Similar Questions

• Reuse of blocks released upward in partitioned table

  Hello world

  I have a database creates a new partition and after a certain retention counter old partitions are deleted. This is done every day. We had some problem sometime before, so we disabled the partition drop script until it is revised for the partitions are not deleted for now. But we're going to delete all the files in the old partitions on a daily basis so that later, when we allow rear drop partition script it will not push the system.

  Now the problem is though that we delete the lines of the old partitions that continues to increase the size of the storage space, which means that it does not use the freed block. Please correct me if I'm wrong here, to my knowledge when the records in a table is removed the blocks going to the freelist for subsequent data inserts.

  The method above does not apply to the partitioned table is? If the space is released only if the partition is deleted? Thank you for you're your time and your answers. Very much appreciated.

  Database: Oracle Database 11 g Enterprise Edition Release 11.2.0.3.0

  OS: Oracle Solaris 10 SPARC 64 bit

  I have a database creates a new partition and after a certain retention counter old partitions are deleted. This is done every day. We had some problem sometime before, so we disabled the partition drop script until it is revised for the partitions are not deleted for now. But we're going to delete all the files in the old partitions on a daily basis so that later, when we allow rear drop partition script it will not push the system.

  If possible, you should use TRUNCATE so that high tide is reset. It will be much faster to delete lines.

  What PROBLEM you encounter who pushed you to 'disable' the drop?

  Now the problem is though that we delete the lines of the old partitions that continues to increase the size of the storage space, which means that it does not use the freed block. Please correct me if I'm wrong here, to my knowledge when the records in a table is removed the blocks going to the freelist for subsequent data inserts.

  Correct - but only "freelist" will NOT be used for removable direct-path.

  My guess is that your new data every day are INSERTED in BULK. If, then Oracle just formats new blocks and fills them directly and does not use the free list.

  Do you use inserts of direct-path for new data?

• partition table

  When we need to use the partition table?

  Partitioning defined

  The concept of divide and conquer has been around since the time of Sun Tzu (500 BC). Recognizing the wisdom of this concept, Oracle applied to management of large tables and indexes. Oracle has continued to evolve and refine its capabilities of partitioning since its first implementation of the Oracle 8 range partitioning. Oracle 8i and 9i, Oracle has continued to add features and new methods of partitioning. The current version of Oracle 9i Release 2 continues this tradition by adding new features to the list partitioning and partitioning method new range-list.

  When the partition

  There are two main reasons to use partitioning in an environment of VLDB (very large database). These reasons are related to improving the management and performance.

  Partitioning offers:

  • Management at the level of individual partition for data loads, creating indexes and reconstruction, and backup/restore. This may cause more quickly down, because only actively managed individual partitions are not available.
  • Performance of the queries increased by selecting only the relevant partitions. This process of weeding out the partitions that do not contain the data required by the application via a technique called the size of the partition.

  Use the partitioning:

  • When a table reaches a size "large". Large being defined according to your environment. Greater than 2 GB should always be considered for partitioning of tables.
  • What performance benefits outweigh the extra management issues related to partitioning.
  • When the archive data is on a calendar and repetitive. For example, data warehouses are usually given for a certain period of time (rolling window). Old data is then output to archive.

• Conversion of non partitioned in partitioned table

  
  Hi gurus,

  I need to convert the partition table not in the partition.  More flexible way is to use the DBMS_REDEFINITION package for this.
  I do not have access to run this package, when I asked the EXECUTE permission for my dev
  CUSTOMER rejected suggestion that
  "DBMS_REDEFINITION is a method very slow migration that has never been used before here for these migrations.
  so I do not recommend using it as it can trigger bugs and unexpected side effects.

  
  is this true?

  What will be the alternative method, I can go away?

  Please suggest

  S

  I don't think DBMS_REDEFINITION has bugs.  However, you (and the client) should be familiar with the steps involved.

  Other that that, you will need to create a partitioned table and insert data to the existing table.  You can speed up integration through parallel direct-path insert.

