Total number of disk drivers in win2000

Hi there
Can somebody show me a java example, how I can find out the numbers of drive in the system in win2000 an XP.
Thanks in advance.
Cheers
FF

File.listRoots()

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  • Find Total Number of Rows in Database

    I am trying to find total number of rows in Database? Thanks in advance

    I'd come back to a question of "why". What possible reason could you have to want to know the number of rows in every table in the database?
    This seems like it is a request from someone non-technical where they're not sure exactly what they need to ask to get the information they want. I'm forseeing something along the lines of
    User: How many rows are in the database
    DBA: Ummm.... <<goes off and spends a lot of resources to count>>> ... 1 billion.
    User: Wow, that's a lot... How big are those rows...
    DBA: Hmmm... <<goes off and spends more time trying to find an average row size>>... 100 bytes
    User: <<Plugs numbers into Excel>>... OK, our database is 93 GB, so we need to order 100 GB of disk for our new environment.
    DBA: <<Months later, after 100 GB of disk has been budgeted, purchased, and delivered>> User, you didn't order enough disk for our disaster recovery environment.
    User: You said our database was only 93 GB!!
    DBA: 93 GB of data. Plus 100 GB of indexes, TEMP, UNDO, space for archived logs, ...
    When "odd" requests come down, it's generally the case that the person asking the question doesn't know what question they really want to ask because they don't have the domain knowledge to know how to ask the question. Walking over and asking them what they're hoping to accomplish/ what they're using the data for generally leads to a case where
    - The user gets exactly the information they need, not the information they think they want
    - The DBA does a lot less work, since the right question is almost always easier to answer
    - The user thinks of the DBA as a really helpful sort of guy and remembers how helpful you were in the past when you have to push back on other things later
    - No one ends up playing the blame game when the user gets just what they asked for and not what they want.
    Justin
    Distributed Database Consulting, Inc.
    http://www.ddbcinc.com/askDDBC

  • ORA-15063: ASM discovered an insufficient number of disks for diskgroup

    Hello DBAs,
    I have encountered this problem. I am using storage vendor snapshot capability. On my first node, I have 2 disks for ASM diskgroups. ORCL:DATA belongs to DATA diskgroup and ORCL:FLASH belongs to FLASH diskgroup. After making the snapshots of these two disks, I mapped them to my second host which already has ASM instance. I ran /etc/init.d/oracleasm scandisk and listdisks I see two disks. But when I start up ASM instance I get this error:
    SQL> startup
    ASM instance started
    Total System Global Area 130023424 bytes
    Fixed Size 2019032 bytes
    Variable Size 102838568 bytes
    ASM Cache 25165824 bytes
    ORA-15032: not all alterations performed
    ORA-15063: ASM discovered an insufficient number of disks for diskgroup "FLASH"
    ORA-15063: ASM discovered an insufficient number of disks for diskgroup "DATA"
    Of course, I have the asm_diskstring and asm_diskgroups parameter set in my init+ASM.ora file. The ASM version is identical across both hosts. I guess my question would be, is it possible to do this? Do I need to re-create the ASM disks using /etc/init.d/oracleasm? Any help is much appreciated.
    Thanks,
    TD

    What is your configuration?
    If you have AIX and EMC san, then refer to this metalink note: 467702.1
    In my case, it was HP-UX and HP san EVA8200. But our problem was two different disks were presented to the two nodes, with same names by the unix admin/san admin guys.
    Before they could figure out the problem, I had done a lot of research for few days to figure this out. Another problem I noticed in some cases was improper permissions. Either the db software owner didn't have permissions to the asm disks, or if the owner of both software was same, then the actual permissions/ownership of the shared raw devices was different on the two nodes.
    Hope this helps. And I will appreciate once you fix the problem if you can also update this forum with your solution. That will help the community in future.
    Thanks

