NULL values in partition Table of Partition key

Hi all

I'm using Oracle 11.2.0.3.

I can add a null value on partition of Partition table key.

If so, how?

Kind regards.

You cannot specify NULL in the VALUES clause.

Referring to clauses PARTITION you set - not the data that you insert.

If you cannot define a range partition that uses the VALUES LESS THAN and use null to this division. But you CAN use NULL values in the VALUES of integration clause.

DROP TABLE mytable;

CREATE TABLE MyTable

(

CREATION_DATE DATE,

Identification number

)

PARTITION BY RANGE (CREATION_DATE)

(

PARTITION prior_to_2016 VALUES LESS THAN (TO_DATE ('2016-01-01', 'YYYY-MM-DD'));

LESS THAN PARTITION of VALUES undesirable (MAXVALUE)

)

Insert into myTable values (null, null)

You can use a VALUES clause to insert NULL values - this line will go into the MAXVALUE partition.

Tags: Database

Similar Questions

How to find null values in the table
Hi experts,

my version is oracledb10g:

I am doubted simple.

for example:

create table ex1 (e varchar2 (20));

Insert ex1 (null).

insert into ex1 (e)

number of will be: 2

I can easily find the value "e".

so my question is:

How can I find null values in the table and I want to remove the null values from the table.
ADF 7 wrote:
Hi experts,

my version is oracledb10g:

I am doubted simple.

for example:

create table ex1 (e varchar2 (20));

Insert ex1 (null).

insert into ex1 (e)

number of will be: 2

It depends on what you expect. If you issue count (*) you ask Oracle "how many lines there. If you issue a County () you ask Oracle 'how non null values for are there. "

ADF 7 wrote:

I can easily find the value "e".

so my question is:

How can I find null values in the table and I want to remove the null values from the table.
```
delete from 
where  is null;
```
Ideally you would simply place a NOT NULL constraint on the column in question, thus prohibiting NULL values into the system (assuming that your business needs, it is that the column MUST always have a value).

Number of rows in each partition is displayed with a NULL value for a table partitioned in user_tab_partitions. Why?

I created a table and partitioned on the date of the entry and added a local partitioned index.
Now, I use a query to extract "num_rows" of user_tab_partitions to know the number of rows in each partition.
Getting this value as null num_rows, wonder why?
After looking to explain the Plan after interrogation ("select * from my_table1 where entry_date = 1 January 2015" ;))
to find out if she actually partitioned table and its data in different partitions, I interpreted in effect because the query plan had a line like Partition_range (Single).
My Question is:
(a) is actually partitioned data (have I misinterpreted the Explain plan)
(b) why is the num_rows null column in the query (Pasted below)
(c) also in addition what difference it would have been if I had created a Global Index instead of the Local Index in my case?
The following code Snippet:
----------------------------------------------------------------------------------------------
create the table my_table1
(
roll_no number constraint my_table1_pk primary key,
date of entry_date
)
partition of range (entry_date)
(
PARTITION data_p1 VALUES LESS THAN (TO_DATE (December 31, 2014 ',' DD-MM-YYYY ""));
PARTITION data_p2 VALUES LESS THAN (MAXVALUE)
);

create an index only my_table1_indx on my_table1 (entry_date) local;
----------------------------------------------------------------------------------------------
I now insert two lines:
insert into my_table1 values (1, to_date ('01-01-2015', ' dd-mm-yyyy'));
insert into my_table1 values (2, to_date('01-02-2015','dd-mm-yyyy'));
----------------------------------------------------------------------------------------------
These have been inserted successfully, now using the query below shows num_rows column as null. I don't know why?
SELECT table_name, num_rows, high_value, nom_partition
Of user_tab_partitions
where table_name = 'MY_TABLE1 '.
ORDER BY table_name, nom_partition;
----------------------------------------------------------------------------------------------

(a) is actually partitioned data (have I misinterpreted the Explain plan)

Yes, it is partitioned. You can query this particular partition SELECT * FROM my_table1 (data_p1) PARTITION to check that.

(b) why is the num_rows null column in the query (Pasted below)

As already mentioned that you have not collected statistics.

(c) also in addition what difference it would have been if I had created a Global Index instead of the Local Index in my case?

