ORACLE\PRODUCT\10.2.0\ORADATA\IRINA\ARC00008_0594204176.001 8 NO
    Luda 8 NO
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    C:\ORACLE\PRODUCT\10.2.0\ORADATA\LUDA\ARC00003_0594204176.001 3 NO
    C:\ORACLE\PRODUCT\10.2.0\ORADATA\LUDA\ARC00005_0594204176.001 5 NO
    C:\ORACLE\PRODUCT\10.2.0\ORADATA\LUDA\ARC00006_0594204176.001 6 NO
    C:\ORACLE\PRODUCT\10.2.0\ORADATA\LUDA\ARC00007_0594204176.001 7 NO
    C:\ORACLE\PRODUCT\10.2.0\ORADATA\LUDA\ARC00008_0594204176.001 8 NO
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    irina.__db_cache_size=79691776
    irina.__java_pool_size=4194304
    irina.__large_pool_size=4194304
    irina.__shared_pool_size=75497472
    irina.__streams_pool_size=0
    *.audit_file_dest='C:\oracle\product\10.2.0\admin\luda\adump'
    *.background_dump_dest='C:\oracle\product\10.2.0\admin\luda\bdump'
    *.compatible='10.2.0.1.0'
    *.control_files='C:\oracle\product\10.2.0\oradata\luda\luda.ctl'
    *.core_dump_dest='C:\oracle\product\10.2.0\admin\luda\cdump'
    *.db_block_size=8192
    *.db_domain=''
    *.db_file_multiblock_read_count=16
    *.db_file_name_convert='luda','irina'
    *.db_name='irina'
    *.db_unique_name='luda'
    *.db_recovery_file_dest='C:\oracle\product\10.2.0\flash_recovery_area'
    *.db_recovery_file_dest_size=2147483648
    *.dispatchers='(PROTOCOL=TCP) (SERVICE=irinaXDB)'
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    *.log_archive_config='DG_CONFIG=(irina,luda)'
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    *.log_archive_dest_2='SERVICE=irina LGWR ASYNC VALID_FOR=(ONLINE_LOGFILES, PRIMARY_ROLE) DB_UNIQUE_NAME=irina'
    *.log_archive_dest_state_1='ENABLE'
    *.log_archive_dest_state_2='ENABLE'
    *.log_archive_max_processes=30
    *.log_file_name_convert='C:/oracle/product/10.2.0/oradata/irina/','C:/oracle/product/10.2.0/oradata/luda/'
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    *.pga_aggregate_target=16777216
    *.processes=150
    *.remote_login_passwordfile='EXCLUSIVE'
    *.sga_target=167772160
    *.standby_file_management='AUTO'
    *.undo_management='AUTO'
    *.undo_tablespace='UNDOTBS1'
    *.user_dump_dest='C:\oracle\product\10.2.0\admin\luda\udump'
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    irina.__db_cache_size=79691776
    irina.__java_pool_size=4194304
    irina.__large_pool_size=4194304
    irina.__shared_pool_size=75497472
    irina.__streams_pool_size=0
    *.audit_file_dest='C:\oracle\product\10.2.0/admin/irina/adump'
    *.background_dump_dest='C:\oracle\product\10.2.0/admin/irina/bdump'
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