on the Clause type

Similar Questions

• SQL query help needed in the Clause type

  Hello

  I'm currently learning Clause type. How can I write in the clause type to get the result below:

  I have a table like this

  DEPTNO ENAME

  -------           ------

  20 SMITH

  CELINE 10

  BOND 20

  ALLEN 30

  WARD 30

  20 JONES

  30 MARTIN

  30 BLAKE

  10 CLARK

  SCOTT 20

  10 KING

  30 TURNER

  20 ADAMS

  30 JAMES

  20 FORD

  10 MILLER

  I want the output should look like this

  DEPTNO ENAME

  ------     -----

  CELINE 10, CLARK, KING, MILLER

  20 JONES, ADAMS, BOND, FORD, SCOTT, SMITH

  30 MARTIN, JAMES BLAKE, WARD, TURNER, ALLEN

  Hear is the script I am currently working on.

  with emp_group as

  (

  Select 20 deptno, ename 'SMITH' Union double all the

  choose 10, 'CÉLINE' from dual union all

  Select 20, "BOND" of all the double union

  Select 30, 'ALLEN' from dual union all

  Select 30, 'WARD' of all the double union

  Select 20, "JONES' from dual union all

  Select 30, 'MARTIN' from dual union all

  Select 30, "BLAKE" from dual union all

  choose 10, 'CLARK' from dual union all

  Select 20, 'SCOTT' from dual union all

  choose 10, 'KING' of the dual union all

  Select 30, 'TURNER' from dual union all

  Select 20, 'ADAMS' from dual union all

  Select 30, 'JAMES' of the dual union all

  Select 20, 'FORD' Union double all the

  choose 10, 'MILLER' from dual

  )

  Select

  DEPTNO

  A1

  of emp_group

  model

  ignore the nav

  partition (deptno)

  dimension (ROW_NUMBER() over (PARTITION BY deptno ORDER BY ename DESC) rn)

  measures (ename, LPAD(' ', 100) A1)

  rules)

  A1 [rn > 0] = ename [cv ()],

  A1 [0] = ename [cv ()]

  )

  order by deptno

  ;

  your help is very appreciated.

  Thank you in advance.

  Try this.

  Select

  *

  Of

  (

  with emp_group as

  (

  Select 20 deptno, ename 'SMITH' Union double all the

  choose 10, 'CÉLINE' from dual union all

  Select 20, "BOND" of all the double union

  Select 30, 'ALLEN' from dual union all

  Select 30, 'WARD' of all the double union

  Select 20, "JONES' from dual union all

  Select 30, 'MARTIN' from dual union all

  Select 30, "BLAKE" from dual union all

  choose 10, 'CLARK' from dual union all

  Select 20, 'SCOTT' from dual union all

  choose 10, 'KING' of the dual union all

  Select 30, 'TURNER' from dual union all

  Select 20, 'ADAMS' from dual union all

  Select 30, 'JAMES' of the dual union all

  Select 20, 'FORD' Union double all the

  choose 10, 'MILLER' from dual

  )

  Select

  DEPTNO

  names

  rn

  of emp_group

  model

  ignore the nav

  partition (deptno)

  dimension of)

  ROW_NUMBER() over (PARTITION BY deptno ORDER BY ename) rn

  )

  measures (ename, cast (null as names of varchar2 (40)))

  rules

  (

  names [any] order by desc = ename [cv ()] rn | «, » || names [cv () + 1]

  )

  order by deptno

  )

  where rn = 1

  Best regards, Andrei

• How to use the clause type conditionally create new lines

  This question is just for learning - no object other than to start real world to create versions of some queries known to learn how to measure the performance of the MODEL clause clause of the MODEL versions.

  The question is: How can you write a STANDARD clause that pivots (updates) ONLY the lines that actually need to rotate and do NOT change the lines that have no value and cannot be rotated.

  I found a good site that has about two dozen example articles on different pieces. This first link is the beginning of the series.

  SQL features tutorials - Clause TYPE

  http://www.sqlsnippets.com/en/topic-11663.html

  This link is to a clause of FAQ of BluShadow MODEL version "how to convert rows to columns."

  Line - MODEL method string

  http://www.sqlsnippets.com/en/topic-11987.html

  The solution to this link uses this line of source:

  POSITION KEY VAL

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

  R08 0 v1, v2, v3,.

  and this result set:

  POSITION KEY VAL

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

  R08 1 v1

  R08 2 v2

  R08 3 v3

  The reason why it produces that result set is this clause of the statement of MODEL

  RETURN THE UPDATED LINES

  If you comment on this article, you will see that the original lines are all in position 0. This means that the solution reproduced EACH SOURCE LINE even if there is only one item in the list of values.

  Thus, even a line source with a single value (for example, the r01 'a') will be updated and updated this line is returned.

  For a large number of data sources with only a few lines that actually need to rotate it would be a great performance of infringement.

  This is the sample data source

  with t as (button 1, 'a' value 'abc' col2, col3 'def' of the double
  Select Union all 2, 'b', 'ghi', 'jkl' from dual
  Union all select 3, 'c, d, e, 'mno', 'pqr' from dual.
  Union all select 4, 'f', 'stu', 'vwx' from dual
  )
  Select the key, value, col2, col3
  t

  KEY, VALUE, COL2, COL3
  1, a, abc, def
  2, b, GHI, jkl
  3, "c, d, e, mno, pqr.
  4, f, stu, vwx

  Lines 1, 2 and 4 only have a value of (a, b, c respectively.