  You will also need to build indexes on the new (partitioned) table.  Define constraints if necessary.  Run grants to other patterns, if necessary.

  Hemant K Collette

• Try to convert the partitioned Table of interval in the range... Swap partition...

  Requirement:
  
  Interval of replacement partitioned Table by range partitioned Table
  

  DROP TABLE A;
  
  CREATE TABLE A
  (
     a              NUMBER,
     CreationDate   DATE
  )
  PARTITION BY RANGE (CreationDate)
     INTERVAL ( NUMTODSINTERVAL (30, 'DAY') )
     (PARTITION P_FIRST
         VALUES LESS THAN (TIMESTAMP ' 2001-01-01 00:00:00'));
  
  
  INSERT INTO A
       VALUES (1, SYSDATE);
  
  INSERT INTO A
       VALUES (1, SYSDATE - 30);
  
  INSERT INTO A
       VALUES (1, SYSDATE - 60);

  I need to change this partitioned Table apart to a partitioned range Table. I can do using the EXCHANGE PARTITION. Like if I use the classic method to create another table range partitioned, then:
  
  

  DROP TABLE A_Range
  CREATE TABLE A_Range
  (
  a NUMBER,
  CreationDate DATE
  )
  PARTITION BY RANGE (CreationDate)
     (partition MAX values less than (MAXVALUE));
  
  Insert  /*+ append */  into A_Range Select * from A; --This Step takes very very long..Trying to cut it short using Exchange Partition.

  Problems:
  
  I can't do
  

   ALTER TABLE A_Range
    EXCHANGE PARTITION MAX
    WITH TABLE A
    WITHOUT VALIDATION;
   

  ORA-14095: ALTER TABLE CHANGE requires a not partitioned table nonclustered
  This is because the tables are partitioned. So it does not allow me.
  
  If I instead:
  
  
  Create a table that is not partitioned for exchanging data by partition.
  

    Create Table A_Temp as Select * from A;
    
     ALTER TABLE A_Range
    EXCHANGE PARTITION MAX
    WITH TABLE A_TEMP
    WITHOUT VALIDATION;
     
    select count(*) from A_Range partition(MAX);
   

  -The problem is that all the data is in MAX Partition.
  Even after the creation of a large number of partitions by walls of separation, the data is still in MAX Partition only.
  
  So:
  
  -What we cannot replace a partitioned Table to the Table partitioned using the EXCHANGE PARTITION range interval. that is, we have to insert in...
  -We can do it, but I'm missing something here.
  -If all the data is in MAX Partition due to "WITHOUT VALIDATION", can say us be redistributed in the right type of range partitions.

  You must pre-create the partitions in a_range and then swap one for one for a tmp, and then to arange. With the help of your sample (thanks to proviing code, incidentally).

  SQL> CREATE TABLE A
    2  (
    3     a              NUMBER,
    4     CreationDate   DATE
    5  )
    6  PARTITION BY RANGE (CreationDate)
    7     INTERVAL ( NUMTODSINTERVAL (30, 'DAY') )
    8     (PARTITION P_FIRST
    9         VALUES LESS THAN (TIMESTAMP ' 2001-01-01 00:00:00'));
  
  Table created.
  
  SQL> INSERT INTO A VALUES (1, SYSDATE);
  
  1 row created.
  
  SQL> INSERT INTO A VALUES (1, SYSDATE - 30);
  
  1 row created.
  
  SQL> INSERT INTO A VALUES (1, SYSDATE - 60);
  
  1 row created.
  
  SQL> commit;
  
  Commit complete.
  