In fact, you have created two types of indexes without knowing (can be)! One is not partitioned (although this column is not partition key) and another is partitioned (LOCAL). They are MY_TABLE1_PK and MY_TABLE1_INDX. You can check that USER_INDEXES.

You can read this article to get an early jump on the partitioning of decision. Partition: Partition decisions

NPE for rows with null values in the table of the adv
I created the table adv, 2 columns, equipped with two text fields. Initially the col2 is empty,
say that the structure is as follows:
```
              col1          col2
Hdr         Question    Answer
row1       abc           zxc
row2       pdd           null (user doesnt enter anything)
row3       zxc            lkj

APPLYBUTTON
```
When the user clicks on apply button I want to update these values in the table.

I don't know how to do that, but that's what I'm trying to do. Please guide me the right way if its wrong.

in LICS, I took the APPLYBUTTON action, got all lines of the object corresponding to the view, consulted with everyrow and I'm checking if there are values in the fields in the table, if the value zero, are simply print null. (we would not update on this line). . I could get o the part update, due to null pointer exception
I get on the row line null pointer exception [1].getAttribute("Answer") for row2, it prints the value of row1 however.

Here is the code I wrote:
Set of rows from the rowset = (RowSet) vo.getRowSet ();
Rank rank [] = vo.getAllRowsInRange ();
System.out.println ("row.length" + row.length);
for (int i = 0; i < row.length; i ++) {}
String question = row.getAttribute("Question").toString ();
String response = row [i].getAttribute("Answer").toString ();
If (question! = null & &!("".)) Equals (question. Trim()))
{
System.out.println ("Question line" + i + "-" + question);
}
on the other
{
System.out.println ("Question line" + i + "is null");
}

If (response! = null & &!("".)) Equals (Answer.Trim ()))
{
System.out.println ("Response line" + i + "-" + response);
}
on the other
{
System.out.println ("Response line" + i + "is null");
}

Why the NPE survey code, I'm checking for null value and you want to print the value null in the logfile... I can not to getAttribute() on the null field, why can not just give me null in the variable I'm assigning to?

You can guide me what approach to take, if it is not correct.
I must examine DataDictionary and collect data there?
Thanks in advance.
Hello

handle the exception of null on these two statements like this pointer

String question = "";
String response = "";

Try
{
question = row.getAttribute("Question").toString ();
}
catch (NullPointerException e)
{
question = "";
}

Try
{
answer = row.getAttribute("Answer").toString ();
}
catch (NullPointerException e)
{
response = "";
}

Si ( !("". Equals (question. Trim()))
{
System.out.println ("Question line" + i + "-" + question);
}
on the other
{
System.out.println ("Question line" + i + "is null");
}

Si ( !("". Equals (Answer.Trim ()))
{
System.out.println ("Response line" + i + "-" + response);
}
on the other
{
System.out.println ("Response line" + i + "is null");
}

Thank you
Gerard

Return of NULL values in the table of results
OK the guru! Your genius has helped me before (see this post... ( Help to return the query results table ) but now I would return null results 0 to the number. Below is the structure of the table (again) and my desired result (new question):

The structure of the table is:
Year of sport
Baseball 10-11
Football 10-11
Football 11-12
Football 11-12
Football 11-12
Baseball 11-12

The result should be sports in each year and should be grouped by year but shown as columns so:
(GROUP OF the year)
1 hockey, ice 0, 1, Soccer baseball
2, 1, Football Soccer 1, Baseball

The new portion of this question is the "0, Soccer.

Thanks for any help!
Hello

For counties of 0, you must join of a table that has one row per sport.
If you do in fact have such a table, you can build a game on the fly that will work just as well of results:
```
WITH     all_sports     AS
(
     SELECT DISTINCT  sport
     FROM              t
)
,     sport_year_totals     AS
(
     SELECT       s.sport
     ,       t.year
     ,       COUNT (t.sport)     AS cnt
     FROM             all_sports  s
     LEFT OUTER JOIN                t  PARTITION BY (year)
                                      ON  s.sport  = t.sport
     GROUP BY  s.sport
     ,            t.year
)
SELECT   year
,       LISTAGG ( cnt || ', ' || sport
            , ' '
           ) WITHIN GROUP (ORDER BY  sport)     AS sports
FROM      sport_year_totals
GROUP BY  year
ORDER BY  year
;
```
However, you may find it useful to have a table with one line per sport really. In this case, you wouldn't need the first auxiliary request. Simply use your name of the table where I used all_sports.