  Only line 3 must rotate. It contains "c, d, e" and which must become 3 lines in the result set

  Value of the key, col2 col3
  1 an abc def
  2B ghi jkl
  3 c mno, pqr
  3 d mno, pqr
  3 e mno, pqr
  4 f stu vwx

  The nut of the problem is that if the clause 'RETURN UPDATED ROWS' is used, then the lines 1, 2 and 4 must be updated in order to be returned in the result set. And if this clause is NOT used, then line 3 must be updated by 'c', d, e 'c' and two new product lines: one line for a ' and one for 'e '.

  How can you do this with the TYPE clause without the help of the clause "Return LINES UPDATE"? I don't have a solution to sample showing what I tried because I can't understand what it takes to even try.

  I have a solution that uses "RETURN UPDATED ROWS" but I want to compare this performance to the exercise, when this clause is NOT used.

  We ask BluShadow to add the solution of MODEL clause to this FAQ with other similar solutions.

  Like this?

  SQL > with t as
  () 2
  3 select the 1 key, 'a' val, 'abc' col2, col3 'def' of the double

  
  Select 4 Union all 2, 'b', 'ghi', 'jkl' from dual
  5 union all select 3, 'c, d, e, 'mno', 'pqr' from dual.
  Select 6 Union all the 4 'f', 'stu', 'vwx' from dual
  7)
  8. Select the key
  9, key_1
  10, val
  11, regexp_substr (val, ' [^,] +', 1, key_1) val_new
  12, col2
  13, col3
  14 t
  model 15
  16 partition by (key)
  dimension (1 key_1) 17
  18 measures (val, col2, col3, (length (val) - length (replace (val, ",")) + 1) as len)
  19 rules
  (20)
  21 val [for 1 to increment of len key_1 [1] 1] = val [1]
  22, col2 [for 1 to increment of len key_1 [1] 1] = col2 [1]
  23, col3 [for 1 to increment of len key_1 [1] 1] = col3 [1]
  24)
  25 order
  26 by key
  27, key_1;

  KEY KEY_1 VAL VAL_N NECK NECK
  ---------- ---------- ----- ----- --- ---
  1 1 has an abc def
  2 b 1 b ghi jkl
  3 1 c, d, e c mno pqr
  3 2 c, d, e d mno pqr
  3 3 c, d, e e mno pqr
  4 1 f f stu vwx

  6 selected lines.

• Cell in the clause type assignments

  Hello
  
  I have a table TEST_DATA with the YEAR, FIELD VALUE, in the form of columns.
  Value1, value2, VALUE 3 columns are determined based on field values. If the data is as follows
  
  THE YEAR FIELD VALUE
  ------------------------------------------
  11111 1 2010
  22222 2 2010
  33333 3 2010
  . . .
  . . .
  . . .
  Can I get the following output using clause type.
  
  YEAR VALUE1 VALUE2 VALUE3
  ------ ----------- ----------- -----------
  2010 11111 222222 333333
  2009 23243 533635 343434
  2008--34343 364653 343434
  2007 34345 789879 343434
  
  I don't want to use pivot, because I have to do some calculations on columns of value above. So I tried the following query using the type clause
  and value1, value2, value3 to come in 3 different lines.
  How can I write cell assignments for the value 1, a value of 2, the value 3 in the same line?
  
  SELECT YEAR, VALUE OF 1, THE VALUE 2, VALUE 3
  OF TEST_FIN_DATA
  MODEL
  RETURN THE UPDATED LINES
  Single KEY
  PARTITION OF (YEAR)
  DIMENSION BY (FIELD)
  MEASURES (VALUE1, VALUE VALUE2, VALUE3 VALUE VALUE)
  RULES
  (value1 ['1'] = value1 ['1'],)
  value [('2 '] = value ['2'],
  value ['3'] = value ['3']);

  Here is an example that you might find interesting:
  http://technology.AMIS.nl/2004/12/16/pivoting-in-SQL-using-the-10G-model-clause/

• help on the clause type

  I have a query as below
  
  Select empno, deptno, sum (sal) on tot_sal (partition deptno) from scott.emp;
  
  
  will give the result like this
  
  EMPNO DEPTNO TOT_SAL
  ---------- ---------- ----------
  7934 10 8750
  7782 10 8750
  7839 10 8750
  7902 20 10875
  7876 20 10875
  7566 20 10875
  7369 20 10875
  7788 20 10875
  
  8 selected lines.
  
  but as you can see the total sal and deptno repeats, what I want is wise dept total salary as shown below
  
  EMPNO DEPTNO TOT_SAL
  ---------- ---------- ----------
  7934
  7782
  7839
  Total for deptno 10 = 8750
  7902
  7876
  7566
  7369
  7788
  Total for deptno = 20 10875
  
  I want to use the type clause and get the above result

  970426 wrote:
  Karthik replied

  Published by: 970426 on January 24, 2013 04:13

  You are not yet able to determine good projection on your own?

  select
   nvl(to_char(empno),'Total for deptno '||deptno||': '||sal)
   empno
  from emp
  model
  partition by (deptno)
  dimension by (row_number() over(
                partition by deptno
                order by empno) rn)
  measures (sal, empno)
  rules(
  sal[0]=sum(sal) [any]
  )
  order by
  deptno, empno
  

  Katrick solution uses my idea of deptno [0]. It has a useless subquery, a useless measure, two unnecessary rules.
  Please don't give any of my points.
  Jiust learn a thing about it.
  Introduction to thinking on your own.

