Similar Questions

• How to write columns unique combinatins in oracle?

  Hi friends,

  We organize a three-sided football tournament (played by three teams on a hexagonal field with three football goals). We have a table with 5 teams participating in the tournament: teams meet in all possible combinations, so we need to find all of the matches (combination of three teams) which must be played in the tournament. The order is not relevant. As long as the query returns 10 games with combinations of three teams, the order of the teams and the lines does not matter.

  I have a (tab_name) table containing a single column of data number type...

  Select * from table_name;

  Col
  1
  2
  3
  4
  5

  I have to get unique combination of output when I play joins.

  T1.Col	T2.col	T3.Col

  123124125145134135234235245345

  Hello

  Here's a way that scales to any number of teams playing at a time (P) and any number of total teams (T > = P):

  SELECT teams SYS_CONNECT_BY_PATH (col, ',')

  FROM t1

  WHERE LEVEL = 3 - or no matter what P

  CONNECTION BY neck > neck PRIOR

  AND LEVEL<= 3   ="" --="" or="" whatever="" p="">

  ;

• How to write several unique java class in the java file?

  Hello
  I use Jdev 11.1.2.0.0.
  
  When I write the java class, I wrote in my file of multiple class. This time Jdeveloper advise that, move this class into distinct java file.
  But I don't want to go, without moving this class it works perfectly. But it is what tells the error indicator 'red '. How to solve this problem.
  
  Kind regards
  Rami

  Hello

  its an audit rule that is facing you. You can go to tools-->--> Audit preferences and disable it if you are more disturbed.

  Frank

• How to divide each unique number in a matrix by the number with an index in another matrix?

  Hello

  Basically, the calculation, I tried to do is called "dividing element by element". It is a transaction between two matrices with few passes and lines. The result matrix also has the same demension and each element is obtained by dividing the element with the same index in a matrix of the element with the index even in a different matrix. for examlpe,.

  A = [2 4 6

  3 9 6]

  B = [2 2 3

  3 3 2]

  the result will be

  C = [1, 2-2

  1 3 3]

  Is there an easy way to do another going out each element using the table to index?

  Thank you very much

  Hao Liu

  Use the 2D tables instead of the matrix data type, then just use a simple primitive division. the operation will be done by item automatically.

  

  (Generally you should always use table 2D instead of the special matrix data type. The matrix data type is most useful for linear algebra)

  (Die, he must be very careful, for example, if you multiply two 2D tables, you get a multiplication of element by element, but if you do the same operation on two matrices, LabVIEW will substitute for a real matrix multiplication, which is not the same. For the division, it seems that the matrices are divided piece-by-piece, so it might work for you directly. You should just be aware that the matrices are often treated differently. "They are a very special type of data).

• How to write to multiple databases in a single transaction?

  I have a phisical file that contains several design databases. Now, I try to write in several databases in this type of file in a single transaction. I have made some efforts to achieve but failed. In the documentation of Berkeley, it is said "to ensure that all transactions are resolved (committed or aborted) before closing the database. ', I think that if I close the first database operation, putting the second database can be influenced and files will not correctly put in the second database. How do I do?

  Please check this page of documentation: atomicity. You have also some good examples in:
  DB-4.7.25\examples_c
  DB-4.7.25\examples_cxx
  DB-4.7.25\examples_java

  If it's not helping you, please post a test program.

  Thank you
  Bogdan

• How to write data I read frequency bands in excellent using the tools for the generation of reports.

  Hi all

  I wrote the code to read data from the Analyzer of spectrum and also for writing in excel using report generation tools. When I run the program values get read and excellent sheet also opens but I don't see anything written in.

  So if anyone can help me in that I'm very gald. I am attaching the screenshot of my program and also attach file .vi. As I use the drivers you can see some of vi.

  Thanking you,

  Patricia

  @tobid,

  Thank you and sorry long back only I got my error and its job well now... again thanks and sorry that I have not mentioned here...

