Histogram - the size of the interval?

Similar Questions

• Partition wise joined possible with partitions of the interval?

  Hello

  I want to know the score wise join (NTC) is possible with interval partitioning - I can't find an explicit statement that he isn't, but I can't make it work - I did a simple test case to illustrate the issue.

  below, I have 2 create table scripts - 1 for the case of interval and 1 for the case of hash - I then a simple query on these 2 objects which should produce a NTC.

  In the case of hash, it works very well (see screenshot 2nd with a set of slaves), the first screenshot shows the case of the interval where I find myself with 2 sets of slaves and no NTC.

  No idea if this is possible and I just missed something?

  (for the test case choose the names of schema/storage appropriate for your system)

  Oh and version (I almost forgot... :-))-East 11.2.0.4.1 SLES 11)

  See you soon,.

  Rich

  -case interval

  CREATE TABLE 'SB_DWH_IN '. "' TEST1 '.

  TABLESPACE "SB_DWH_INTEGRATION".

  PARTITION BY RANGE ("OBJECT_ID") INTERVAL (10000)

  (PARTITION 'LESS_THAN_ZERO' VALUES LESS THAN (0) TABLESPACE "SB_DWH_INTEGRATION")

  in select * from DBA_OBJECTS where object_id is not null;

  CREATE TABLE 'SB_DWH_IN '. "" TEST2 ".

  TABLESPACE "SB_DWH_INTEGRATION".

  PARTITION BY RANGE ("OBJECT_ID") INTERVAL (10000)

  (PARTITION 'LESS_THAN_ZERO' VALUES LESS THAN (0) TABLESPACE "SB_DWH_INTEGRATION")

  in select * from DBA_OBJECTS where object_id is not null;

  -case of hash

  CREATE TABLE 'SB_DWH_IN '. "' TEST1 '.

  TABLESPACE "SB_DWH_INTEGRATION".

  8 partitions PARTITION OF HASH ("OBJECT_ID")

  store in ("SB_DWH_INTEGRATION")

  in select * from DBA_OBJECTS where object_id is not null;

  CREATE TABLE 'SB_DWH_IN '. "" TEST2 ".

  TABLESPACE "SB_DWH_INTEGRATION".

  8 partitions PARTITION OF HASH ("OBJECT_ID")

  store in ("SB_DWH_INTEGRATION")

  in select * from DBA_OBJECTS where object_id is not null;

  -query to run

  Select / * + PARALLEL(TEST2,8) PARALLEL(TEST1,8) * / *.

  of 'SB_DWH_IN '. "" TEST2 ","SB_DWH_IN ". "' TEST1 '.

  where TEST1.object_id = test2.object_id

  

  

  It is planned and a consequence of the estimate of the number of parallel slaves.

  To the parallel 41 each slave made 3 passes (i.e. sleeves 3 partitions).

  Add a partition (by table), and a set of slaves will have to manage a 4th pass: the cost of the query using NTC would increase from 33 percent even if the modification of the data is less than 0.8%.

  I guess that in the production Oracle distributes your lines of 1 M for a hash join.

  Because the decision is encrypted, it is possible that a very extreme tilt in partition in the table sizes billion line might overthrow the optimizer in a non - NTC join - but I have not tested that.

  If you want to force the plan John Watson suggestion for a hint of pq_distribute is relevant.  To cover all the bases and call your tables SMALL and LARGE

  /*+

  leading (FAT kid)

  USE_HASH (large)

  no_swap_join_inputs (large)

  PQ_DISTRIBUTE (wide none none)

  */

  If it's legal, that should do it.

  Concerning

  Jonathan Lewis

• CUIC statement limits - the interval length

  Hello

  Customer a CCX 11.0 deployed using the resident Co cuic for reporting.

  We had several cases where the absence of fields available in the reports of stock is frustrating customer engineer and deployment (me).

  For example, the length of the interval field is not widely available.  So, we can not report on an interval of a day and length of interval of 30 minutes when it is desired.  Simple counting of the number of calls directly to a CCX trigger cannot be done for intervals of 30 minutes.

  Are there licenses cuic that allows custom reports where we could take a report on stocks and add a field to it, and then download on CUIC CCX.    I thought that maybe the stand-alone CUIC option would be appropriate but as I read this is only available for UCCE.