I copied the example of data that MichaelS posted in your previous thread {message identifier: = 10388879} below.
You should always post sample data yourself when you ask a question, especially if everything you have to do is copy and paste from another thread.
```
CREATE TABLE t as (
 select 'Baseball' sport, '10-11' year from dual union all
 select 'Football', '10-11' from dual union all
 select 'Football', '11-12' from dual union all
 select 'Football', '11-12' from dual union all
 select 'Soccer',   '11-12' from dual union all
 select 'Baseball', '11-12' from dual
);
```

Conversion of NULL values to 0 in graphics

Hello again everyone,
I use Oracle Business Intelligence 11.1.1.7.131017.
In our chart, we have 2 lines with made cumulative information, but whenever this information is equal to 0, it will not draw the line, only for items > 0.
I tried to change the data format to: #0;-#; 0 but it does not work...
Although the table indicates 0 where NULL values are supposed to be, when he tries to draw the graph, the line appears only for values > 0...
Is it possible to convert NULL values to 0 in graphics?
Thanks a lot for all the help and have a nice day!
Kind regards
Frederico Barracha.

Hi Fred,.

It's not because you have NULL values in the table. This is because there is no entry in the warehouse.

For example, you wise date data for March 1 to March 5. in this March 4 isn't any data in the table. That is to say, there is no entry for March 4. In this case, you can see graphs broken. You can follow one of the below approach.

Combine a similar request and add the column date dimension of time where all date entries are there and all the other Null columns. It will be visible from the graph in uninterrupted.

Another thing to do is cross join with a date column so that the data will come for all dates.

creating table of partition with null values
Hello

I have a table with 20 GB of data. I changed this table in the table to partition with partition of the range
the column with which I took for range (date), holding the values null

can I have a partition for null values?
Hello

CREATE TABLE ICBS. KNET_STMT_EXTRACT_NEW_P
(
SRVC_TYPE VARCHAR2 (200 BYTE),
FILLER_1 VARCHAR2 (200 BYTE),
RTE_STAT VARCHAR2 (200 BYTE),
TRAN_DATE VARCHAR2 (200 BYTE),
TRAN_TIME VARCHAR2 (200 BYTE),
POST_DATE VARCHAR2 (200 BYTE),
DATE OF REC_INSERT_DT,
DATE OF PROCESSED_DATE,
KNET_RECON_RESSON VARCHAR2 (100 BYTE),
NEW_TRAN_DATE VARCHAR2 (10 BYTE),
REASON VARCHAR2 (200 BYTE)
REASON_IND NUMBER (5),
NODE_ID VARCHAR2 (8 BYTE),
PROGRAM_ID VARCHAR2 (8 BYTE),
USER_ID VARCHAR2 (8 BYTE),
TIME_STAMP DATE
)
ICBS TABLESPACE
PCTUSED 0
PCTFREE 10
INITRANS 1
MAXTRANS 255
STORAGE)
64K INITIALS
MINEXTENTS 1
MAXEXTENTS UNLIMITED
PCTINCREASE 0
DEFAULT USER_TABLES
)
LOGGING
PARTITION BY RANGE (REC_INSERT_DT)
(

KNET_STMT_EXTRACT_NEW_P_DEC11 PARTITION VALUES LESS THAN (TO_DATE (' 2012-01-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIAN '))
LOGGING
NOCOMPRESS
ICBS TABLESPACE
PCTFREE 10
INITRANS 1
MAXTRANS 255
STORAGE)
INITIAL 1 M
MINEXTENTS 1
MAXEXTENTS UNLIMITED
DEFAULT USER_TABLES
),
KNET_STMT_EXTRACT_NEW_P_JAN12 PARTITION VALUES LESS THAN (TO_DATE (' 2012-02-01 00:00:00 ',' SYYYY-MM-DD HH24:MI:SS ',' NLS_CALENDAR = GREGORIAN '))
LOGGING
NOCOMPRESS
ICBS TABLESPACE
PCTFREE 10
INITRANS 1
MAXTRANS 255
STORAGE)
INITIAL 1 M
MINEXTENTS 1
MAXEXTENTS UNLIMITED
DEFAULT USER_TABLES
),
PARTITION KNET_STMT_EXTRACT_NEW_P_MVAL VALUES LESS THAN (MAXVALUE)
LOGGING
NOCOMPRESS
TABLESPACE TBL_MAXVALUE
PCTFREE 10
INITRANS 1
MAXTRANS 255
STORAGE)
INITIAL 1 M
MINEXTENTS 1
MAXEXTENTS UNLIMITED
DEFAULT USER_TABLES
)
NOCOMPRESS
NOCACHE
NOPARALLEL
MONITORING
ALLOW THE MOVEMENT OF THE LINE;