  Patricia

• HP - 50g written problem complex number. How to write (3 + i * 3 * sqr (3)) to RPN? (3, i3sqr (3))

  Hello everyone, I don't know how to write (i 3, * (3 * sqr (3))) RPN mode.

  Can someone help me please? I already searched the gug pdf guide help, without result.

  Thank you.

  Hello

  A few notes in the representation of a complex number as a vector:

  (1) the i is omitted, the second number is supposed to be the imaginary part.

  (2) the 50 g accepts only decimal numbers for vector representation

  Thus, at the entrance there. I guess you have the calculator in complex mode and exact mode (about not checked in "MODE"-> "SAE"), i.e. "C =" displayed at the top of the screen.

  RPN mode:

  Press 3 then ENTRY (3 is at level 1)

  Press ENTER again (3 is now duplicated and on both levels 1 and 2 - one advantage of RPN)

  Press ENTER again (3 is now duplicated and on levels 1, 2 and 3)

  Press the button of sqrt (x). Level 1 shows a 3 with a sign of the square root.

  Press on multiply key, level 1 shows 3· SQRT (3)

  Press LeftShift (white), then I (above the TOOL key), level 1 indicates an i

  Press on multiply key, level 1 shows 3· SQRT (3)· I have

  Press + key, level 1 shows 3 + 3· SQRT (3)· I have (or 3· SQRT (3)· i + 3 according to the parameter Flag - 27)

  If you want to keep the symbolic result (with the symbol of the 'SQUARE root'), leave it in this form, if you want to convert to vector format press (red) RightShift, then -> NUM (above the ENTER key).

  The result will be a decimal vector shape (3, 5.19615242271).

  Kind regards

  BartdB

• How to write a function to estimate the number of rows returned SQL?

  How to write a function to estimate the number of rows returned SQL through SQL Execution Plan?
  My idea is
  Call dbms_sql.parse to create the SQL PLAN, then ask the PLAN for the number of estimated return lines.
  But how to get SQL plan through "id cursor?
  Thank you.

  You can use EXECUTE IMMEDIATE to explain plan statement. About the STATEMENT_ID generation, it could be anything. Even a SYSTIMESTAMP cast as TANK would work.

• How to write VBA code that is compatible DEP

  Hello

  We have a number of VBA macros, Excel. Some users report an error

  

  After some research, I believe that it is related to the Microsoft Data execution prevention feature. I presume our adds that something that does not like the DEP. Is there a guide on how to write/re-write macros to avoid problems with DEP?

  Thank you

  Emmet

  This issue is beyond the scope of this site and must be placed on Technet or MSDN

  http://social.technet.Microsoft.com/forums/en-us/home

  http://social.msdn.Microsoft.com/forums/en-us/home

• How to write a decryption function?

  Hi! everyone ,

  
  

  I see one encryption function in my database.

  
  

  Select f_pwd_encrypt ('password') of double

  
  

  -> 12412913141313139139130121

  
  

  My question is

  
  

  How to write a decryption function?

  
  

  As:

  
  

  Select f_pwd_decrypt ('12412913141313139139130121') of double

  
  

  ->password

  CREATE OR REPLACE FUNCTION EPADM."F_PWD_ENCRYPT" ( vpwd in varchar2)
  return varchar2
  is
      vother_p   varchar2(9);
      vtr_pwd  varchar2(2048);
      i       number;
      j         number;
      vsubstr varchar2(9);
      vtemp1 varchar2(08);
      vvalue number := 0;
      vdb_pwd varchar2(100);
  
  
      function str_2_bit(vstring in varchar2)
      return varchar2
      is
         i number;
         vtemp number;
         v1 varchar(2048);
  