  Any guidance is appreciated.

  Thank you

  Eddie

  Are there licenses cuic that allows custom reports where we could take a report on stocks and add a field to it, and then download on CUIC CCX.    I thought that maybe the stand-alone CUIC option would be appropriate but as I read this is only available for UCCE.

  You can certainly do it for UCCX as well. Deploy a standalone cuic server and install Premium license on it and after that, you can export a report on stocks of CCX CUIC, import in standalone cuic modifications required in terms of fields, report definition etc. and then import it to CCX CUIC. This thing is mentioned in below document so it is for UCCX:

  http://www.Cisco.com/c/en/us/TD/docs/voice_ip_comm/cust_contact/contact _...

  Concerning

  Deepak

• Impact of setting the interval partitioning off and then on again

  Hi all

  Have discovered that to split the partitions of a partitioned table Beach-interval, in that we need to define the partitioning interval all issue the command split then on again.

  e, g.

  ALTER TABLE retailer_transaction SET INTERVAL ();

  ALTER TABLE POR2_retailer_transaction

  SPLIT PARTITION PART_400012 AT (TO_DATE (' 2020-09-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIAN '))

  INTO (PARTITION PART_202008,

  PART_400012 PARTITION)

  ALTER TABLE retailer_transaction SET INTERVAL (NUMTOYMINTERVAL(1,'MONTH'))

  Everything we need to pay attention to that?

  Using oracle 11.2.0.3

  Thank you

  You can NOT split the LAST (most HIGH_VALUE) partition of a partitioned table interval interval. You can only separate partitions BEFORE the last. Which converts also all partitions from the bottom (split and below) for the partitions of the RANGE.

  If interval partitioning will only apply to data added above the transition point (the score higher.

  To return to your original question. To split the LAST interval follow the procedure described in your original post

  1. range partitioning

  2 split the partitioning

  3 set the partitioning interval on

  Conclusion with this however when adding data back to a nine month doesn't automatically create the partition again.

  This "new month" MUST BE above the point of passage for a new partition to create. In your code example, this point of transition is for the year 4000.

  This works very well for me:

  ALTER TABLE sales PARTITION RENAME SYS_P1122 to SALE_400012

  ALTER TABLE sale SET INTERVAL ();

  ALTER TABLE SPLIT PARTITION for sale

  SALE_400012 TO (TO_DATE('01/02/2020','DD/MM/YYYY')) IN

  (

  SALE_202001 PARTITION,

  SALE_400012 PARTITION

  );

  ALTER TABLE sale SET INTERVAL (NUMTOYMINTERVAL(1,'MONTH'))

  INSERT INTO sales

  (SALES_DATE, STORE_NUM, QTY_SOLD)

  values (to_date('31/12/4001','DD/,MM/YYYY'), 1, 10);

  This last INSERT statement above creates a NEW partition because the date (31/12/4001) is above the transition point.

  Looks like you want to move the transition point downwards. That is - you were using this year there date of 4000 as a 'catch all '.

  1. set the partitioning interval

  2 split that last partition into two NEW partitions (new names) by using a VALUE that is high enough to contain ALL the data you have in this last partition.

  3. remove the last partition - it should now be empty

  4. turn on range partitioning

  Now, any data that you insert above this NEW point of transition (the limit that you used for the new partition) it will create a new partition of the interval. Make sure the border is really an end of month, if that's what you want.

  If your data sample had this

  INSERT INTO sales

  (SALES_DATE, STORE_NUM, QTY_SOLD)

  values (to_date('05/10/2016','DD/,MM/YYYY'), 1, 10);

  INSERT INTO sales

  (SALES_DATE, STORE_NUM, QTY_SOLD)

  values (to_date('31/12/4000','DD/,MM/YYYY'), 1, 10);

  So for the step #2 above use split date 06/2016 so that the line above for 10/05/2106 goes to the new partition and 12/31/4000 stays in the old.

  Then, drag this old and lines in it (31/12/4000).

  You can NOT split the last partition if interval is always on because it is the last partition interval.

  And as long as you have given for 12/31/4000, you will get no new partitions unless you insert was more TOP than this.

  So if you want partitioning interval to activate again 06/2016 that higher data would be deleted and the score he falls.

