SQL : Select clause

Hi all..
In open sql, is it impossible..?
  Select a~*
    into table gt_mara
    from mara as a inner join marc as b
    ~~~
For performance, I want to fetch the fields of table mara. But, while compiling, occur error..
How to do..? What to do..? tell me plz..
Thank you!!
regards..

Hi Kil,
Just an addition to the prev post..
Avoid the "into corresponding fields" for performance reasons... as long as the fields in the internal table are in the same order, use the below clause..
Select *
into table gt_mara
from mara as a inner join marc as b
ON amatnr = bmatnr.
Rgds,
Prash.

Similar Messages

  • SQL Query : Getting specific datatype in select clause

    Hi
    I want to write one query that returns me the values inside a table based on the datatype of column(s) of the table. For e.g. I need only those columns of the table in select clause where the datatype of the column is VARCHAR.
    Can you please help me on this?

    Use the "user_tab_columns" Data Dictionary to get the columns and build the SQL Dynamically.
    Thanks,
    Karthick.

  • In a SQL statement, the SELECT clause is used to

    In a SQL statement, the SELECT clause is used to select
    (a) columns
    (b) rows
    (c) tables
    (d) none of the above
    can any one help Immediately

    Is used to select rows of specified column...
    SELECT column_name(s) FROM table_name

  • Decimal Separator in SELECT Clause

    Hi
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    SQL> select value
    2 from v$nls_parameters
    3 where parameter = 'NLS_NUMERIC_CHARACTERS';
    VALUE
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    SQL> select 10/100 from dual;
    10/100
    ,1
    But if I use a decimal separator in the SELECT clause I get:
    SQL> select 100 * 1,1 from dual;
    100*1 1
    100 1
    It doesn't work. But using a period as the decimal separator works:
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    110
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    Hi,Néstor,
    user594312 wrote:
    ... I thought that the numeric format applied to the sql results and also the numbers that you used in the sql clauses.No; it applies to results, and it can affect implicit conversions, but it doesn't apply to SQL code.
    The period (or dot, '.') is always the decimal separator in numeric literals. There is no way to change that.
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    If you really wanted to use comma as the decimal separator, you could have to use strings, not numbers, and that could be a lot less efficient.
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    SELECT  100 * TO_NUMBER ('1,1')    -- This assumes your NLS settings are correct
    FROM    dual;Of course, efficiency won't be an issue when you're selecting 1 row from dual.