  You can find the form of existing partitions assistance:

  SQL> select table_name, partition_name, high_value
    2  from user_tab_partitions
    3  where table_name = 'A';
  
  TABLE_NAME PARTITION_NAME HIGH_VALUE
  ---------- -------------- --------------------------------------------------------------------------------
  A          P_FIRST        TO_DATE(' 2001-01-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIA
  A          SYS_P44        TO_DATE(' 2013-01-28 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIA
  A          SYS_P45        TO_DATE(' 2012-12-29 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIA
  A          SYS_P46        TO_DATE(' 2012-11-29 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIA
  

  You can then create the table a_range with apporopriate partitions. Note that you may need to create additional in a_range partitions because the partitioning interval does not create the partitions has no data for, even if that leaves 'holes' in the partitioning scheme. So, on that basis:

  SQL> CREATE TABLE A_Range (
    2     a NUMBER,
    3     CreationDate DATE)
    4  PARTITION BY RANGE (CreationDate)
    5     (partition Nov_2012 values less than (to_date('30-nov-2012', 'dd-mon-yyyy')),
    6      partition Dec_2012 values less than (to_date('31-dec-2012', 'dd-mon-yyyy')),
    7      partition Jan_2013 values less than (to_date('31-jan-2013', 'dd-mon-yyyy')),
    8      partition MAX values less than (MAXVALUE));
  
  Table created.
  

  Now, create a regular table to use in the constituencies:

  SQL> CREATE TABLE A_tmp (
    2     a              NUMBER,
    3     CreationDate   DATE);
  
  Table created.
  

  and all partitions in Exchange:

  SQL> ALTER TABLE A
    2    EXCHANGE PARTITION sys_p44
    3    WITH TABLE A_tmp;
  
  Table altered.
  
  SQL> ALTER TABLE A_Range
    2    EXCHANGE PARTITION jan_2013
    3    WITH TABLE A_tmp;
  
  Table altered.
  
  SQL> ALTER TABLE A
    2    EXCHANGE PARTITION sys_p45
    3    WITH TABLE A_tmp;
  
  Table altered.
  
  SQL> ALTER TABLE A_Range
    2    EXCHANGE PARTITION dec_2012
    3    WITH TABLE A_tmp;
  
  Table altered.
  
  SQL> ALTER TABLE A
    2    EXCHANGE PARTITION sys_p46
    3    WITH TABLE A_tmp;
  
  Table altered.
  
  SQL> ALTER TABLE A_Range
    2    EXCHANGE PARTITION nov_2012
    3    WITH TABLE A_tmp;
  
  Table altered.
  
  SQL> select * from a;
  
  no rows selected
  
  SQL> select * from a_range;
  
           A CREATIOND
  ---------- ---------
           1 23-NOV-12
           1 23-DEC-12
           1 22-JAN-13
  

  John

• migration in partitioned table

  Hello world
  
  I need to migrate on a two table range-list partitioned table, one of them is partitioned list, the other is not. The largest contains 400 million lines (partitioned), the smallest contains 90 million lines. The database cannot insert all of the table at a time, so I started to migrate the tables into pieces. I have two options:
  
  1 create pieces using a date_field (IDO_ID)
  2. create the pieces with the help of a generated field (MIGR_RANK) using the list field (TAB_KOD) and a number generated randomly, which limits the lines to migrate a line no more than 1 million of partitioning
  
  My questions are the following:
  1. is it possible that the insert is slow due to the large number (about 750) list partitions? Can I use the 2. method, inserting only one partition? Or a kind of simple is enough for partitioning columns? How does the insert works in Oracle for a partitioned table? Especially in a range-list partitioned table?
  2. How can I adjust the database to migrate two tables in one step (if possible)?
  
  Thanks in advance,
  Gabor
  
  Published by: csiszgab on Sep 28, 2012 07:47

  Whenever you provide post your Oracle version 4-digit (result of SELECT * FROM V$ VERSION).
  >
  I need to migrate on a two table range-list partitioned table, one of them is partitioned list, the other is not. The largest contains 400 million lines (partitioned), the smallest contains 90 million lines. The database cannot insert all of the table at a time, so I started to migrate the tables into pieces. I have two options:
  >
  -Oracle is certainly able to process the entire table in one so why do you say otherwise?