Published by: hitgon on April 30, 2012 10:56

Published by: hitgon on April 30, 2012 10:58

Published by: hitgon on April 30, 2012 10:59

TIMESTAMP (6) partitioned key-> range partitioned table ddl needed
What is the DDL TIMESTAMP syntax (6) partitioned key, range partitioned table

Published by: oracletune on January 11, 2013 10:26
>
What is the DDL TIMESTAMP syntax (6) partitioned key, range partitioned table
>
Don't know what you're asking. Are you asking how to create a table partitioned using a TIMESTAMP column (6) for the key?
```
CREATE TABLE TEST1
(
    USERID                 NUMBER,
    ENTRYCREATEDDATE     TIMESTAMP(6)
)
PARTITION BY RANGE (ENTRYCREATEDDATE) INTERVAL(NUMTOYMINTERVAL(1, 'MONTH'))
(
    PARTITION P0 VALUES LESS THAN (TO_DATE('1-1-2013', 'DD-MM-YYYY'))
)
```
See my answer posted: January 10, 2013 21:56 if you need to do on a TIMESTAMP with TIME ZONE column. You must add a virtual column.
Creating scores of auto range

What Oracle Table contains the name of the Partition key field?
What Oracle table/view retains the name of partition key field?

All_Tab_Partitions does not seem to keep this information.

When I use the toad-> pattern-> Tables-> Partitions, it lists the name of the key field of partition, that partition is based.

Thank you
all_part_key_columns
or
USER_part_key_columns

Published by: OrionNet on December 5, 2008 15:56

Partition key would exist in local index, if so it should be first or last or is not serious?

Hello

I'm trying to find the rule for:

-J' have a partitioned table every day and each partition has about 10 GB.

-J' need to keep the data in 3 months.

-I need to partition cutting happens.

I'm trying to find if I need to use the local index the partition key or not.

I am currently adding as the first column of the index.

In my research on the web and with colleagues, I find three different answers:

-It is not required to add local index the partition key.

-It is compulsory and must be the first position.
-It is necessary and should be the last position.

I share below the criation of table as an example:

CREATE TABLE MY_BIG_TABLE
(
  MY_SOURCE             VARCHAR2(50 CHAR)       NOT NULL,
  MY_FILENAME           VARCHAR2(255 CHAR)      NOT NULL,
  MY_DATE               DATE                    NOT NULL,
  MY_ID                 VARCHAR2(50 CHAR)       NOT NULL,
  MY_ORIG_GROUP         VARCHAR2(50 CHAR)       NOT NULL,
  MY_DEST_GROUP         VARCHAR2(50 CHAR)       NOT NULL,
  MY_DURATION           NUMBER,
  MY_NR_EVENT           NUMBER
)
TABLESPACE MY_BIG_TABLESPACE_1
PARTITION BY RANGE (MY_DATE)
INTERVAL (NUMTODSINTERVAL(1,'DAY'))
(  
  PARTITION P_FIRST VALUES LESS THAN (TO_DATE(' 2014-01-01 00:00:00', 'SYYYY-MM-DD HH24:MI:SS', 'NLS_CALENDAR=GREGORIAN'))
    LOGGING
    NOCOMPRESS 
    TABLESPACE MY_BIG_TABLESPACE_1
)
LOGGING 
NOCOMPRESS 
NOCACHE
NOPARALLEL
MONITORING
/
CREATE UNIQUE INDEX I_MY_BIG_TABLE
 ON MY_BIG_TABLE (
   MY_DATE,           
   MY_SOURCE,    
   MY_FILENAME,       
   MY_ID,             
   MY_ORIG_GROUP,     
   MY_DEST_GROUP     
 )
 TABLESPACE MY_BIG_IDX_TABLESPACE_1
 LOGGING
 NOCOMPRESS
 NOPARALLEL
 LOCAL
/

Best regards

Ricardo Tomas

Well, I don't like this example because I would tell you to use on your text string to_date and to_char on your column.