  • Help with SQL MODEL Clause

    I have the privilege of performing a very tedious task.
    We have some home grown regular expressions in our company. I now need to expand these regular expressions.
    Samples:
    a = 0-3
    b = Null, 0, 1
    Expression: Meaning
    1:5: 1,2,3,4,5
    1a: 10, 11, 12, 13
    1b: 1, 10, 11
    1[2,3]ab: 120, 1200, 1201, ....
    It get's even more inetersting because there is a possibility of 1[2,3]a.ab
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    Any help will be greatly appreciated.
    CODE:
    Reference:          http://www.sqlsnippets.com/en/topic-11663.html
    Objective:          Expand a range with ITERATE
    WITH t AS
    (SELECT '2:4' pt
    FROM DUAL
    UNION ALL
    SELECT '6:9' pt
    FROM DUAL)
    SELECT pt AS code_expression
    -- , KEY
    -- , min_key
    -- , max_key
    , m_1 AS code
    FROM t
    MODEL
    PARTITION BY (pt)
    DIMENSION BY ( 0 AS KEY )
    MEASURES (
                        0 AS m_1,
                        TO_NUMBER(SUBSTR(pt, 1, INSTR(pt, ':') - 1)) AS min_key,
                        TO_NUMBER(SUBSTR(pt, INSTR(pt, ':') + 1)) AS max_key               
    RULES
    -- UPSERT
    ITERATE (100000) UNTIL ( ITERATION_NUMBER = max_key[0] - min_key[0] )
    m_1[ITERATION_NUMBER] = min_key[0] + ITERATION_NUMBER
    ORDER BY pt, m_1
    Explanation:
    Line numbers are based on the assupmtion that "WITH t AS" starts at line 5.
    If you need detailed information regarding the MODEL clause please refer to
    the Refrence site stated above or read some documentation.
    Partition-
    Line 18:     PARTITION BY (pt)
                   This will make sure that each "KEY" will start at 0 for each value of pt.
    Dimension-
    Line 19:     DIMENSION BY ( 0 AS KEY )     
                   This is necessary for the refrences max_key[0], and min_key[0] to work.
    Measures-
    Line 21:      0 AS m_1
                   A space holder for new values.
    Line 22:     TO_NUMBER(SUBSTR(pt, 1, INSTR(pt, ':') - 1)) AS min_key
                   The result is '1' for '1:5'.
    Line 23:     TO_NUMBER(SUBSTR(pt, INSTR(pt, ':') + 1)) AS max_key                                        
                   The result is '5' for '1:5'.
    Rules-
    Line 26:     UPSERT
                   This makes it possible for new rows to be created.
    Line 27:     ITERATE (100000) UNTIL ( ITERATION_NUMBER = max_key[0] - min_key[0] )
                   This reads ITERATE 100000 times or UNTIL the ITERATION_NUMBER = max_key[0] - min_key[0]
                   which would be 4 for '1:5', but since the ITERATION_NUMBER starts at 0, whatever follows
                   is repaeted 5 times.
    Line 29:     m_1[ITERATION_NUMBER] = min_key[0] + ITERATION_NUMBER
                   m_1[ITERATION_NUMBER] means m_1[Value of Dimension KEY].
                   Thus for each row of KEY the m_1 is min_key[0] + ITERATION_NUMBER.
    Reference:          http://www.sqlsnippets.com/en/topic-11663.html
    Objective:          Expand a range using FOR
    WITH t AS
    (SELECT '2:4' pt
    FROM DUAL
    UNION ALL
    SELECT '6:9' pt
    FROM DUAL)
    , base AS
    SELECT pt AS code_expression
    , KEY AS code
    , min_key
    , max_key
         , my_increment
    , m_1
    FROM t
    MODEL
    PARTITION BY (pt)
    DIMENSION BY ( CAST(0 AS NUMBER) AS KEY )
    MEASURES (
                        CAST(NULL AS CHAR) AS m_1,
                        TO_NUMBER(SUBSTR(pt, 1, INSTR(pt, ':') - 1)) AS min_key,
                        TO_NUMBER(SUBSTR(pt, INSTR(pt, ':') + 1)) AS max_key,     
                        .1 AS my_increment     
    RULES
    -- UPSERT
              m_1[FOR KEY FROM min_key[0] TO max_key[0] INCREMENT 1] = 'Y'
    ORDER BY pt, KEY, m_1
    SELECT code_expression, code
    FROM base
    WHERE m_1 = 'Y'
    Explanation:
    Line numbers are based on the assupmtion that "WITH t AS" starts at line 5.
    If you need detailed information regarding the MODEL clause please refer to
    the Refrence site stated above or read some documentation.
    Partition-
    Line 21:     PARTITION BY (pt)
                   This will make sure that each "KEY" will start at 0 for each value of pt.
    Dimension-
    Line 22:     DIMENSION BY ( 0 AS KEY )     
                   This is necessary for the refrences max_key[0], and min_key[0] to work.
    Measures-
    Line 24:      CAST(NULL AS CHAR) AS m_1
                   A space holder for results.
    Line 25:     TO_NUMBER(SUBSTR(pt, 1, INSTR(pt, ':') - 1)) AS min_key
                   The result is '1' for '1:5'.
    Line 26:     TO_NUMBER(SUBSTR(pt, INSTR(pt, ':') + 1)) AS max_key                                        
                   The result is '5' for '1:5'.
    Line 27:     .1 AS my_increment     
                   The INCREMENT I would like to use.
    Rules-
    Line 30:     UPSERT
                   This makes it possible for new rows to be created.
                   However seems like it is not necessary.
    Line 32:     m_1[FOR KEY FROM min_key[0] TO max_key[0] INCREMENT 1] = 'Y'
                   Where the KE value is between min_key[0] and max_key[0] set the value of m_1 to 'Y'
    */

    Of course, you can accomplish the same thing without MODEL using an Integer Series Generator like this.
    create table t ( min_val number, max_val number, increment_size number );
    insert into t values ( 2, 3, 0.1 );
    insert into t values ( 1.02, 1.08, 0.02 );
    commit;
    create table integer_table as
      select rownum - 1 as n from all_objects where rownum <= 100 ;
    select
      min_val ,
      increment_size ,
      min_val + (increment_size * n) as val
    from t, integer_table
    where
      n between 0 and ((max_val - min_val)/increment_size)
    order by 3
       MIN_VAL INCREMENT_SIZE        VAL
          1.02            .02       1.02
          1.02            .02       1.04
          1.02            .02       1.06
          1.02            .02       1.08
             2             .1          2
             2             .1        2.1
             2             .1        2.2
             2             .1        2.3
             2             .1        2.4
             2             .1        2.5
             2             .1        2.6
             2             .1        2.7
             2             .1        2.8
             2             .1        2.9
             2             .1          3
    15 rows selected.--
    Joe Fuda
    http://www.sqlsnippets.com/