  You can speed up insert PARALLEL aid (for the source and target tables), using the direct-path load (APPEND indicator) and NOLOGGING mode on the target table.
  >
  1. is it possible that the insert is slow due to the large number (about 750) list partitions? Can I use the 2. method, inserting only one partition? Or a kind of simple is enough for partitioning columns? How does the insert works in Oracle for a partitioned table? Especially in a range-list partitioned table?
  >
  Your "2" is not just insert in a partition. You said your target table was partitioned ' range-list' but your source is only 'list' partitioned. Therefore a partition target: data from the "one" won't a single list.

  Load using the entire table at once without (other than the partition key) index on the target table.

• Disable LOGGING for partition table based on the automatic INTERVAL

  Hello
  
  I created a database table of paritioned range interval with NOLOGGING as the default attribute for the table as well as the tablespace definition.
  
  When ORACLE automatically creates a new partition for this table, it activates the LOGGING for this automatic partition even if the table definition says anything else.
  
  How can I go about changing this behavior? Is it still possible?
  
  FYI my platform is 11 GR 1 (11.1.0.7) material on SUN SOLARIS 10 SPARC.
  
  Any help will be appreciated.

  Right, so refer to a new feature of Oracle 11 g partitioning called 'partitioning interval. "

  +"+
  + The interval partitioning: A new strategy of partitioning in Oracle Database 11g, +.
  + Interval partitioning extends the functionality of the method range to define equipartitioned +.
  + ranges using a definition of the interval. Rather than specify individual +.
  + going to explicitly, Oracle will create any partition automatically as needed.
  + every time the data of a partition are inserted for the first time. Interval +.
  + a lot of partitioning improves maneuverability of a partitioned table. For +.
  example, + a partitioned table interval could be set so that Oracle creates a +.
  + new partition for each month in a calendar year; a partition is then automatically +.
  + created for "September 2007" as soon as the first record in this month is inserted.
  + in the database. +
  + The techniques available for a partitioned table interval are interval, interval-+.
  + List, range-Hash, and interval range. +
  +"+

  In order to disable interval partitioning on the transactions table, use:
  
  ALTER TABLE transactions SET INTERVAL ();
  

  http://download.Oracle.com/docs/CD/B28359_01/server.111/b32024/part_admin.htm

  Disable partitioning interval on this table, create a procedure to run every day we'll say, to create a new partition and chop the old partition of the appropriate table. I presume that the automatic partitioning interval creates the partition with the logging by default option without checking the parameters in the table. I have seen no information on this in the Oracle documentation.

  Hope that helps.

  Ogan

• Partitioned Tables and indexes

  Hello
  
  
  I have a question on the table and index partitioning. My scenario is:
  
  Charge 2 mio records in table T once a month. Loaded records are added to existing records, and once loaded data is never changed.
  At some point, I want to delete the older recordings, so I intend to this partition table.
  
  T table looks like:
  

  create table t (id       number(10) not null  constraint t_pk primary key,
                  period   number(10) not null,
                  contract number(10) not null,
                  attr     number(10) not null);
  
  create unique index t_ux1 on t(contract,period);
  
  create index t_ix2 on t(period);

  My plan is to partition T over the period, and I'm trying to read through the concepts
  http://download.Oracle.com/docs/CD/B19306_01/server.102/b14220/partconc.htm#g471747
  
  
  My question is now, how to manage the indexes, the t_pk, the t_ux1 and the t_ix2. Concepts of say,
  
  «1. If the table partitioning column is a subset of index keys, use a local index.»
  
  "2. If the index is unique, use a global index. If this is the case, you are finished. »
  
  
  So, that's how I read it
  -t_pk is unique, so this should be global
  -t_ux1 of columns is a subset, unless I have misunderstood (?), which should be local
  -index t_ix2 column is the same as the partitioning column, so it must be local
  
  Is this right, this t_ux1 should be a local partioned index, even if the period is the second column in the index?
  
  If true, what will happen when a partion fell?
  
  
  I am new in this area, so please feel the comment as you wish.
  