  
       function single_byte(vin in number)
         return varchar2
         is
           i number;
           vresult varchar2(08);
           vtemp number := vin;
         begin
           for i in 1..8 loop
             vresult := to_char(mod(vtemp,2))||vresult;
             vtemp := trunc(vtemp/2);
           end loop;
           return(vresult);
         end;
    --
      begin
        for i in 1..lengthb(vstring) loop
          select to_number(substrb(dump( vstring ,10,i,1),instr(dump( vstring ,10,i,1),' ',-1)+1))
            into vtemp
          from dual;
          v1 := v1 || single_byte(vtemp);
        end loop;
  
  
        return(v1);
      end;
  
  
  begin
    vtr_pwd := str_2_bit(vpwd);
    vtr_pwd := substrb(vtr_pwd,4)||substrb(vtr_pwd,1,3);
    vvalue := 0;
    vdb_pwd := null;
  
  
     for i in  1..(lengthb(vtr_pwd)/4)  loop
       vtemp1 := substrb(vtr_pwd,(i-1)*4+1,4);
       for j in 1..4 loop
         vvalue :=  vvalue + to_number(substrb(vtemp1,j,1)) * power(2,j-1);
         dbms_output.put_line(j||' '||vvalue);
       end loop;
            vdb_pwd := to_char(vvalue) ||vdb_pwd;
            vvalue := 0;
     end loop;
    return(vdb_pwd);
  
  
  END;
  /
  

  OK, after reviewing the, I don't think you'll be able to write a function of decryption for him.

  The first thing he does is take the ascii value of each character in the password and converts them into a binary string.  The code it uses is far too complex and can be simplified, but which is not a problem here.

  I've recreated the first step of SQL like this...

  SQL > ed

  A written file afiedt.buf

  1 with chr_val like)

  2. Select level l

  3, dump('password',10,level,1) in the dmp

  4, to_number (substrb (dump('password',10,level,1), instr (dump('password',10,level,1),' ', 1) + 1)) as chr_val

  5, ascii (substr('password',level,1)) as chr_val - equivalent of extraction of useless dump

  6 double

  7. connect by level<=>

  8        )

  9, r (l, b, ch, chr_val, result, vtemp) as

  10 (select l, 0 b, chr (chr_val), chr_val)

  11, cast (null as varchar2 (8)) as a result

  12, chr_val as vtemp

  13 of chr_val

  14 union of all the

  15 select l, b + 1, b, ch, chr_val

  16, to_char (mod(vtemp,2)) | result as a result

  17, trunc(vtemp/2) as vtemp

  18 r

  where the 19 b + 1<=>

  (20) depth search first by l, defined b seq

  21, as)

  22 select l, ch, chr_val, str_to_bit result

  23 r

  where the 24 b = 8

  25 arrested by l, seq

  26            )

  27 select listagg (ch) within the Group (order) as password

  28, listagg (chr_val, ',') within the Group (order) byte_vals

  29, listagg (str_to_bit) within the Group (order) bit_vals

  30 sec.

  SQL > /.

  PASSWORD BYTE_VALS BIT_VALS

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

  password 0111000001100001011100110111001101110111011011110111001001100100 112,97,115,115,119,111,114,100

  Then he takes this string binary ("bit_vals" in my example) and does the following:

  1. take the first 3 bits of left and to transpose on the right end of the string.

  2 chops the resultant bit string upward into sections of 4 bits (which is known as a 'nibble' inside)

  3. for each bit in the nibble, he treats the bits in binary reverse to normal and gives them a value of 1,2,4 or 8 from left to right for each bit set to 1

  4. for each nibble it adds the value of 1,2,4,8 bits to give a value from 0 to 15

  To show that the use SQL...