  • ASM discovered an insufficient number of disks for diskgroup

    Running RH4U3 and Oracle 10gr2
    Every time I reboot my system, the minor number changed and i can't startup the ASM instance. Does any one know why this is happening? Any advise would be appreciated.
    the minor number was 4 and 5, now after reboot i got::
    # ls -l /dev/oracleasm/disks
    total 0
    brw-rw---- 1 ora dba 253, 4 Feb 21 10:10 VOL1
    brw-rw---- 1 ora dba 8, 65 Feb 21 10:10 VOL2
    I can't startup my ASM instance:
    $ sqlplus "/ as sysdba"
    SQL*Plus: Release 10.2.0.1.0 - Production on Wed Feb 21 10:13:03 2007
    Copyright (c) 1982, 2005, Oracle. All rights reserved.
    Connected to an idle instance.
    SQL> startup
    ASM instance started
    Total System Global Area 83886080 bytes
    Fixed Size 1217836 bytes
    Variable Size 57502420 bytes
    ASM Cache 25165824 bytes
    ORA-15032: not all alterations performed
    ORA-15063: ASM discovered an insufficient number of disks for diskgroup "OARCH"
    y ASM instance:

    I have no problem reading the disks.
    $ dd if=/dev/oracleasm/disks/VOL1 of=/dev/null
    485336+0 records in
    485336+0 records out
    $ dd if=/dev/oracleasm/disks/VOL2 of=/dev/null
    741760+0 records in
    741760+0 records out

  • Count number of disk accesses

    Hi,
    I'm facing a problem while coding some stuff for my thesis. I need to count the number of disk accesses. I know there's clock_t to do that in C++, but I cant seem to find any equivalent in Java. Can someone please help me in this matter? Thank you very much.

    You can't do that in pure java. So if you really wantd it, you would need to learn JNI.Even then: most disc drivers use caching (read ahead) and most OSes
    try to be smart too when it comes to raw disc access. Without any
    kernel access it can't be done.
    OTOH, most disc drivers keep some form of statistics that can be
    queried.
    kind regards,
    Jos

  • How to get the total number of occurrences based on the value of a column.

    Hello everyone,
    This is the first time that I will ask question here on your forum but has been following several threads ever since. I guess that now is my turn to ask a question. So anyway here is the thing, I have a query that should return count the number of rows depending on the value of SLOT. Something like this:
    WIPDATAVALUE          SLOT             N            M
    1-2                   TRALTEST43S1     1            3
    1-2                   TRALTEST43S1     2            3
    3                     TRALTEST43S1     3            3
    4-6                   TRALTEST43S2     1            4
    4-6                   TRALTEST43S2     2            4
    4-6                   TRALTEST43S2     3            4
    7                     TRALTEST43S2     4            4-----
    As you can see above, on the SLOT TRALTEST43S1, there are three occurrences so M (Total number of occurrences) should be three and that column N should count it. Same goes with the SLOT TRALTEST43S2. This is the query that I have so far:
    SELECT DISTINCT
    WIPDATAVALUE, SLOT
    , LEVEL AS n
    , m
    FROM
      SELECT
        WIPDATAVALUE
        , SLOT
        , (dulo - una) + 1 AS m
      FROM
        SELECT
          WIPDATAVALUE
          , SLOT
          , CASE WHEN INSTR(wipdatavalue, '-') = 0 THEN wipdatavalue ELSE SUBSTR(wipdatavalue, 1, INSTR(wipdatavalue, '-')-1) END AS una
          , CASE WHEN INSTR(wipdatavalue, '-') = 0 THEN wipdatavalue ELSE SUBSTR(wipdatavalue, INSTR(wipdatavalue, '-') + 1) END AS dulo
        FROM trprinting
        WHERE (containername = :lotID OR SLOT= :lotID) AND WIPDATAVALUE LIKE :wip
    ) CONNECT BY LEVEL <= m
    ORDER BY wipdatavalue;And that it results to something like this:
    WIPDATAVALUE          SLOT             N            M
    1-2                   TRALTEST43S1     1            2
    1-2                   TRALTEST43S1     2            2
    3                     TRALTEST43S1     1            1
    4-6                   TRALTEST43S2     1            3
    4-6                   TRALTEST43S2     2            3
    4-6                   TRALTEST43S2     3            3
    7                     TRALTEST43S2     1            1-----
    I think that my current query is basing its M and N results on WIPDATAVALUE and not the SLOT that is why I get the wrong output. I have also tried to use the WITH Statement and it works well but unfortunately, our system cant accept subquery factoring.
    I know you guys will be helping out because you are all awesome. Thanks everyone
    Edited by: 1001275 on Apr 19, 2013 8:07 PM
    Edited by: 1001275 on Apr 19, 2013 8:18 PM