  You have discovered only among "traps" with the help of partitioning of the interval. Oracle will happily create new partition needed to keep the data. If ANYONE, even accidentally, between a date as 12/31/4000 that nobody would ever know until the new data seem to DISAPPEAR for some use cases.

  This is because all data will start to enter this new partition ACCIDENTALLY created instead of a partition for the month, it's part.

  And, and you have discovered, you cannot split this partition directly.

  Delayed segment creation can allow you to pre-create the partitions ahead without actually affecting the segment for them until the first row is inserted.

  Interval partitioning will create the appropriate partitions but not if someone accidentally or intentionally, data will be a valuable capitalized in the table.

• dbms_scheduler.create_window fails with ORA-27418: syntax error in the interval of recurrence or calendar

  I tried to create a window to test the activation of plans of RM. But I get this error.

  SQL > EXECUTE dbms_scheduler.create_window.

  > (-)

  > window_name = > 'day ', -.

  > resource_plan = > 'my_plan ', -.

  > start_date = > systimestamp + 1 /(24*60).

  > repeat_interval = > ' freq = > daily '.

  > time = > ' 0 01:00:00 '.

  > Comments = > ' C know if it works without a job "-"

  >);

  BEGIN dbms_scheduler.create_window (window_name = > 'day', resource_p)

  LAN = > 'my_plan', start_date = > systimestamp + 1 /(24*60), repeat_interv

  Al = > ' freq = > daily ', duration = > ' 0 01:00:00 ', comments = > ' C so she

  works without employment '); END;

  *

  ERROR on line 1:

  ORA-27481: 'SYS '. "' DAY ' has an invalid schedule

  ORA-27418: syntax error in the interval of recurrence or calendar

  ORA-06512: at "SYS." DBMS_ISCHED", line 486

  ORA-06512: at "SYS." DBMS_SCHEDULER', line 1295

  ORA-06512: at line 1

  I have tried many tips found on the web, nothing works.

  Any help would be greatly appreciated.

  Also, can you pls tell me if this command run once automatically activates the my_plan after 1 minute, 1 minute, after which, it will disable the my_plan? So, you do not need to create a JOB for this. If you create a window, that is, as a job creation to enable an RM plan?

  2734369 wrote:

  Wow, discovered the issue: the repeat interval is broken (I use a database of 12.1 in Windows 7). It must be uppercase. Very surprising in the windows to be case-sensitive.

  SQL >

  SQL > EXECUTE dbms_scheduler.create_window.

  >  (-

  > window_name-online 'day ', -.

  > resource_plan-online 'my_plan ', -.

  > start_date-online systimestamp + 1 /(24*60).

  > repeat_interval-online ' FREQ = DAILY; BYHOUR = 10; BYMINUTE = 30', -.

  > time-online ' 0 01:00:00 '.

  > Comments => ' C know if it works without a job "-"

  >  );

  PL/SQL procedure successfully completed.

  Problem is, after 1 minute, always that the plan is not activated?

  SQL > SEE THE RESOURCE_MANAGER_PLAN PARAMETER

  VALUE OF TYPE NAME

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

  resource_manager_plan chain

  Why the Plan is not activated after 1 minute. Have we not create a job (I don't think).

  I think you misunderstand the meaning of "start_date".  Please see the documentation on what this means exactly.

  This attribute specifies the first date and time where it is scheduled to open in this window. If the value start_date specified is in the past or is not specified, the window opens, from its inception.

  To repeat windows using an expression of the calendar to specify the repetition interval, the value of start_date is a reference date. The first time the window opens depends on the specified repeat interval and the value of start_date .

  (From Types and Oracle® packages of PL/SQL Database 11g Release 2 (11.2) reference E40758-03); dbms_scheduler, create_window.  DBMS_SCHEDULER

  Once you have done this, you will see that it is pretty useless to specify that you have.

  Your repeat_interval said "every day at 10:30.

• Query for the partitions of the interval, system and reference

  Hi, I want to ask some of range partitions, system and reference information.