  SQL > byte (select ' 0111000001100001011100110111001101110111011011110111001001100100' as pieces of double)

  2, swap3bit as (-take the binary string and put the first bits (high) 3 as a (low) bit of the right hand side)

  3. Select bytes.bits

  4, substr (bit 4) | substr (bits, 1, 3) as init_substr

  5 bytes

  6                   )

  7, split4 as (-chop the string of bits nibbles (half bytes - 4 bits))

  8. Select level l

  9, substr (init_substr, ((level-1) * 4) + 1, 4) as a nibble

  swap3bit 10

  11. connect by level<=>

  12                 )

  13, bitpowers (select l

  14, snack

  15, to_number (substr(nibble,1,1)) * power (2, 1-1) as bitval1

  16, to_number (substr(nibble,2,1)) * power (2, 2-1) as bitval2

  17, to_number (substr(nibble,3,1)) * power (2, 3-1) as bitval3

  18, to_number (substr(nibble,4,1)) * power (2, 4-1) as bitval4

  19, to_number (substr(nibble,1,1)) * power (2, 1-1) +.

  20 to_number (substr (nibble, 2, 1)) * power (2, 2-1) +.

  21 to_number (substr (nibble, 3, 1)) * power (2, 3-1) +.

  22 to_number (substr (nibble, 4, 1)) * power (2, 4-1) as total_val

  23 of split4

  24                   )

  25 select * from bitpowers

  26.

  L NIBB BITVAL1 BITVAL2 BITVAL3 BITVAL4 TOTAL_VAL

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

  1 1000          1          0          0          0          1

  2 0011          0          0          4          8         12

  3 0000          0          0          0          0          0

  4 1011          1          0          4          8         13

  5 1001          1          0          0          8          9

  6 1011          1          0          4          8         13

  7 1001          1          0          0          8          9

  8 1011          1          0          4          8         13

  9 1011          1          0          4          8         13

  10 1011          1          0          4          8         13

  11 0111          0          2          4          8         14

  12 1011          1          0          4          8         13

  13 1001          1          0          0          8          9

  14 0011          0          0          4          8         12

  15 0010          0          0          4          0          4

  16 0011          0          0          4          8         12

  16 selected lines.

  These final values are then re-combination as strings in reverse order so that you get then:

  '12' |' 4'||' 12' |' 9'||' 13'... and so on.

  In SQL...

  SQL > byte (select ' 0111000001100001011100110111001101110111011011110111001001100100' as pieces of double)

  2, swap3bit as (-take the binary string and put the first bits (high) 3 as a (low) bit of the right hand side)

  3. Select bytes.bits

  4, substr (bit 4) | substr (bits, 1, 3) as init_substr

  5 bytes

  6                   )

  7, split4 as (-chop the string of bits nibbles (half bytes - 4 bits))

  8. Select level l

  9, substr (init_substr, ((level-1) * 4) + 1, 4) as a nibble

  swap3bit 10

  11. connect by level<=>

  12                 )

  13, bitpowers (select l

  14, snack

  15, to_number (substr(nibble,1,1)) * power (2, 1-1) as bitval1

  16, to_number (substr(nibble,2,1)) * power (2, 2-1) as bitval2

  17, to_number (substr(nibble,3,1)) * power (2, 3-1) as bitval3

  18, to_number (substr(nibble,4,1)) * power (2, 4-1) as bitval4

  19, to_number (substr(nibble,1,1)) * power (2, 1-1) +.

  20 to_number (substr (nibble, 2, 1)) * power (2, 2-1) +.

  21 to_number (substr (nibble, 3, 1)) * power (2, 3-1) +.

  22 to_number (substr (nibble, 4, 1)) * power (2, 4-1) as total_val

  23 of split4

  24                   )

  25 select listagg (to_char (total_val)) the Group (order of the desc) as pwd

  26 of bitpowers

  27.

  PWD

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

  12412913141313139139130121

  Now, the problem of decryption is that these numbers are concatenated without padding for a fixed number of digits by value, so you don't know if it was

  '12' |' 4'||' 12' |' 9'||' 13'... and so forth as we did it, or whether he was

  '1'||' 2'||' 4'||' 12' |' 9'||' 1'||' 3'... and so on, or any other combination of values from 0 to 15

  There is essentially no information to allow you to divide the string upwards in the correct components to allow the whole process be reversed.