  
  Concerning
  Peter
  
  
  BANNER
  ----------------------------------------------------------------
  Oracle Database 10 g Enterprise Edition release 10.2.0.3.0 - 64bi
  PL/SQL version 10.2.0.3.0 - Production
  CORE Production 10.2.0.3.0
  AMT for IBM/AIX RISC System/6000: Version 10.2.0.3.0 - production
  NLSRTL Version 10.2.0.3.0 - Production

  Peter Gjelstrup wrote:

  My question is now, how to manage the indexes, the t_pk, the t_ux1 and the t_ix2. Concepts of say,

  «1. If the table partitioning column is a subset of index keys, use a local index.»

  "2. If the index is unique, use a global index. If this is the case, you are finished. »

  So, that's how I read it
  -t_pk is unique, so this should be global
  -t_ux1 of columns is a subset, unless I have misunderstood (?), which should be local
  -index t_ix2 column is the same as the partitioning column, so it must be local

  Is this right, this t_ux1 should be a local partioned index, even if the period is the second column in the index?

  A partitioned index locally can only be defined as unique if the partition key is part of the columns in the index. Imagine what the database would have to do if this is not the case: in order to verify if a newly added or updated value violates the uniqueness, it will have to travel all the partitions in a serialized operation - means that no one else could do the same thing at the same time. Since he is a killer of serious scalability in terms of locking and contention, this is not allowed.

  So: Your T_UX1 index can be defined as a unique index that is local because it contains the partition key. Although the index is not prefixed ("Prefix" means that it is divided by the left side of the columns in the index) which means that there may be access patterns where all partitions should be scanned or the optimizer cannot use a method of size of effective partition according to the way the index is reached.

  Your T_PK index cannot be set as local because it must be unique (you can not use a local non-unique index in this case), but does not contain your partition key. It must be a global index. An overall index can be partitioned as well (different from the underlying table) but it doesn't have to be.

  Depends on how you access your data you have not T_IX2 index when partitioning by this key because it corresponds to the partition key and therefore could not actually be used by the mechanism of partition pruning that limit your query to the scores of individuals.

  If you have more than one MAS environment where running queries are used longer, you should be fine with the index the in general (because they could be analyzed in parallel in parallel operations), but if you have an OLTP environment, then you should avoid local no prefix indexes due to the potential problem that you need to analyze all partitions.

  Be borne in mind that with partitioning adds an important layer of complexity to other areas: in particular the options available to the optimizer and analyze cost optimizer statistics. Depends on how you access your statistical data must be maintained on several levels now (level of score and at the global level, in the case of subpartitioning may be still at this level). If your data is important and you rely on "global" level statistics (these are always the case when the optimizer at the time analysis cannot limit access to a single partition) then in the pre - 11 g databases analyze these "global" level statistics can take a lot of time and resources, since actually , you need data several times (once for the partition and even global level).

  Presenting this partitioning may mean other potential problems in terms of execution that change (not for the better sometimes) plans and how to effectively collect statistics. Note that g 11 addresses the issue of 'statistics' by introducing the so-called "extra" global statistics. Greg Rahn wrote a [blog note | http://structureddata.org/2008/07/16/oracle-11g-incremental-global-statistics-on-partitioned-tables/] on this nice feature.

  >

  If true, what will happen when a partion fell?

  Since you're already on 10g, you can specify the database to update the scores of the local index using the UPDATE of the INDEX clause, while 9i could maintain only an overall index and it is up to you to rebuild the local index partitions after the partition DDL on the table (according to the DDL operation).

  Kind regards
  Randolf

  Oracle related blog stuff:
  http://Oracle-Randolf.blogspot.com/

  SQLTools ++ for Oracle (Open source Oracle GUI for Windows):
  http://www.sqltools-plusplus.org:7676 /.
  http://sourceforge.NET/projects/SQLT-pp/

  Published by: Randolf Geist on Sep 30, 2008 16:39

  Added statistics / optimizer warning when you use the partitioning

• How to encrypt a USB, when I get the message "(GPT) GUID Partition Table partitioning scheme is required? '."