  • Need to know the column names in my dynamic select clause

    Dear All,
    Please go through the following code. While executing the following code i am getting an error saying that dbms_sql.describe_columns overflow, col_name_len=35. Use describe_columns2.
    Please guide me how to proceed further. Or please help me, how can i get the column names when i issue a dynamic select clause.
    DECLARE
    CUR INTEGER;
    COL_CNT INTEGER ;
    A INTEGER;
    SEL_CLAUSE VARCHAR2(2000);
    DESC_T DBMS_SQL.DESC_TAB;
    REC DBMS_SQL.DESC_REC;
    b number;
    BEGIN
    SEL_CLAUSE := 'SELECT 1,2,DECODE(1,1,''ONE'',2,''TWO'',3,''THREE'') FROM DUAL';
    --'SELECT ROWID,PARA_SUB_CODE,DECODE('||''''||'ENG'||''''||','||''''||'ENG'||''''||',PARA_NAME,PARA_BL_NAME),NULL,NULL FROM PCOM_APP_PARAMETER';
    --'SELECT 1,2,DECODE(1,1,''ONE'',2,''TWO'') FROM DUAL';
    DBMS_OUTPUT.PUT_LINE( SEL_CLAUSE );
    CUR := DBMS_SQL.OPEN_CURSOR;
    DBMS_SQL.PARSE(CUR,SEL_CLAUSE,DBMS_SQL.NATIVE);
    DBMS_SQL.DESCRIBE_COLUMNS(CUR,COL_CNT,DESC_T);
    B := desc_t.first;
    FOR J IN 1..COL_CNT
    LOOP
    DBMS_OUTPUT.PUT_LINE('J := '||J || ' COL CNT ' || COL_CNT);
    END LOOP;
    BEGIN
    A := DBMS_SQL.EXECUTE(CUR);
    EXCEPTION
    WHEN OTHERS THEN
    NULL;
    END;
    END;
    Regards,
    Balaji

    Is there any way can i have it directly??It does not work with static SQL either. Dynamic SQL is no different.
    SQL> select 1 x from dual where x = 1 ;
    select 1 x from dual where x = 1
    ERROR at line 1:
    ORA-00904: "X": invalid identifier
    SQL>As suggested already, you will need to use alias if you want your select expression to be referred in the where clause.
    SQL> BEGIN
      2      FOR rec IN (SELECT *
      3                  FROM   (SELECT 1,
      4                                 2,
      5                                 DECODE(1, 1, 'ONE', 2, 'TWO', 3, 'THREE') "DECODE(1, 1, 'ONE', 2, 'TWO', "
      6                          FROM   DUAL)
      7                  WHERE  "DECODE(1, 1, 'ONE', 2, 'TWO', " = 'ONE')
      8      LOOP
      9          NULL;
    10      END LOOP;
    11  END;
    12  /
    PL/SQL procedure successfully completed.
    SQL>Message was edited by:
    Kamal Kishore

  • How can I set a variable number of values in a SQL IN clause?

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    Raj

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    regards
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  • &param lexical variable in SELECT clause

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  • Merge can't accept a variable in the select clause?

    oracle 10.2
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    Apparently a merge can't handle variables in the select clause. I get an error. I can get it to work when I hard code a value.
    this is ridiculous...

    merge can't handle variables in the select clauseCare to prove?
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      2   v_first_name varchar2(20) := 'BOSS';
      3  BEGIN
      4  MERGE INTO sun_employees se
      5  USING (SELECT * FROM employees WHERE department_id = 20) e
      6  ON (e.employee_id = se.employee_id)
      7  WHEN MATCHED THEN
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      9  WHEN NOT MATCHED THEN
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    13  /
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  • Using a select clause in the column formula in an Analysis

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    i.e : add more option to sql here u can write SQL query in that column
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  • How to use a function in select clause

    hi gems...good evening...
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      2  /
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      2  /
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      2  /
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      3  is
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      5  begin
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      7
      8  for i in 1..tab_rec.count
      9              loop
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    11                       t_val.EXTEND;
    12                    t_val(i) := obj_typ_bankers(tab_rec(i).businesses * 12.50);
    13              end loop;
    14              return t_val;
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    16  /
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      4                                             ) )
      5  /
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          62.5
           150
           225
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    tab_rec value (2)12
    tab_rec value (3)18Hope this will help you... :)
    Thanks!
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  • How to remove column name in select clause

    Hello Guys,
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    Hi,
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    FROM       scott.dept
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            10 NEW YORK
            40 BOSTON
            20 DALLAS
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    Edited by: Frank Kulash on Feb 26, 2013 10:10 AM
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  • Maximum no. of columns allowed in SELECT clause - Urgent please

    Hi,
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    ORA-01467 sort key too longIt's not the SELECT clause that causes this, it's most likely the GROUP BY clause. Basically, the columns in the GROUP BY clause have to fit comfortably within a single database block. Does that sound like it might be a problem with your query?
    Cheers, APC
    Message was edited by:
    APC

  • Select clause with where part is needs 600 times longer to complete

    Hello.
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    Edited by: 993896 on 14.3.2013 7:24

    Hi,
    welcome to the forum.
    Please read SQL and PL/SQL FAQ
    If you have a performance issue have a look at SQL and PL/SQL FAQ
    Additionally when you put some code or output please enclose it between two lines starting with {noformat}{noformat}
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    {noformat}{noformat}
    SELECT ...
    {noformat}{noformat}
    Regards.
    Al                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               

  • Problems with query with more than 20 values in the select clause

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