  So, you are out of luck... no chance of decrypting it.

• How to get the frame number current timeline using jsx?

  How to get the frame number current timeline using jsx?

  Thanks for the link!

  Clarification: I was looking for the current image on the timeline of the video. OP he mentions that he had found how to get that so I searched a bit more and found this thread: Re: can I inpoint in read/write and out-point of the clips in a group of video with scripts? They have the following functions, I have not tried yet, but it seems like it should work.

  function getCurrentFrame() {}

  try {}

  Var ref = new ActionReference();

  ref.putProperty (charIDToTypeID ('Rprp'), stringIDToTypeID ('currentFrame'));

  ref.putClass (stringIDToTypeID ('timeline'));

  var / / desc = new ActionDescriptor();

  desc.putReference (charIDToTypeID ('null'), ref);

  var TC = executeAction (charIDToTypeID ('getd'), desc, DialogModes.NO);

  Return TC.getInteger (stringIDToTypeID ('currentFrame'));

  } catch (e) {return null ;}

  };

• Automatically generated Unique number

  Hello

  I would like to have some sort of function on my document that creates a unique number for each submitted form.

  The form is both submitted by .xml and .pdf.

  is it possible to do something like that?

  See you soon,.

  Wes

  Hi Wes,

  I don't know how good you are with script/Javascript, but also I have such a function, you may have different ideas 2 based on what you want to do:

  1. you can have a string generator that could be based on information inserted in certain areas of the user.

  for example: you can use the substring function to get some partial information from different fields of the form and concatenate together to create a unique "ID" of the form.

  Unfortunately, this could have IDS that might be the same if a form is filled with more than once by a user, and or (rare) some information could be almost equal to the others...

  2. There is the option which can use the date and time of the form when it is submitted. Using this method is almost perfect never have the same unique ID 2. You can also use a random string generator to concatenate the time with her.

  The best option you have is to have all of the answers above. Unique ID would start with partial information in the form + date/time of the form provided or vice versa.

  If you want more of these 3 options above, you can also add a random string generator in the process. Here is a help for the random string generator, you can modify it as you wish...

  random password generator?

  Mag

• How to write a procedure to call and run the custom package backend

  Hi all
  
  Oracle 10g
  Oracle Apps R12
  
  I work with here oracle order management, we have a package called (Pick Release) to customize. Due to a problem, we have this concurrent program execution manually giving Route_id as parameter. The route_id comes from the road to the Table. By using this query
  
  Select distinct route@DB_LINK_APPS_TO_ROADSHOW route_id
  When trunc (route_date) = trunc (sysdate + 2).
  
  on a daily basis, we have almost 42 routes and we run this simultaneous program manually close times.
  
  so now how to write a procedure for this
  
  Step 1 make the route to the routing table. (By cursor we can get the route_id accordingly)
  
  Step 2 How to trigger custom backend package and run accordingly to this output of the cursor (route_id)
  
  If 40 routes of cursor get is - that the simultaneous program runs 40 times according to this route_id.
  
  
  can some could provide the steps to do this
  
  
  Thanks and greetings
  
  Srikkanth.M

  To submit a competing request from the back - end:

  FND_REQUEST. SUBMIT_REQUEST (Client or server)

  Summary

  function FND_REQUEST. SUBMIT_REQUEST

  (application IN varchar2 default NULL,

  program IN varchar2 NULL by default,

  Description IN varchar2 default NULL,

  start_time IN varchar2 default NULL,

  sub_request IN default boolean FALSE

  Argument1,

  argument2,..., argument99.

  Return to argument100 number);

  Description

  Submits a competing treatment by a simultaneous Manager. If the query is successful, this function returns the ID of the concurrent request; Otherwise, it returns 0.

  ATTENTION: FND_REQUEST needs to know information about the user and accountability whose application is submitted. Therefore, this feature works of concurrent programs or forms within the Oracle Applications.