    Hi,
    Sorry, it's still not clear what you want.
    Are you saying that, given this table:
    CREATE TABLE trprinting
      WIPDATAVALUE       VARCHAR2(255)
    , SLOT               VARCHAR2(255)
    INSERT INTO trprinting (wipdatavalue, slot) VALUES ('1-2',  'TRALTEST43S1');
    INSERT INTO trprinting (wipdatavalue, slot) VALUES ('3',    'TRALTEST43S1');
    INSERT INTO trprinting (wipdatavalue, slot) VALUES ('4-6',  'TRALTEST43S2');
    INSERT INTO trprinting (wipdatavalue, slot) VALUES ('7',    'TRALTEST43S2');you want to produce this output:
    WIPDATAVALUE SLOT                     N          M
    1-2          TRALTEST43S1             1          3
    1-2          TRALTEST43S1             2          3
    3            TRALTEST43S1             3          3
    4-6          TRALTEST43S2             1          4
    4-6          TRALTEST43S2             2          4
    4-6          TRALTEST43S2             3          4
    7            TRALTEST43S2             4          4? If so, here's one way:
    WITH     got_numbers     AS
         SELECT     wipdatavalue
         ,     slot
         ,     TO_NUMBER ( SUBSTR ( wipdatavalue
                               , 1
                           , INSTR ( wipdatavalue || '-'
                                ) - 1
                     )          AS low_number
         ,     TO_NUMBER ( SUBSTR ( wipdatavalue
                               , 1 + INSTR ( wipdatavalue
                     )          AS high_number
         FROM     trprinting
    SELECT       wipdatavalue
    ,       slot
    ,       ROW_NUMBER () OVER ( PARTITION BY  slot
                                 ORDER BY          low_number
                        )                    AS n
    ,       COUNT (*)     OVER ( PARTITION BY  slot )     AS m
    FROM       got_numbers
    CONNECT BY     LEVEL               <= high_number + 1 - low_number
         AND     low_number          = PRIOR low_number
         AND     PRIOR SYS_GUID ()      IS NOT NULL
    ORDER BY  low_number
    ,            n
    ;Much of the complexity here is caused by storing 2 numbers in 1 VARCHAR2 column, wipdatavalue. Relational databases work best when there is no more than 1 item in any given column of any given row. This is so basic to datbase design that it is called First Normal Form. Also, numbers belong in NUMBER columns, not VARCHAR2. If you stored your data like that in the fist place, then you wouldn't need the sub-query I called got_numbers, which is about 60% of the code above. (That could be reduced by replacing SUSTR and INSTR with the less efficient REGEGP_SUBSTR.)

  • Hw to find total number of records

    Hi All,
    Can anyone help from these
    1. how to find total number of reports for a particular cube/ods... need step - step solution
    2. how to find total number of records for a particular Cube and ODS and Aggr's to till date.
    3.what is sandbox,mirror sys,instance of a sys..?
    4.what r TWS(Tivoli Workload Scheduler) jobs? how these r different to standard schedulers?
    Thanks in ADv
    Linda

    Hello Linda,
    As you have lots of answers on first 2 so i'll start from 3rd onward.
    3. Sandbox is mostly practice system where you can do all kind of R&D, mirror sys can be mirror image of any system depends on the organization and instance of system is again mirror image of one system.
    4. TWS is third party tool for scheduling which doesn't come along with SAP like standard scheduler as TWS has been prepared specially for this purpose so it has some more features than standard.
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  • Old   How to find total number of events in an event list?