Let's say if you have a partition of the range with a value of one month of data in the partition and you want to select the data for a given day. Oracle must be able plum of the score, but then you have to complete the partition scanning unless there was an index containing the day (assuming it's the partition key)

Update of the partition key + compression on disk space
Hello

We have a table of partitoned.

With the help of 11.2.0.3

We need to update the partition key column because it has the wrong values.

I take it's all done it automatically - if we upadte the column, the data is moved into the appropriate automatically partitions.

Is this the case?

Also, m we want to create a tab that is compressed; the short title of disc spae.

How small is a pill against the table uncompressed on average?

Thank you
We have just in the research to update the small portion of these data.

Question if it's a small or large portion of data, you have no need to allow the movement of the line.
Yes, it takes a quick operation.

Compression data warehouse - no OLTP.

Outside the Exadata, there are two compression options for table - "fundamental" or "oltp.
http://allthingsoracle.com/compression-Oracle-basic-table-compression/

What is the index that is created automatically on the partitioning key?

CREATE TABLE sales
( prod_id NUMBER(6)
, cust_id NUMBER
, time_id DATE
, channel_id CHAR(1)
, promo_id NUMBER(6)
, quantity_sold NUMBER(3)
, amount_sold NUMBER(10,2)
)
STORAGE (INITIAL 100K NEXT 50K) LOGGING
PARTITION BY RANGE (time_id)
( PARTITION sales_q1_2006 VALUES LESS THAN (TO_DATE('01-APR-2006','dd-MON-yyyy'))
TABLESPACE tsa STORAGE (INITIAL 20K NEXT 10K)
, PARTITION sales_q2_2006 VALUES LESS THAN (TO_DATE('01-JUL-2006','dd-MON-yyyy'))
TABLESPACE tsb
, PARTITION sales_q3_2006 VALUES LESS THAN (TO_DATE('01-OCT-2006','dd-MON-yyyy'))
TABLESPACE tsc
, PARTITION sales_q4_2006 VALUES LESS THAN (TO_DATE('01-JAN-2007','dd-MON-yyyy'))
TABLESPACE tsd
)
ENABLE ROW MOVEMENT;

N ° you must create indexes for yourself.

Index BITMAP Advisor Advisor access SQL on the Partition key
I ran SQL and SQL Access Advisor as below he recommends index bitmap on p_key... Given that the cust table is partitioned on the p_key, does make sense to create indexes on p_key?

Not sure if SQL Access Advisor look just in ' where clause conditions/predicates "and the cardinality of such columns and not look at whether the table is partitioned or not. T it?

Is it wise to create BITMAP indexes on the partition key? If so what scenario would be beneficial?
```
SELECT * FROM cust
WHERE 
    'T' = 'T' AND 
    part_key IN 
    (   2, 3, 4 )) AND 
    (   p_key, act_key) IN ( select p_key, act_key from account where act_type = 'PENDING' and p_key in (2,3,4) )
```
user4529833 wrote:
Jonathan, I have exactly one value per partition for the partition key. However, most SQLs use 'IN' as predicate of partitioning pruning so overall stats are always used and they are always a bit bland compared to the partition level stats. Then this led SQL Access advisor recommendations?

Yes, the Counselor also recommended to create Bitmap indexes on act_type... Given this does make sense to have the Bitmap index on the partition key?

I'll post the execution plan as soon as I have access to the system...

The fact that statistics are "a little stale" was not much of a difference.

The fact that Oracle has probably used statistics at the level of the tables is likely to be the underlying issue. Is the partitioned table list or you have rigged it with partitioning range: If you faked list partitioning using partitioning of the range which MAY have contributed to the issue (but it's an assumption that I have not tested).