  I just tried to encrypt a USB drive with my Mac book (latest version) and got the message "A GUID Partition Table (GPT) partitioning schΘma required." What does this mean and suggestions on how to proceed?

  The drive must be formatted for Mac. Start in the Recvery mode. Erase the drive using disk utility (select the top icon which represents the entire disk and click clear. Partition scheme must be Mac OS (Extended) first partition GUID, which is the default value.

  If you want to install a system on it follow these steps:

  How to reinstall OS X on your Mac - Apple Support

• Qosmio x 500 - 14W - invalid partition table

  Hay guys,

  I wanted to start my Toshiba Quosmio x 500 - 14W PQX33E (Product Code), but the only thing I found on my desk: invalid partition table.

  In another Forum I found a threat, which means I have to restart my computer. So I tried a Reboot Cd, but found nothing. Then I tried to call the Call Center of Toshiba, but they are closed...

  For me, is the Toshiba Service miserably and I hope that some guys from the Forum can help me.

  Thank you
  Paul

  Hello

  You must install the operating system once more.
  Invalid partition table means that something is wrong with the HARD drive and the operating system cannot start.
  So if you n t have a recovery disk, you must order one:
  http://backupmedia.Toshiba.EU/landing.aspx

  PS: I made good experience with the Toshiba service comparing the service with another laptop from the manufacturer I have to say that s OK and I meet a lot of people who said that other manufacturers are not really better
  Which; s my opinion

• Satellite L630-00V - need to partition table

  Hello to you guys...

  So I really screwed up here, I did my girlfriend buy a laptop (Toshiba Satellite L630) and also a new hard drive to replace the one on the inside because they have slow (5400 RPM)

  I've used ghost to clone the drive inside the laptop to the new one I mounted in a housing that is connected via USB to the computer.

  Although I double checked everything and that the destination drive was the good the bad thing happened, the blank disc has been cloned on the full.

  I don't think that all my data are deleted, because there were at least 70 GB of commands on the computer (3 recovery partitions and Windows 7 one) and I doubt enough that it can all be erased within 5 seconds.

  I used GetBackData for NTFS witch scanned the hard drive for about 05:30 (Yes, 5 hours and a half) and well found a lot of files... If right now I copy all partitions said based on another hard drive...

  I'll have to do the scan a 2nd time to get Fat32 partitions...

  Something that would be useful for me is to have the partition table of a released factory Toshiba L630-00v (500 GB HDD) *.

  
  
  
  

  Beginning and end clusters would be really useful, I would be able to rebuild my much easier partition table

  
  And oh, no idea how to get the OEM Windows 7 cd to a new facility? They were paying for a copy of windows with their computer, their must be a way to install it...

  Thank you very much! ;)

  Hello

  > And oh, no idea how to get the OEM Windows 7 cd to a new facility? They were paying for a copy of windows with their computer, their must be a way to install it.

  Now, you can get money from Toshiba Recovery disk.

  For European models:

  https://backupmedia.Toshiba.EU/landing.aspx

  American models:

  https://www.CSD.Toshiba.com/cgi-bin/TAIS/support/JSP/serviceUnitVerification.jsp?orderType=SFF

  > my girlfriend did buy a laptop (Toshiba Satellite L630) and also a new hard drive to replace the one on the inside because they have slow (5400 RPM)

  What happens if put you the old hard drive to make a recovery disk and replace it with a new one, then... Then you will be able to restore the factory settings on the new hard drive with a recovery disk. Or you have not previously burned?

• after image restore, backup partition table is different

  Hello

  He is a 2012 Macbook mid Pro with El Capitan and Bootcamp.

  After repeated with Bootcamp (Windows 10 Pro) problems I started using Image of Terabyte for Linux in order to safeguard this Macbook. Forums terabyte that people know that the Image of Linux (such as Image for Windows) is able to create a reliable backup, but also to restore - restore an entire disk image or restoring individual partitions.