• need help on the clause type

  I have a temp table as below

  LOC X CNT

  20 20 C

  80 20 N

  60 30 C

  40 30 N

  100 40 N

  the rule is for each column of the LOC, there are 2 records of C, N and sum (NTC) must be = 100. If there is no value for C for a LOC (LOC = 40) then the value of the CNT corresponding to C should appear as null or 0 if the value of N is > 0.

  I need a model clause to do this, see below,

  LOC X CNT

  20 20 C

  80 20 N

  60 30 C

  40 30 N

  100 40 N

  0 TO 40 C

  
  

  
  

  Why why why? I am not able to understand the model clause. How did you learn model clause?

• How to use the clause type without hard-coding the values inside?

  Query

  Select acct_no,

  gl_code,

  CASE

  WHEN entry_type_label ("earned INCOME") THEN

  "Recipes".

  ON THE OTHER

  "Deferred income of credit."

  END as entry_type_label,

  CASE

  WHEN entry_type_label IN ("opening balance") THEN

  "Opening balance".

  WHEN entry_type_label ("deferred revenue CREDIT") THEN

  "The amount of the invoice.

  WHEN entry_type_label ("earned INCOME") THEN

  "Recipes".

  WHEN entry_type_label IN ('closing balance") THEN

  "Closing balance.

  ON THE OTHER

  "Deferred income of credit."

  END tag as,

  entry_type_no,

  orig_chg_start_date,

  period_no,

  -amt as amt

  of revrec_test

  WHERE acct_no = 1788562

  AND IN (2, 4) entry_type_no (acct_no model dimension,

  gl_code,

  entry_type_label,

  entry_type_no,

  orig_chg_start_date,

  period_no) measures (amt) rules upsert

  all (amt 1788562,

  ' NON-10011561',

  "Opening balance"

  2,

  2 OCTOBER 17 08.30.00 AM'.

  190 = 0,

  AMT 1788562,

  ' NON-10011561',

  "Closing balance."

  2,

  2 OCTOBER 17 08.30.00 AM'.

  190 = amt 1788562,

  ' NON-10011561',

  "Deferred income of credit."

  2,

  2 OCTOBER 17 08.30.00 AM'.

  190 - amt 1788562,

  "WON-10011561',

  "Revenue."

  4,

  2 OCTOBER 17 08.30.00 AM'.

  190,

  AMT 1788562,

  ' NON-10011561',

  "Opening balance"

  2,

  2 OCTOBER 17 08.30.00 AM'.

  191 = amt 1788562,

  ' NON-10011561',

  "Closing balance."

  2,

  2 OCTOBER 17 08.30.00 AM'.

  190,

  AMT 1788562,

  ' NON-10011561',

  "Deferred income of credit."

  2,

  2 OCTOBER 17 08.30.00 AM'.

  191 = 0,

  AMT 1788562,

  ' NON-10011561',

  "Closing balance."

  2,

  2 OCTOBER 17 08.30.00 AM'.

  191 =.

  (amt 1788562, "UNEARNED-10011561',"Opening balance", 2,")

  2 October 17 08.30.00 AM', 191 + amt 1788562, "UNEARNED-10011561',

  ("'Revenue deferred credit', 2, 08.30.00 October 2, 17 h", 191)-amt

  1788562,

  "WON-10011561',

  "Revenue."

  4,

  2 OCTOBER 17 08.30.00 AM'.

  191,

  AMT 1788562,

  ' NON-10011561',

  "Opening balance"

  2,

  2 OCTOBER 17 08.30.00 AM'.

  192 = amt 1788562,

  ' NON-10011561',

  "Closing balance."

  2,

  2 OCTOBER 17 08.30.00 AM'.

  191,

  AMT 1788562,

  ' NON-10011561',

  "Deferred income of credit."

  2,

  2 OCTOBER 17 08.30.00 AM'.

  192 = 0,

  AMT 1788562,

  ' NON-10011561',

  "Closing balance."

  2,

  2 OCTOBER 17 08.30.00 AM'.

  192 =.

  (amt 1788562, "UNEARNED-10011561',"Opening balance", 2,")

  2 October 17 08.30.00 AM', 192 + amt 1788562, "UNEARNED-10011561',

  "Revenue deferred credit", 2, 08.30.00 2 October 17: "(, 192)-amt"

  1788562,

  "WON-10011561',

  "Revenue."

  4,

  2 OCTOBER 17 08.30.00 AM'.

  192,

  AMT 1788562,

  ' NON-10011561',

  "Opening balance"

  2,

  2 OCTOBER 17 08.30.00 AM'.

  193 = amt 1788562,

  ' NON-10011561',

  "Closing balance."