Unofficially, the optimizer has said something like:

+ "There are 25 possible values for pkey, there are 4 possible values for act_type, there seems to be 100 different combinations - so your query will pick up X lines. 80% of these lines will be packed in Y blocks and 20% of them will be scattered through the Z blocks if we have two bitmap index, and we will have to do sequential reading of the db file SSS. If we do an analysis through the affected partitions, we do MMM db file scattered reads. What are those cheaper. » +

You know how false it's - if you know that you need not the pkey index. You also know what is the efficiency of the column act_type is in the identification of data, so you can decide whether or not an index on act_key may be useful.

Concerning
Jonathan Lewis
http://jonathanlewis.WordPress.com
http://www.jlcomp.demon.co.UK

"The premature temptation to theories of shape on the lack of data is the scourge of our profession."
Sherlock Holmes (Sir Arthur Conan Doyle) in "the Valley of fear."

Add the constraint not NULL in the existing table that has null values

Hello
I want to add a constraint not null to and an existing table, but the table already contains values null in this column.
EMP
Emp_id name
1 axada
2
3 sdkdd
Here is already the data IE 2 empid is Null as name. I must add a fool of constraint not null which new values will not be null, but I don't want to change the data of exisitng alreadt which is null.

Hello

"The opposite": NOVALIDATE does not validate the data that is ALREADY in the table, but do not allow the insertion of a NULL value.

Have you tried my sample code?

CREATE TABLE MaTable (x NUMBER PRIMARY KEY, y NUMBER);

INSERT INTO myTable VALUES (1, 123);

INSERT INTO myTable VALUES ( 2, NULL );

INSERT INTO myTable VALUES (3, 456);

ALTER TABLE mytable MODIFY (y NOT NULL NOVALIDATE );

INSERT INTO myTable VALUES (4, 678);

INSERT INTO myTable VALUES ( 5, NULL );

SELECT * FROM MyTable;

'2' line was inserted with null before the creation of the NOT NULL constraint, this line remains "as what" at the end of the trial.

'5' line trying to insert a NULL value after creating the NOT NULL constraint, which is denied.

Best regards

Bruno.

Insert some values into multiple tables. One of the tables has a complex primary key.

Hello, I am using Jdeveloper version 11.1.2.3.0
I have a FORM_TYPE table and another table Str. The table of Str is used to translate the strings from the name of the table which are read from another program from the database and used to generate forms...
The FORM_TYPE table has a relationship with the table of STR via formtype.title_str_id = str.str_id.
STR. table has a primary key which is based on the mf_language and str_id. columns I generated the entity for the table of STR, but impossible to insert new values of add-on application for some reason any... It is said that recording invalidates its own entity...
I am trying to create a table that will be editable and will have the column Description, name, cdes in English, Str in French etc...
To create a Form_Type and at the same time insert new values for each language. But can not do the inserts of the new work of STR...
Also the FormType.TITLE_STR_ID already has a lot of null values in the database.
SELECT FormType.DESCRIPTION,
FormType.FORM_TYPE_ID,
FormType.NAME,
FormType.PAPER_REPORT_FORMAT,
FormType.PAPER_REPORT_NAME,
FormType.TITLE_STR_ID,
in. En_str str,
in. STR_ID,
en.object_name en_object_name,
Fr. Str,
Fr. STR_ID,
Fr.object_name,
of the. Str,
of the. STR_ID,
du.object_name,
BG.Str,
BG. STR_ID,
BG.object_name
OF FORM_TYPE FormType.
(select str.str_id, str.mf_language_id, str.str, str.object_name
of mf_language ml, str
where ml.mf_language_id = str.mf_language_id
and ml.code = 'fr') en
(select str.str_id, str.str, str.object_name
of mf_language ml, str
where ml.mf_language_id = str.mf_language_id
and ml.code = 'fr') en,.
(select str.str_id, str.str, str.object_name
of mf_language ml, str
where ml.mf_language_id = str.mf_language_id
and ml.code = 'from'),
(select str.str_id, str.str, str.object_name
of mf_language ml, str
where ml.mf_language_id = str.mf_language_id
and ml.code = 'bg') bg
WHERE
FormType.title_str_id = en.str_id and
FormType.title_str_id = fr.str_id and
FormType.title_str_id = du.str_id and
FormType.title_str_id = bg.str_id

Each table that you want to update must have its own entity object. You need to connect it to an another VO/OS using the associations/view links

NULL values in partition Table of Partition key

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