  I followed their instructions precisely and activated the two options "restore first Beach" and "TPG hidden OS" (as one terabyte official, Member of the forum wrote) when restoring a partition.

  Unfortunately, it did not work! Bootcamp was no longer visible in the start screen or wasn't visible, but not start.

  What I found was that I had to keep the two options (see above) with DISABILITIES.

  Then I could restore some partition images - and 10 Windows in Bootcamp is bootable.

  Well, it was the long introduction.  I'm sorry.

  Now my question is:

  Is this something that has changed my partition table?

  Before restoring an Image for backup of Linux for the first time, I had this partition table:

  localhost: ~ root # fdisk/dev/disk0

  Disk: geometry / dev/disk0: 60801/255/63 [976773168 sectors]

  Signature: 0xAA55

  From end

  #: cyl hd s - cyl hd s id [Start - size]

  ------------------------------------------------------------------------

  1: EA 1023 254 63 - 1023 254 63 [1-409639]

  2: AF 1023 254 63 - 1023 254 63 [409640-857906448] HFS +.

  3: AB 1023 254 63 - 1023 254 63 Boot Darwin [858316088-1269536]

  * 04:07 1023 254 63-1023 254 63 [859586560-117186560] HPFS/QNX / TO THE

  BUT:

  After failed attempts to restore and work finally restore my partition table operation is different:

  Fdisk/dev/disk0

  Disk: geometry / dev/disk0: 60801/255/63 [976773168 sectors]

  Signature: 0xAA55

  From end

  #: cyl hd s - cyl hd s id [Start - size]

  ------------------------------------------------------------------------

  1: EA 0 0 2 - 1023 255 63 [1-409639]

  2: AF 25 102 15 - 1023 255 63 [409640-857906448] HFS +.

  3: AB 1023 0 57 - 1023 255 63 Boot Darwin [858316088-1269536]

  * 04:07 1023 199 8-1023 255 63 [859586560-117186560] HPFS/QNX / TO THE

  Everything seems to work fine, but I wonder if the different appearance of the partition table can be a problem?

  For example, compare:

  1: EA 1023 254 63 - 1023 254 63]

  3 o

  409639]

  TO

  1: EA 0 0 2 - 1023 255 [63 3 o 409639]

  It's the output of 'diskutil list ':

  localhost: ~ root # diskutil list

  / dev/disk0 (internal, physical):

  #: NAME SIZE TYPE IDENTIFIER

  0: GUID_partition_scheme * GB 500,1 disk0

  1: disk0s1 EFI EFI 209.7 MB

  2: Apple_HFS Macintosh HD 439,2 GB disk0s2

  3: disk0s3 Apple_Boot Recovery 650.0 MB HD

  4: Microsoft database BOOTCAMP 60.0 GB disk0s4

  And the output of 'TPG - r - vv show/dev/disk0':

  See the TPG: / dev/disk0: mediasize = 500107862016; SectorSize = 512; blocks = 976773168

  See the TPG: / dev/disk0: MBR suspicious to sector 0

  See the TPG: / dev/disk0: Pri GPT to sector 1

  See the TPG: / dev/disk0: GPT Sec at sector 976773167

  start index size summary

  0 1 MBR

  1 1 Pri GPT header

  2 32 table GPT Pri

  34 6

  40 409600 1 part TPG - C12A7328-F81F-11D2-BA4B-00A0C93EC93B

  409640 857906448 2 part TPG - 48465300-0000-11AA-AA11-00306543ECAC

  858316088 1269536 3 part TPG - 426F6F74-0000-11AA-AA11-00306543ECAC

  859585624 936

  859586560 117186560 4 part TPG - EBD0A0A2-B9E5-4433-87C0-68B6B72699C7

  976773120 15

  976773135 32 table dry GPT

  976773167 1 dry GPT header

  Out of 'gdisk/dev/disk0:

  localhost: ~ root # gdisk/dev/disk0

  GPT fdisk (gdisk) version 1.0.1

  WARNING: Open with shared lock devices will not have their

  table partition automatically reloaded!