  2,

  2 OCTOBER 17 08.30.00 AM'.

  192,

  AMT 1788562,

  ' NON-10011561',

  "Deferred income of credit."

  2,

  2 OCTOBER 17 08.30.00 AM'.

  193 = 0,

  AMT 1788562,

  ' NON-10011561',

  "Closing balance."

  2,

  2 OCTOBER 17 08.30.00 AM'.

  193 =.

  (amt 1788562, "UNEARNED-10011561',"Opening balance", 2,")

  2 October 17 08.30.00 AM', 193 + amt 1788562, "UNEARNED-10011561',

  ("Revenue deferred credit", 2, 08.30.00 2 October 17: ', 193)-amt

  1788562,

  "WON-10011561',

  "Revenue."

  4,

  2 OCTOBER 17 08.30.00 AM'.

  193)

  ORDER BY period_no, entry_type_no;

  The query above works fine. But I hardcoded the values. I want to do the same operation for the other account number which will have different non periodical. How can I reach it?

  Thanks in advance.

  And why do you need a model for that? If I understand the reqs OK:

  with t as)

  Select acct_no,

  "U-11561' gl_code,

  "Deferred income credit ' entry_type_label,.

  2 entry_type_no,

  orig_chg_start_date,

  period_no,

  Max)

  case entry_type_no

  When 2 then - amt

  0 otherwise

  end

  ) amt,.

  Description of the "invoice."

  2 weight

  from table_one

  Acct_no group,

  orig_chg_start_date,

  period_no

  Union of all the

  Select acct_no,

  gl_code,

  entry_type_label,

  entry_type_no,

  orig_chg_start_date,

  period_no,

  -amt,

  Description "paid the Bill."

  3 weight

  from table_one

  where entry_type_no = 4

  Union of all the

  Select acct_no,

  "U-11561' gl_code,

  "Deferred income credit ' entry_type_label,.

  2 entry_type_no,

  orig_chg_start_date,

  period_no,

  AMT 0,.

  case lvl

  When 1 then 'account opening. "

  of other "close account."

  description of the end,

  case lvl

  When 1 then 1

  another 99

  end weight

  from table_one,.

  (select level lvl from dual connect by level<=>

  Acct_no group,

  orig_chg_start_date,

  period_no,

  LVL

  )

  Select acct_no,

  gl_code,

  entry_type_label,

  entry_type_no,

  orig_chg_start_date,

  period_no,

  case

  what weight (sum) (1.99) can

  case weight

  When 2 then amt

  When 3 then - amt

  0 otherwise

  end

  )

  During)

  acct_no partition

  order of period_no,

  weight

  )

  AMT else

  AMT of end,

  Description

  t

  order of acct_no,

  period_no,

  weight

  /

  ACCT_NO GL_CODE ENTRY_TYPE_LABEL ENTRY_TYPE_NO ORIG_CHG_START_DATE PERIOD_NO AMT DESCRIPTION
  ------- ---------------------------- ----------------------- ------------- --------------------- --------- ---------- ---------------
  1001 U-11561 deferred revenue credit 2 2 October 17 08.30.00 AM 100 0 opening account
  1001 U-11561 deferred revenue credit 2 Bill October 2, 17 08.30.00 AM 100 44
  E-11561 1001 recipes 4 08.30.00 2 October 17: 100 13.87 paid invoice

  1001 U-11561 deferred revenue credit 2 2 October 17 08.30.00 AM 100 30.13 closing account
  1001 U-11561 deferred revenue credit 2 2 October 17 08.30.00 AM 101 30.13 opening account
  1001 U-11561 deferred revenue credit 2 2 October 17 08.30.00 AM 101 0 Bill
  E-11561 1001 recipes 4 08.30.00 2 October 17: 101 14,35 paid invoice

  1001 U-11561 deferred revenue credit 2 2 October 17 08.30.00 AM 101 15.78 closing account
  1001 U-11561 deferred revenue credit 2 2 October 17 08.30.00 AM 102 15.78 opening account
  1001 U-11561 deferred revenue credit 2 2 October 17 08.30.00 AM 102 Bill 0
  E-11561 1001 recipes 4 08.30.00 2 October 17: 102 14.83 invoice paid

  1001 U-11561 deferred revenue credit 2 2 October 17 08.30.00 AM 102.95 close account
  1001 U-11561 deferred revenue credit 2 2 October 17 08.30.00 AM 103.95 opening account
  1001 U-11561 deferred revenue credit 2 2 October 17 08.30.00 AM 103 Bill 0
  E-11561 1001 recipes 4 08.30.00 2 October 17: 103.95 invoice paid

  1001 U-11561 deferred revenue credit 2 2 October 17 08.30.00 AM 103 0 closing account
  1002 U-11561 deferred revenue credit 2 17 November 13 09.30.00 AM 104 0 opening account
  1002 U-11561 deferred revenue credit 2 17 November 13 09.30.00 AM 60 104 Bill
  1002 fixed costs recurring revenue COA Code 4 09.30.00 November 17, 13 h 104 9.13 invoice paid