  The FND_REQUEST. SUBMIT_REQUEST function returns the ID of the concurrent application after successfully. It is up to the caller to issue a commit to complete the application.

  Your code should retrieve and handle the error message generated if there is a problem of presentation (the ID of the concurrent request returned is 0). Use FND_MESSAGE. RETRIEVE and FND_MESSAGE. ERROR to retrieve and display the error (if the application is made on the client side).

  Related essays: overview of the Message dictionary (see page)

  You must call FND_REQUEST. SET_MODE before calling FND_REQUEST. SUBMIT_REQUEST of a database trigger.

  If FND_REQUEST. SUBMIT_REQUEST fails to go anywhere but a database trigger, database changes are cancelled until the time of the function call.

  After a call to the FND_REQUEST. SUBMIT_REQUEST function, installation of all parameters are reset to their default values.

  Arguments (input)

  short name of the application associated with the concurrent request for enforcement.
  short simultaneous program (not the executable) name of the program for which the application must be made.
  Description Description of the application that appears in the form of concurrent requests (optional).
  start_time time during which demand is expected to start running in the (optional) HH24 or HH24:MI:SS format.
  sub_request set to TRUE if the request is made by another application and should be treated as a subquery.
  From version 11, this parameter can be used if you submit requests for in a concurrent program of PL/SQL stored procedure.
  argument1... 100 arguments for the concurrent request; up to 100 arguments are allowed. If the Oracle Forms submitted, you must specify all arguments of 100.

• Put a unique number on each form is submitted

  My Department is new to Live Cycle Designer.  We created a PDF using LCD ES form that is used to approve the purchase of certain types of equipment.  I was prompted to change to include a unique identification number, and I need advice on where to start.

  The user goes on the web opens the PDF homologation form, fills out and clicks a button to send (on the form) that fits over the filled PDF an email that is sent to our office.  Once the form comes to us, it must be distributed to the other 4 or 5 oversight offices, and we want to identify each shape with a unique number, so we can monitor his progress.

  So what I was asked to do is to find a way to put a unique number generated automatically on the form at one point before it is emailed to us.  The number may have a number any numbers or characters (for as long as it is readable by humans) and does not need to be sequential.  I think a 5 or 6 digit would be good.

  I thought that the right approach would be to create a small Access database to generate/store identification numbers and then get the form of capture/read a new number every time that you click the submit button.  While I can connect to my database of test very well, what I do understand not how to execute SQL of LCD, or how to do it after the events, as when the form is opened or when the user clicks the submit button.  I didn't see any chance of finding similar good one so far.

  Is this the right way to go, or is there a better way to do it in LCD?

  Thanks for your time,

  -Max

  I created an example showing the various options that are available. Note that if you connect an external source such as a database and your users use player then you have to drive to extend the form using LiveCycle to allow this.

  The file that I was talking about is attached. If you need the process and controls for the DB I mention just ask and I'll post those as well. You will need LiveCycle server for this part.

  Paul

• Generation of a unique number on the columns.

  Hello
  
  Please check this query,
  
  < pre >
  
  SELECT DISTINCT A.LKP_ID AS ID
  D.PART_CNDTN_CD as a CONDITION
  D.PART_DISPOSTN_CD AS PROVISION
  C.LKP_NM AS DESCRIPTION
  OF FWOWNER. CLM_LKP_VALUE HAS
  FWOWNER. CLM_LKP_USAGE B
  FWOWNER. C CLM_LKP_VALUE_LANG
  IWOWNER. WC_CLM_SRVC_PART D
  WHERE B.LKP_USED_AS IN ('RR_CAUSAL_PART_CNDTN_CODE', 'rr_parts_disposition')
  AND A.LKP_REF BETWEEN B.LKP_REF AND B.LKP_REF + '.' 99999
  and A.LKP_ID = C.LKP_ID
  AND C.LANG_CD = 'ENG_USA. '
  AND A.LKP_LVL = 3
  and A.LKP_VALUE = BOX
  WHEN B.LKP_USED_AS IN ('RR_CAUSAL_PART_CNDTN_CODE')
  THEN D.PART_CNDTN_CD
  WHEN B.LKP_USED_AS IN ('rr_parts_disposition')
  THEN D.PART_DISPOSTN_CD
  END
  