  Scanning partition table:

  MBR: hybrid

  BSD: absent

  APM: absent

  TPG: present

  Found a valid GPT with hybrid MBR; using GPT.

  Command (? for help): p

  Disk/dev/disk0: 976773168 sectors, GiB 465.8

  Logical sector size: 512 bytes

  Identifier (GUID) of disc: 38A2E667-6E00-4C0F-BD78-C95979E3CFAE

  Partition table contains up to 128 entries

  First usable sector is 34, last usable area is 976773134

  The partitions will be aligned with the boundaries of sector 8

  Space is free total 957 sectors (478,5 KiB)

  Starting number (sector) end (sector), Code name of size

  409639 40 1 MiB EF00 EFI 200.0

  2 409640 858316087 409.1 giB AF00 Macintosh HD

  3 858316088 859585623 619.9 MiB AB00 Recovery HD

  4 859586560 976773119 55.9 giB 0700 BOOTCAMP

  Any ideas?

  Another question:

  Is it possible to save the table to partition like that?

  DD if = / dev/sda of = gptandmbr bs = 512 count = 34

  Thank you very much

  NEUMAC

  In Fdisk, the CHS numbers are arbitrary on modern disks. They are the time of BIOS and old disks with CHS geometry. Puts BCA from Mac 0-1023 (1024 cylinders), 0-255 (256 heads) and 0-63 (64 sectors). Modern disks far exceed those numbers and would drain them. Must match what is the triplet of start/end/size and it fits in GPT and MBR. You can put arbitrary 0,0,0 for CHS and you will see all issues on GPT disks.

  Please do not use TBIL to restore W10 to circa 1970-80 disks.

• invalid partition table

  invalid partition table is displayed on my screen, pl help out me...

  Exactly what happened before you saw the error message?

• After reinstalling Windows on Linux invalid partition table

  I got an operating system linux on my pc, then I tried to install windows along the side with linux, so I put my windows cd in the drive and follow the steps, the first steps were well made, but when the pc restarts, it is a message (invalid partition table). p.s. I formatted the c partition and it did not work, so I deleted and made a new partition in unallocated space and there do not have too much work, help?
  Another PS it is no os can I enter on my pc.

  If you deleted all the partitions of the hard disk using Linux, then you should be able to put your Windows install disk, boot from it and install Windows.  The installation procedure needs to configure your hard drive correctly for Windows.

  -JW

• Partition table / no entry of safe mode

  HERE'S MY QUESTION:

  I'M TRYING TO RECOVER MY XP DESKTOP COMPUTER. WHENEVER I TRY not to enter MY "SAFE MODE" VIA THE XP CD I GET NO PASSAGE. NORMALLY I CAN GO THROUGH LOG BUT I'VE NOW DAMAGED A PARTITION. I USED THE HELP FILE (FIXMBR) HERE IS THE MESSAGE I HAVE,

  * THIS COMPUTER SEEMS TO HAVE A NON STANDARD OR INVALID MASTER BOOT RECORD *.

  FIXMBR may damage your partition tables if you continue (of course I did). This could cause all of your scores on the current hard disk to become inaccessible.

  If you have problems to access your drive, DO NOT CONTINUE (of course I did)

  Are you sure you want to write a new MBR?       (of course I checked yes)

  Then... If I'm screwed?   I have to buy a new hard drive?   I can solve this problem?  Can I use the old as a ghost player to get my files?

  HELP ME OB 1... YOUR MY ONLY HOPE THOMAS

  Hi Thomas,

  ·         Are able to connect in normal mode?

  ·         What happens when you try to connect in safe mode?

  ·         You get the error message?

  I suggest you to run the disk check to check for possible errors.

  How to perform disk error checking in Windows XP

  http://support.Microsoft.com/kb/315265

  Note: If bad sectors are found in the hard drive, then it could try to fix this particular sector. If you have any data on that, it can get lost.