  1002 U-11561 deferred revenue credit 2 17 November 13 09.30.00 AM 104 50.87 closing account
  1002 U-11561 deferred revenue credit 2 17 November 13 09.30.00 AM 105 50.87 opening account
  1002 U-11561 deferred revenue credit 2 December 20, 13 09.30.00 AM 105 50.87 opening account
  1002 U-11561 deferred revenue credit 2 09.30.00 December 20, 13 h 105 19.24 invoice
  1002 U-11561 deferred revenue credit 2 17 November 13 09.30.00 AM 105 0 Bill
  1002 fixed costs recurring revenue COA Code 4 09.30.00 November 17, 13 h 105 44,91 paid invoice

  1002 fixed costs recurring revenue COA Code 4 December 20, 13 09.30.00 AM 105 - 7.82 Bill paid

  1002 fixed costs recurring revenue COA Code 4 09.30.00 December 20, 13 h 105 3.93 invoice paid

  1002 U-11561 deferred revenue credit 2 17 November 13 09.30.00 AM 105 29.09 closing account
  1002 U-11561 deferred revenue credit 2 December 20, 13 09.30.00 AM 105 29.09 closing account
  1002 U-11561 deferred revenue credit 2 17 November 13 09.30.00 AM 106 29.09 opening account
  1002 U-11561 deferred revenue credit 2 December 20, 13 09.30.00 AM 106 29.09 opening account
  1002 U-11561 deferred revenue credit 2 December 20, 13 09.30.00 AM 106 0 Bill
  1002 U-11561 deferred revenue credit 2 17 November 13 09.30.00 AM 106 0 Bill
  1002 fixed costs recurring revenue COA Code 4 09.30.00 November 17, 13 h 106 44,91 paid invoice

  1002 fixed costs recurring revenue COA Code 4 09.30.00 December 20, 13 h 106 11.75 invoice paid

  1002 fixed costs recurring revenue COA Code 4 December 20, 13 09.30.00 AM - 23.47 106 paid

  1002 U-11561 deferred revenue credit 2 17 November 13 09.30.00 AM 106-4.1 closing account
  1002 U-11561 deferred revenue credit 2 December 20, 13 09.30.00 AM 106-4.1 closing account
  1002 U-11561 deferred revenue credit 2 December 20, 13 09.30.00 AM 107 - 4.1 opening account
  1002 U-11561 deferred revenue credit 2 17 November 13 09.30.00 AM 107 - 4.1 opening account
  1002 U-11561 deferred revenue credit 2 17 November 13 09.30.00 AM 107 Bill 0
  1002 U-11561 deferred revenue credit 2 December 20, 13 09.30.00 AM 107 Bill 0
  1002 fixed costs recurring revenue COA Code 4 December 20, 13 09.30.00 AM - 13.69 107 paid invoice

  1002 fixed costs recurring revenue COA Code 4 November 17, 13 09.30.00 AM 107 35.91 invoice paid

  1002 fixed costs recurring revenue COA Code 4 09.30.00 20 December 13 AM 107 6,86 invoice paid

  1002 U-11561 deferred revenue credit 2 17 November 13 09.30.00 AM 107 - 33.18 closing account
  1002 U-11561 deferred revenue credit 2 December 20, 13 09.30.00 AM 107 - 33.18 closing account

  47 selected lines.

  SQL >

  SY.

• Clause type SQL how can the others cannot?

  Hi all

  I met a topic called the type Clause when I was reading Pro Oracle SQL by Karen Morton, Kerry Osborne, Robyn SandsRiyaj Ntanguand Jared still. I just read the chapter and did some practice on my local machine. It is obvious that it is a very powerful feature. And I also reachered on the web and found out that it is used for spreadsheets (excel etc.). Can someone please give me a simple example which, in which case one must use the clauses types SQL Analytic Functions or group Advanced functions cannot cure?

  Either said by the way, I recommend strongly Pro SQL Oracle book to all who want to learn the knowledge of SQL.

  Thanks in advance

  Hello

  In short Yes, the TYPE clause allows you to perform calculations of "spreadsheet-like". With the help of the clause TYPE is quite rare. Now with the introduction of the WITH recursive clause in 11g 2, you can almost everything in SQL, do not even talk about the pattern match feature introduced in 12 c.

  You can probably do now with analytical functions / recursive clause the clause of MODEL can do. And I must say that it is a shame that the TYPE clause has never been improved since its introduction in 10g.

  In my view, the clause TYPE can come in handy when you need cross reference calculations (typically dealing with periods such as dates, years, weeks). For example, I personally used a MODEL clause for an analysis of sales time, where I watch the sales of previous years to find a pattern for the current. References of cells in a MODEL clause allows to easily, something like [my_year, my_week] model = local_pattern [(y ENTRE CV()-2 ET CV) (-1, w BETWEEN (CV-2) RESUME AND () + 2)]

• Include as the value of the clause in the list of results

  I have several similar terms on the same column in my query. Is it possible to include the value of the clause type in the result set?

  For example

  SELECT * ADDR ADDRESS WHERE

  addr. Line LIKE '% 1% '.

  OR addr.line LIKE '% 2 '.

  OR addr.line LIKE '% 3% '.