  In the partition of the interval, the script looks like this

  CREATE TABLE interval_sales

  (prod_id NUMBER (6))

  cust_id NUMBER

  time_id DATE

  )

  PARTITION OF RANGE (time_id)

  INTERVAL (NUMTOYMINTERVAL (1, 'MONTH'))

  (PARTITION p0 VALUES LESS THAN (TO_DATE (' 01/01/2005 ',' DD-MM-YYYY "")));

  I would like to write a query to find the expression of the interval "MONTHS" but could not find what he makes.

  In the reference partition, the script looks like this

  CREATE TABLE child_tab)

  ID NUMBER NOT NULL,

  parent_tab_id NUMBER NOT NULL,

  code VARCHAR2 (10),

  Description VARCHAR2 (50).

  CREATED_DATE DATE,

  CONSTRAINT child_tab_pk PRIMARY KEY (id),

  CONSTRAINT child_parent_tab_fk FOREIGN KEY (parent_tab_id)

  REFERENCES parent_tab (id)

  )

  PARTITION BY REFERENCE (child_parent_tab_fk);

  I want to write a query to find the foreign key name 'child_parent_tab_fk' after the PARTITION BY REFERENCE to the last line, but have no luck so far.

  In the end, I want to achieve this goal:

  If (interval_partitioning_type)

  Assign the interval expression to a string variable

  Another yew (reference_partitioning_type)

  Set the foreign key to a variable of type of foreign key

  Is there a way to get the type of partitioning? (interval and reference here)

  Please advise,

  Jack

  You can query ALL_PART_TABLES

  REF_PTN_CONSTRAINT_NAME	VARCHAR2(30)		Name of the referential partitioning constraint for partitioned tables reference
  INTERVAL	VARCHAR2(1000)		The value of the interval chain

• Partitions of the interval

  Hi guys,.

  I use DBMS_REDEFINITION to partition tables not partitioned in order to facilitate better performance. My question is; These tables already have clues about them. I read scores interval may not work on nested index tables. I need advice here. Can I used partitions of the interval on tables with indexes on the?

  by default, oracle creates a table as an array of heap and stores the rows in its blocks without a special order. On a heap table (which can be partitioned), you can create indexes. An index organized table (IOT) is basically an index without an additional table of the segment (but you can also define a segment of overflow, but it is another story). There are some cases of use of IOT, but in most situations, a heap table is quite sufficient.

  By the way: here's an example for the redefinition of a table with partitions of the interval: http://uhesse.com/2010/02/15/partitioning-a-table-online-with-dbms_redefinition/

• How to create the INTERVAL partitioning using EM?

  Hi guys

  I'm still not able to use the INTERVAL partitioning using EM. Is it possible to do?

  Oracle 11 g 2 on Linux

  Thank you

  John

  Hi John,.

  I don't think it's possible, that the option of partition of the range seems to have things that make you think of it would generate the definition as the partitioning interval but it doesn't.

  I'm on 12.1.0.3 of the SGC and 12.1.0.5 DB plugin so I'm sure it's 'missing' features for the moment.

  See you soon,.

  Rich

• Change the interval between slides in the carousel of bootstrap

  How can I change the interval between slides in the carousel "bootstrap"?  He is currently at 6 seconds, and I wish it were more than 15-30 seconds.

  Add below AFTER the link to the file bootstrap.js

• What is the interval of time between the virtual machine Linux and ESXi host synchronization?

  I have ESX5.0 with a centos6 installed on this linux virtual machine.

  After the installation of vmware for Linux virtual machine tools, I activated the time synchronization between the virtual machine linux and ESXi host through vmware-toolbox-cmd.

  I wonder what is the interval of time between the virtual machine linux and ESXi host synchronization? If I quickly time OS system, when the OS system time will sync back to the ESXi host?

  According to http://www.vmware.com/files/pdf/techpaper/Timekeeping-In-VirtualMachines.pdf:

  By default, the demon checks the guest only once per minute operating system clock.

  However, depending on whether the guest is behind or in front of the result is different. So, please read the section "periodic synchronization" (at least) in the document mentioned above.

  André

• Using the interval function works well for the end of the month

  
  I need to be able to get 1 year, 2 years, etc and also 1 month 2 months ago for each date in the calendar.  I can use the function of the INTERVAL for the most part, but it does not work if there is 1 month (year) would produce an invalid date.

  For example, the monitoring generates an error.