  < / pre >
  
  
  top query let me out like,
  
  < pre >
  AVAILABLE FOR THE ID CODE DESCRIPTION
  ----- ------- ---------- ---------------------------------
  CP07 25774 REVAMPED bad calibration
  25789 CP22 leak (null)
  25795 CP28 scars / covered
  25796 CP29 (null) Warped/distorted/curly
  25768 CP01 PROCESSED Bent/pleated/Fold/bent/blister
  CP10 25777 chipped
  Contamination CP12 25779
  25786 CP19 grooved/crosses (null)
  25793 pitted CP26 (null)
  25768 CP01 (null) Bent/Creased/Fold/twisted/Crinkled
  25777 chipped CP10 (null)
  CP11 25778 Circuit (null) Bad
  25783 dented CP16 (null)
  25790 obsolete CP23 / supplant (null)
  25797 CP30 (null) worn too
  25798 CP31 (null) bad part
  * 25799 CP32 (null) other *.
  * Other CP32 25799 *.
  28225 CP07 REVIEW REVIEW
  25771 CP04 Broken/fractured (null)
  25779 CP12 Contamination (null)
  25780 CP13 corroded/eroded (null)
  25781 CP14 Cracked (null)
  25785 CP18 Galled/frettees (null)
  25788 CP21 (null) failure to point
  25795 CP28 Scarred/Spalled (null)
  CP30 25797 door too
  28225 CP01 REVIEW REVIEW
  
  28 response (s) selected [extract metadata: 0/ms] [extract data: 0/ms]
  
  
  < / pre >
  
  Please see the part "BOLD", I get two duplicates with a vertical drop of NULL and empty space I want to generate a unique number on the CONDITION column and available how to do this.
  
  
  (Please check my previous thread of reference
  MERGE TWO COLUMNS?
  )

  This is not Oracle...

  In any case:

  SELECT  DISTINCT A.LKP_ID          AS ID
  ,       D.PART_CNDTN_CD          as CONDITION
  ,       CASE WHEN LENGTH(D.PART_DISPOSTN_CD)=0 or D.PART_DISPOSTN_CD IS NULL THEN null ELSE D.PART_DISPOSTN_CD END  AS DISPOSITION
  ,       C.LKP_NM AS DESCRIPTION
  FROM     FWOWNER.CLM_LKP_VALUE          A
  ,     FWOWNER.CLM_LKP_USAGE          B
  ,     FWOWNER.CLM_LKP_VALUE_LANG      C
  ,     IWOWNER.WC_CLM_SRVC_PART     D
  WHERE     B.LKP_USED_AS          IN ( 'RR_CAUSAL_PART_CNDTN_CODE', 'rr_parts_disposition')
  AND     A.LKP_REF          BETWEEN     B.LKP_REF     AND     B.LKP_REF + '.99999'     -- No quotes
  and     A.LKP_ID          = C.LKP_ID
  AND     C.LANG_CD          = 'ENG_USA'
  AND     A.LKP_LVL          = 3
  and      A.LKP_VALUE          = CASE
                           WHEN  B.LKP_USED_AS     IN ( 'RR_CAUSAL_PART_CNDTN_CODE')
                           THEN  D.PART_CNDTN_CD
                           WHEN  B.LKP_USED_AS     IN ('rr_parts_disposition')
                           THEN  D.PART_DISPOSTN_CD
                        END
  

  Max
  [My Italian blog Oracle | http://oracleitalia.wordpress.com/2010/02/07/aggiornare-una-tabella-con-listruzione-merge/]