    Hi,
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    Not the length of EVENT LISTS i want to find the Length of EVENTS.
    Thanks,
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    Hi,
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    call function 'REUSE_ALV_FIELDCATALOG_MERGE'
    EXPORTING
       I_PROGRAM_NAME               = sy-repid
       I_INTERNAL_TABNAME           = 'ITAB'
       I_INCLNAME                   = sy-repid
      changing
        ct_fieldcat                  = IT_FIELDCAT
    EXCEPTIONS
       INCONSISTENT_INTERFACE       = 1
       PROGRAM_ERROR                = 2
       OTHERS                       = 3
    if sy-subrc <> 0.
    MESSAGE ID SY-MSGID TYPE SY-MSGTY NUMBER SY-MSGNO
             WITH SY-MSGV1 SY-MSGV2 SY-MSGV3 SY-MSGV4.
    endif
    now describe your fieldcat . and find no of columns.
    and their order also..
    regards
    vijay

  • Need to Find Total number of InfoPart form in our Web application

    Hello,
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    The above will list all of the files. Do you need counts by library, by site or other?
    Mike Smith TechTrainingNotes.blogspot.com

  • How to find total number of records in a BDoc?

    Dear all,
    I have replicated about BP 1088 records from ISU into CRM system with block size 100. Technically on SMW01, for each successfully processed BDoc, there will be 100 records (corresponds to 100 block size). But due to some failed BDocs, not all "successfully" BDocs will have 100 records each, some may have only 1 record inside...or 30...or 88 for example. So, may i know how to find or is there a report i can look into to find the total number of records clearly shown for each of the successfully processed green status BDocs???
    Please help and points will be rewards!!
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    CK

    I am just showing this to show how to get the rowcount along with the cursor, if the program has so much gap of between verifying the count(*) and opening the cursor.
    Justin actually covered this, he said, oracle has to spend some resources to build this functionality. As it is not most often required, it does not makes much sence to see it as a built-in feature. However, if we must see the rowcount when we open the cursor, here is a way, but it is little bit expensive.
    SQL> create table emp_crap as select * from emp where 1 = 2;
    Table created.
    SQL> declare
      2   v_cnt     number := 0;
      3   zero_rows         exception;
      4  begin
      5    for rec in (select * from (select rownum rn, e.ename from emp_crap e) order by 1 desc)
      6     loop
      7        if v_cnt = 0 then
      8           v_cnt := rec.rn;
      9        end if;
    10     end loop;
    11     if v_cnt = 0 then
    12        raise zero_rows;
    13     end if;
    14   exception
    15    when zero_rows then
    16      dbms_output.put_line('No rows');
    17   end;
    18  /
    No rows
    PL/SQL procedure successfully completed.
    -- Now, let us use the table, which has the data
    SQL> declare
      2   v_cnt     number := 0;
      3   zero_rows         exception;
      4  begin
      5    for rec in (select * from
      6          (select rownum rn, e.ename from emp e)
      7          order by 1 desc)
      8     loop
      9        if v_cnt = 0 then
    10           v_cnt := rec.rn;
    11           dbms_output.put_line(v_cnt);
    12        end if;
    13     end loop;
    14     if v_cnt = 0 then
    15        raise zero_rows;
    16     end if;
    17   exception
    18    when zero_rows then
    19      dbms_output.put_line('No rows');
    20   end;
    21  /
    14
    PL/SQL procedure successfully completed.Thx,
    Sri

  • To find total number of files inside folder

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    Hi Kiruthika,
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    Robert

  • To find total number of pages in XML publisher

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    Exactly what my requirement is- In my template i have set header and footer page setup as "different first page'" and also printing the user text at the bottom of the last page using <?start@last-page-first:body?> <?end body?>.When i run the template, for multiple pages output user text is displaying at the bottom of the last page that is working fine but for single page output user text is not displaying at the bottom of the page because of the ""different first page'" header and footer.If you have any idea please suggest.

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    Hi,
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    refer below screen shot:
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  • How can I view the total number of songs in my iTunes?

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