  Select TO_DATE (February 29, 2012 ')- INTERVAL of '1' YEAR  FROM DUAL; 

  Select TO_DATE('31-MAR-2011') - MONTH INTERVAL '1' FROM DUAL;

  Is there a way to get chosen to return on February 28, 2011?

  Thank you for your help.

  Sandy

  Hi, Sandy,

  ADD_MONTHs always returns a valid DATE.

  ADD_MONTHS (TO_DATE ((29 février 2012 ', «DD-MON-YYYY»), 12 * n))

  will return on February 29, n years after 2012, if this year is a leap year and February 28 this year, if it is not a leap year.

  In general

  ADD_MONTHS (d, m)

  Returns the date months after DATE m d.  If d is located near the end of his month and months m months in the future is for several days, then ADD_MONTHS returns the last day of that month.  By example, if d is on 31 March and 1 m, then ADD_MONTHS returns the last day in April (1 month after March), because there is no day 31 in April.

• Range of partitioning of the interval

  Hello
  
  
  Using Oracle 11.20.3
  We want to have a table partitioned composite range-hash, a partition for each month.
  
  However, want to esnure are still a tehrfore partition would use range interval partitioning.
  
  Have a few questions, if I say 201301 to 201312 but get intellectualization rosw partition with a value date in June 2014 it create
  a new partition for each month IE 201401, 201402 etc. or just a 201406?
  
  Wanyt esnure system can automatically dial with a scenario
  
  Daty1 have partitions 201301 to 201312
  
  Day 2 get record for 201406 (June 2014)
  
  Day 3 get record for 201403 (March 2014)
  
  
  We have no control over the date range, we were able to get but don't want to tpo have to manually create partitions BOF tehse all dates in advance - want system she managed automatically.
  
  Also you can just rename it the names of system partition generated to something more useful later.

  Yes, the creation of partitions of the interval is 'rare' - a given partition will be created only when necessary. (This should be easy to try, by the way).

• assign the interval

  BANNER
  Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit Production
  PL/SQL Release 11.2.0.3.0 - Production
  CORE     11.2.0.3.0     Production
  TNS for HPUX: Version 11.2.0.3.0 - Production
  NLSRTL Version 11.2.0.3.0 - Production

  is it possible to assign a data type of a table variable interval?
  

  CREATE TABLE mytable
  AS
     (SELECT '4' num_of_days FROM DUAL);
  
  DECLARE
     myinteverval   INTERVAL DAY TO SECOND;
  BEGIN
     myinterval := INTERVAL '4' DAY;  --happy
  
     SELECT num_of_days INTO myinteverval FROM mytable;
  
     myinteverval := interval myinteverval day; -- not happy
  end;

  Hello

  Pollywog wrote:

  BANNER
  Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - 64bit Production
  PL/SQL Release 11.2.0.3.0 - Production
  CORE     11.2.0.3.0     Production
  TNS for HPUX: Version 11.2.0.3.0 - Production
  NLSRTL Version 11.2.0.3.0 - Production
  

  is it possible to assign a data type of a table variable interval?

  CREATE TABLE mytable
  AS
  (SELECT '4' num_of_days FROM DUAL);
  
  DECLARE
  myinteverval   INTERVAL DAY TO SECOND;
  BEGIN
  myinterval := INTERVAL '4' DAY;  --happy
  

  What is the reason why he is happy

  INTERVAL '4' DAY
  

  a perfectly formed INTERVAL is literal. By itself

  '4'
  

  would be a VARCHAR2 interval, but is not in itself. immediately before the first quotation mark is the INTERVAL keyword, which means that the expression is a literal INTERVAL, and a literal INTERVAL must contain a certain number in single quotes; No matter what kind of expression.

  SELECT num_of_days INTO myinteverval FROM mytable;
  
  myinteverval := interval myinteverval day; -- not happy
  end;
  

  Use the NUMTODSINTERVAL function:

  myinterval := NUMTODSINTERVAL (4, 'DAY');
  

• SQL error: ORA-01873: the interval leading precision is too small

  Hello
  
  My requirement is to get the current timestamp in the format microsec Epoch.
  