  OR addr.line LIKE '% 4% ';

  The result set contains all the addresses including ONE, TWO, THREE, FOUR. Now, I would like to see the result defined as

  Address1ONE | ONE

  Address2TWO | TWO

  THREEaddr | THREE

  any clue?

  R

  Use CASE:

  SELECT addr.line,

  CASE

  WHEN addr.line LIKE '% 1% ' THEN 'ONE '.

  WHEN addr.line LIKE '% 2% ', THEN 'TWO '.

  WHEN addr.line LIKE '% 3% ' THEN 'THREE '.

  ELSE 'FOUR '.

  END

  The ADDRESS ADDR

  WHERE addr.line LIKE '% 1% '.

  OR addr.line LIKE '% 2 '.

  OR addr.line LIKE '% 3% '.

  OR addr.line LIKE '% 4% '.

  /

  SY.

• Clause type and its Alternative to Oracle 10 g

  Hello
  I went through a few examples on the clause type in oracle 10g, but not very clear on this subject. can someone please explain it to me in detail and when to use it.
  Also, how can we create the same performance as that of the model clause by using other means.

  [http://stanford.edu/dept/itss/docs/oracle/10g/server.101/b10736/sqlmodel.htm#i1006326]

  Reference Models
  In addition to the multi-dimensional array on which rules operate, which is called the main model, one or more read-only multi-dimensional arrays, called reference models, can be created and referenced in the MODEL clause to act as look-up tables. Like the main model, a reference model is defined over a query block and has DIMENSION BY and MEASURES clauses to indicate its dimensions and measures respectively. A reference model is created by the following subclause:
  
  REFERENCE model_name ON (query) DIMENSION BY (cols) MEASURES (cols)
     [reference options]
  
  Like the main model, a multi-dimensional array for the reference model is built before evaluating the rules. But, unlike the main model, reference models are read-only in that their cells cannot be updated and no new cells can be inserted after they are built. Thus, the rules in the main model can access cells of a reference model, but they cannot update or insert new cells into the reference model. References to the cells of a reference model can only appear on the right side of rules. You can view reference models as look-up tables on which the rules of the main model perform look-ups to obtain cell values. The following is an example using a currency conversion table as a reference model:
  
  CREATE TABLE dollar_conv_tbl(country VARCHAR2(30), exchange_rate NUMBER);
  INSERT INTO dollar_conv_tbl VALUES('Poland', 0.25);
  INSERT INTO dollar_conv_tbl VALUES('France', 0.14);
  ...
  
  Now, to convert the projected sales of Poland and France for 2003 to the US dollar, you can use the dollar conversion table as a reference model as in the following:
  
  SELECT country, year, sales, dollar_sales
  FROM sales_view
  GROUP BY country, year
  MODEL
    REFERENCE conv_ref ON (SELECT country, exchange_rate FROM dollar_conv_tbl)
    DIMENSION BY (country) MEASURES (exchange_rate) IGNORE NAV
  MAIN conversion
  DIMENSION BY (country, year)
     MEASURES (SUM(sales) sales, SUM(sales) dollar_sales) IGNORE NAV
  RULES
  (dollar_sales['France', 2003] = sales[CV(country), 2002] * 1.02 *
     conv_ref.exchange_rate['France'],
     dollar_sales['Poland', 2003] =
        sales['Poland', 2002] * 1.05 * exchange_rate['Poland']);
  
  Observe in this example that:
  
  A one dimensional reference model named conv_ref is created on rows from the table dollar_conv_tbl and that its measure exchange_rate has been referenced in the rules of the main model.
  
  The main model (called conversion) has two dimensions, country and year, whereas the reference model conv_ref has one dimension, country.
  
  Different styles of accessing the exchange_rate measure of the reference model. For France, it is rather explicit with model_name.measure_name notation conv_ref.exchange_rate, whereas for Poland, it is a simple measure_name reference exchange_rate. The former notation needs to be used to resolve any ambiguities in column names across main and reference models.
  
  Growth rates, in this example, are hard coded in the rules. The growth rate for France is 2% and that of Poland is 5%. But they could come from a separate table and you can have a reference model defined on top of that. Assume that you have a growth_rate(country, year, rate) table defined as the following:
  
  CREATE TABLE growth_rate_tbl(country VARCHAR2(30),
     year NUMBER, growth_rate NUMBER);
  INSERT INTO growth_rate_tbl VALUES('Poland', 2002, 2.5);
  INSERT INTO growth_rate_tbl VALUES('Poland', 2003, 5);
  ...
  INSERT INTO growth_rate_tbl VALUES('France', 2002, 3);
  INSERT INTO growth_rate_tbl VALUES('France', 2003, 2.5);
  
  Then the following query computes the projected sales in dollars for 2003 for all countries:
  
  SELECT country, year, sales, dollar_sales
  FROM sales_view
  GROUP BY country, year
  MODEL
    REFERENCE conv_ref ON
            (SELECT country, exchange_rate FROM dollar_conv_tbl)
             DIMENSION BY (country c) MEASURES (exchange_rate) IGNORE NAV
    REFERENCE growth_ref ON
            (SELECT country, year, growth_rate FROM growth_rate_tbl)
             DIMENSION BY (country c, year y) MEASURES (growth_rate) IGNORE NAV
    MAIN projection
     DIMENSION BY (country, year) MEASURES (SUM(sales) sales, 0 dollar_sales)
    IGNORE NAV
    RULES
    (dollar_sales[ANY, 2003] = sales[CV(country), 2002] *
     growth_rate[CV(country), CV(year)] *
     exchange_rate[CV(country)]);
  

• Can we use type multiset and with the clause the two

  Please let me know how to use the multiset type and with the clause.