  I am trying to execute the following query
  
  Select (CAST (((current_timestamp-TO_DATE (1er janvier 1970 00:00:00 ', ' 'JJ-MM-AAAA HH24:MI:SS)) * 24 * 60 * 60 * 1000000) AS varchar (32))) as "dt" from dual;))
  
  I get the following error
  Error from the 1 in the command line:
  Select (CAST (((current_timestamp-TO_DATE (1er janvier 1970 00:00:00 ', ' 'JJ-MM-AAAA HH24:MI:SS)) * 24 * 60 * 60 * 1000000) AS varchar (32))) as double 'dt'))
  Error report:
  SQL error: ORA-01873: the interval leading precision is too small
  01873 00000 - "the interval leading precision is too small."
  * Cause: The precision of the interval leading is too small to store the
  specified interval.
  * Action: Increase the accuracy of foreground of the interval or specify a
  meantime, with smaller leading precision.
  
  But if I tried to do
  Select (CAST (((sysdate-TO_DATE (1er janvier 1970 00:00:00 ', ' 'JJ-MM-AAAA HH24:MI:SS)) * 24 * 60 * 60 * 1000000) AS varchar (32))) as "dt" from dual;))
  works fine, but I get the bad epochtime sysdate has no time zone information.
  
  Please help me how to do the first query to work so that I can get the current_timestamp change to gmt and convert it in time.
  Please let me know if there is another way to get the Epochtime.

  Depending on whether you want to or not to include the fraction of a second use:

  with t as (
             select  current_timestamp at time zone 'gmt' - to_timestamp_tz('01/01/1970 00:00:00 GMT','MM-DD-YYYY HH24:MI:SS TZR') interval_diff,
                     cast(current_timestamp at time zone 'gmt' as date) - to_date('01/01/1970 00:00:00', 'MM-DD-YYYY HH24:MI:SS') num_diff
               from  dual
            )
  select  cast(
               (
                (
                 (
                  extract(day from interval_diff) * 24 + extract(hour from interval_diff)
                 ) * 60 + extract(minute from interval_diff)
                ) * 60 + extract(second from interval_diff)
               ) * 1000000
               AS varchar(32)
              ) as "ts",
          cast(num_diff * 24 * 60 * 60 * 1000000 AS varchar(32)) as "dt"
    from t
  / 
  
  ts                   dt
  -------------------- --------------------
  1336237158093000     1336237158000000
  
  SQL>  
  

  SY.

• overall index using partitioning of the interval.

  Can we have an index that is partitioned across the world using range partitioning on the number field in the database table.

  >
  Can we have an index that is partitioned across the world using range partitioning on the number field in the database table.
  >
  No - a global index cannot be paritioned of the interval.
  You can create a range partitioned or Hash-Partitioned Global index, but it cannot use partitioning interval. The index may be on a table is partitioned apart, but the index itself cannot be partitioned apart.

  See Creation of Partitions in the VLDB and partitioning Guide database
  http://docs.Oracle.com/CD/E18283_01/server.112/e16541/part_admin001.htm#i1006455

  and the CREATE INDEX SQL language doc section
  http://docs.Oracle.com/CD/E11882_01/server.112/e26088/statements_5012.htm#i2062403

• find the interval of time between the generation of logs archives of v$ log_history

  Hi all:
  
  I want to find the interval of generation of newspaper archives of v$ log_history,.
  
  SEQUENCE # FIRST_TIME
  11349 10/17/2009 12:08:58 AM
  11350 10/17/2009 12:13:10 AM
  11351 10/17/2009 12:15:46 AM
  11352 10/17/2009 12:18:32 AM
  11353 10/17/2009 12:21:32 AM
  11354 10/17/2009 12:23:06 AM
  11355 10/17/2009 12:24:59 AM
  11356 10/17/2009 12:26:10 AM
  11357 10/17/2009 12:27:18 AM
  11358 10/17/2009 12:34:34 AM
  11359 10/17/2009 12:37:43 AM
  11360 10/17/2009 12:48:36 AM
  11361 10/17/2009 12:52:10 AM
  
  Add a column to display the time difference (in minutes) between 11350 and 11349. It is actually 5 minutes, even between 11351 and 11350 nearly 2 minutes.
  
  Kind regards
  
  Hassan

  select sequence#,
         to_char(first_time, 'DD/MM/YYYY HH:MI:SS AM'),
         (first_time - lag(first_time) over (order by first_time)) * (24*60) as mins
  from v$log_history