  You need more inlining, as for example in:

  SQL> select *
  from table (cast (multiset (select *
                              from (with t as (select deptno from dept)
                                    select * from t)) as sys.dbms_debug_vc2coll))
  /
  COLUMN_VALUE
  ----------------------------------------
  10
  20
  30
  40
  50                                      
  
  5 rows selected.
  

• Define a map or ORDER method for the object type

  Hi gurus

  I created an object and then its type and then I use this object and type based on line, see below:

  Create the object

  CREATE OR REPLACE

  TYPE test_object

  IS

  OBJECT

  (

  next_appearance_dt DATE, - next_appearance_dt

  youth_adult VARCHAR2 (5) - youth_adult

  ) ;

  
  /

  Create the object Type

  CREATE or REPLACE TYPE t_docket_object IS TABLE OF THE test_object;

  /

  Create function Pipeline

  FUNCTION to CREATE or REPLACE f_report (p_dt date, p_c_cd VARCHAR2)
  return t_test_object pipeline
  IS
  BEGIN
  FOR J IN)
  Select distinct test_object)
  next_appearance_dt,--862,
  'YOUTH '.
  ) AS test_object
  Jen.next_appearance base
  WHERE 1 = 1
  AND (base.next_appearance_dt = p_dt)
  AND (base.circuit_point_cd = p_c_cd)
  - and cse.information_id = 322
  -ORDER 15 - alias_name
  )

  loop
  PIPE ROW (J.test_object);
  END loop;
  END;

  /

  Run function

  SELECT * FROM TABLE (F_REPORT (TO_DATE('25-sep-2015','dd-mon-yyyy'),'1 '))

  Error

  ORA-22950: cannot ORDER objects without map or ORDER method

  ORA-06512: at "F_REPORT", line 5

  22950 00000 - "cannot ORDER objects without map or ORDER method.

  * Cause: an object type must have a defined for map or ORDER method

  all comparisons other than equality and inequality comparisons.

  * Action: Define a map or ORDER method for the object type

  I know the reason of this error and the reason is that I use the clause separate in my pipeline service, but do not know how to get rid of this error...

  Confuse you the type of table with the object type. He forge

  Select test_object (next_appearance_dt) test_object

  Not:

  Select t_test_object (next_appearance_dt) test_object

  Then:

  SQL > CREATE OR REPLACE
  FUNCTION f_report (DATE p_dt 2,
  3 P_C_CD VARCHAR2
  4                   )
  5 t_test_object of RETURN
  6 IN PIPELINE
  7 EAST
  BEGIN 8
  9 FOR (IN) v_rec
  10 WITH () DID
  11. SELECT TO_DATE('01-jan-2015','dd-mon-yyyy') NEXT_APPEARANCE_DT
  the double 12
  13                                  )
  14 select test_object (next_appearance_dt) test_object
  15                          from  t
  16                      ) LOOP
  PIPE 17 ROW (v_rec.test_object);
  18 END OF LOOP;
  END 19;
  20.

  The function is created.

  SY.

• Nested set tables in select query "in the clause of" take long time

  create or replace type t_circuitids is table of the varchar2 (100);

  -Under anonymous block continues to run away and never ends

  DECLARE
  v_circuitid t_circuitids;
  number of v_count;
  l_circuitids VARCHAR2 (4000)
  : = "Value1, value2, value3, value4, Value5";
  BEGIN
  -Query below converts the output concatinated with commas to the list and stores it in the nested table collection v_circuitids
  WITH an ACE
  (SELECT ',' | l_circuitids |) ',' AS circuitid
  THE DOUBLE)
  SELECT DISTINCT TRIM (SUBSTR (circuitid,
  INSTR (circuitid, "," 1, LEVEL) + 1.
  INSTR (circuitid, "," 1, LEVEL + 1)
  -INSTR (circuitid, "," 1, LEVEL)
  -1
  )
  ) cid
  LOOSE COLLECTION v_circuitid
  A
  CONNECT BY LEVEL <
  LENGTH (circuitid)
  -LENGTH (REPLACE (circuitid, ','));

  SELECT COUNT (1)
  IN v_count
  TABLE
  WHERE name IN (SELECT COLUMN_VALUE
  TABLE (v_circuitid));
  END;
  /

  -I had the question, query "SELECT COLUMN_VALUE FROM TABLE (v_circuitid)" that is used in code above is responsible for this.

  -Same code works fine in development and Test environments, but prod it continues to work on

  -I solved this problem by creating a temporary table, loading of all values in the collection in the temporary table and using this temporary table "in the clause" "

  -Can answer why his behavior like this when I use the collection where clause?

  -I use Oracle 9i

  Here is a summary of the question and the solution for this.

  -Nested type to collect multiple values

  CREATE or REPLACE the TYPE t_circuitid IS TABLE OF VARCHAR2 (100);

  Below the code will simply on the run.

  DECLARE
  v_circuitid t_circuitid;
  v_count NUMBER;
  BEGIN
  SELECT nal_name
  LOOSE COLLECTION v_circuitid
  OF fs_head
  WHERE groupid = 10;

  SELECT COUNT (1)
  IN v_count
  OF fs_attrib
  WHERE NAME IN (SELECT COLUMN_VALUE
  TABLE (v_circuitid));

  Dbms_output.put_line (v_count);
  END;
  /

  Cause:-SELECT COLUMN_VALUE TABLE (v_circuitid)); -This request is causing problem

  Why?  : - Because the CBO does not know how many lines is present in the collection, by default it takes 8168 lines

  Note:-always happens, it depends on the volume of data in the table, the path chosen by CBO. In my case, the table is huge on prod compared to DEV so he was causing a problem.

  -The following modified code works very well

  DECLARE
  v_circuitid t_circuitid;
  v_count NUMBER;
  BEGIN
  SELECT nal_name
  LOOSE COLLECTION v_circuitid
  OF fs_head
  WHERE groupid = 10;

  SELECT COUNT (1)
  IN v_count
  OF fs_attrib
  WHERE NAME IN (SELECT / * + cardinality (20 t) * /)
  COLUMN_VALUE
  TABLE (v_circuitid) t);

  Dbms_output.put_line (v_count);
  END;
  /

  Solution:-used as cardinality hint below.

  SELECT / * + cardinality (20 t) * / t COLUMN_VALUE TABLE (v_circuitid);

  Using cadinality I am saying CBO which dataset contains 20 lines or less.

  If it is version of Oracle 10 g and more, we can use below code (CARD utility).

  SELECT COLUMN_VALUE TABLE (CARD ((v_circuitid))

  For more information reach me at [email protected]

  Thank you best regards &,.

  Amarnath a. Reddy.

• Problem with the case where the clause

  Hello

  I want to use cases in which clause. I wrote a very simple query to test how it works. However, I always got an error message. Here's what I did.

  Select eid, firstname, lastname from employee where eid = 1 works fine

  Select eid, name, name of the employee

  where

  cases where 1 = 1 then eid = 1

  When 1! = 1 then eid = 2

  end

  I ORA-00905: lack of keyword.

  Did I miss something?

  Thank you

  Jennifer

  Hi, Jennifer.

  Why do you want to use BOX in a WHERE clause?  The wonderful thing about the CASE is that it allows you to do IF-THEN-ELSE logic.  In a SELECT clause, or an ORDER BY clause, this is really useful, but the WHERE clause already allows you to use the IF-THEN-ELSE logic.

  However, if you use the BOX in a WHERE clause, it is as it does nowhere else.  Always BARGAIN instead of a single expression in a SQL data types.  All the clauses THEN (and ELSE, if there is one) must return a single expression, such as the NUMBER.  (There is no BOOLEAN data type in SQL).  So, you can use a CASE statement like this:

  ...

  CASES WHERE

  WHEN 1 = 1 THEN 1

  WHEN 1! = 1 THEN 2

  END = eid

  Here, the CASE expression takes the place of a NUMBER.

  However, if I was not sure if 1 = 1 or not, I would write the condition in this way, without a CASE expression:

  WHERE (1 = 1 AND eid = 1)

  OR (1! = 1 AND eid = 2).

• As for the Clause on Oracle 10 g db

  Hi experts,
  
  I had little doubt about my simple request
  
  When I go through the black book of Oracle for db10g article on the clause
  They stated something like this

  SELECT * FROM EMPLOYEES (EMPLOYEE_ID = '1981');

  on the clause simliar to the place where clause.
  
  SELECT * FROM EMPLOYEES WHERE EMPLOYEE_ID = '1981'.
  
  When I run this statement .it fine.
  
  ' using ' ON clause ' instead of ' where clause.
  
  He didn't accept the clause on, run the command not properly ended sql.
  
  where I'm not kidding...?
  
  
  Thank you & & looks
  ADF7.
  
  Version
  Oracle Database 10g Enterprise Edition Release 10.2.0.1.0 - Production

  Oracle provides the clause TO specify join conditions in the SELECT statements involving multiple tables. It is independent of data types and column names. As the USING clause, the clause and NATURAL JOIN are mutually exclusive.

  In cases where the tables participating are more two CLAUSES, Oracle uses the first specification of the clause to join the two tables. He uses the second ON the specification of the clause to combine the next table with the result of the join of the first.
  The key word THERE is also used in other contexts, such as in GRANT and INDEX statements.

  Notes:

  After the keyword ON the join condition is written, in general, it can contain predicates connected with Boolean AND, OR, and NOT
  The order of the tables is not relevant
  The use of parentheses are optional in syntax, but the readability of the increases

  Syntax example:

  SELECT

  FROM TABLE_1

  JOIN TABLE_2

  ON (join condition)

  Usage example:

  The SQL query below shows the employee's name and the name of Department of their work:

  SQL > SELECT E.ENAME, D.DNAME

  EMPLOYEE E JOIN DEPT D

  WE (E.DEPTNO = D.DEPTNO)