internal table concatenation

[Former Member]

How can I concatenate the data where CUDC = 1. Sample data:

CUDC RECORD DATA

1 1 002232:2:1-002223:2:3-002346:2:2-002306:2:3-002314:6:8-002321:6:8-002346:6:6-002333:7:3-002433:8:3-

002433:9:3-002413:10:10-002435:10

1 2 :3-002431:19:3-002430:20:3-002431:21:3-002431:22:3

expected result

1 1 [0022]32:2:1-23:2:3-46:2:2-[0023]06:2:3-14:6:8-21:6:8-46:6:6-[0024]33:7:3-33:8:3-33:9:3-13:10:10-35:10:3-

31:19:3-30:20:3-31:21:3-31:22:3

thanks in advance.

Edited by: kapishr on Jan 26, 2012 3:22 PM

[Former Member]

let itab1 and itab2 both be copies of same data.

sort itab1 by cudc and delete adjacent duplicates comparing cudc.

loop at itab1.

loop at itab2 where cudc = itab1-cudc.
   concatenate itab1-record itab2-record to itab1-record.
endloop.

modify itab1. or use field symbols.
endloop.

above code is psuedocode, check respective syntaxes while implementing.

[Former Member]

let itab1 and itab2 both be copies of same data.

sort itab1 by cudc and delete adjacent duplicates comparing cudc.

loop at itab1.

loop at itab2 where cudc = itab1-cudc.
   concatenate itab1-record itab2-record to itab1-record.
endloop.

modify itab1. or use field symbols.
endloop.

above code is psuedocode, check respective syntaxes while implementing.

[Former Member]

hi kapishr,

can you make your tables more readable?

[Former Member]

The three column of internal table are CUDC RECORD and DATA

CUDC

1

1

2

2

3

RECORD

1

2

1

2

1

Data

002232:2:1-002223:2:3-002346:2:2-002306:2:3-002314:6:8-002321:6:8-002346:6:6-002333:7:3-002433:8:3-002433:9:3-002413:10:10-002435:10

:3-002431:19:3-002430:20:3-002431:21:3-002431:22:3

002232:2:1-002223:2:3-002346:2:2-002306:2:3-002314:6:8-002321:6:8-002346:6:6-002333:7:3-002433:8:3-002433:9:3-002413:10:10-002435:10

:3-002431:19:3-002430:20:3-002431:21:3-002431:22:3:002431:19:3-000030:20:3-0000031:21:3-000031:22:3:000031:19:3-000030:20:3-000031:21:3

002431:19:3-002430:20:3-002431:21:3-002431:22:3

I have to concatenate the data based on CUDC and sort the record column accordingly

Thanks and regards

Edited by: kapishr on Jan 26, 2012 4:43 PM

Edited by: Rob Burbank on Jan 26, 2012 11:47 AM

[former_member777515]

Assuming the DATA has a field length of 20, and for each CUDC. its corresponding DATA has values of maximum 20... Therefore, may I ask how should your concatenated result be like?

I simplify your internal table as such(using ~ as delimiter), correct me if I'm wrong

CUDC(x1)_RECORD(x1)DATA(x6)

1₁AAAAA

1₂BBBBBB

2₁CCCCCC

2₂DDDDDD

3₁EEEEEE

Using the above example, please explain how should the result be?

Edited by: bloodcool on Jan 27, 2012 11:44 AM

[Former Member]

Hello Kapishr,

Here is a code you can use :

data : wa_tab like line of itab, "work area of your itab.
          temp1(50) type c, "itab - your internal table that is containing above fields.
          temp2(50) type c,  "assuming data length can not be more than 50 chars.
           " 
           "
         tempn(50) type c.


Loop at itab into wa_tab.
     Case wa_tab-CUDC.
            When '1'
                      Concatenate temp1 wa_tab-DATA into temp1.
            When '2'
                       Concatenate temp2 wa_tab-DATA into temp2.
             "
             "
             "
            When 'n'
                        Concatenate tempn wa_tab-DATA into tempn.
     Endcase.
Endloop.

After using above code you can use temp1-temp2-......tempn data wherever you want.

Kind Regards,

Kerim.

Edited by: Kerim Merih Kılıç on Jan 27, 2012 7:23 AM

[Former Member]

Thanks for helping me, I am new to programming and so I m having trouble explaining..

This is the data that I have

CUDC_RECORDDATA~LENGTH

1₁002232:2:1-002223:2:3-002246:2:2-002306:2:3-002314:6:8-002321:6:8-002346:6:6-002433:7:3-002433:8:3-002433:9:3-002413:10:10-002435:10~132

1₂:3-002431:19:3-002430:20:3-002431:21:3-002431:22:3~50

2₁002232:2:1-002223:2:3-002246:2:2-002306:2:3-002314:6:8-002321:6:8-002346:6:6-002433:7:3-002433:8:3-002433:9:3-002413:10:10-002435:10~length

2₂:3-002431:19:3-002430:20:3-002431:21:3-002431:22:3-002431:19:3-000030:20:3-0000031:21:3-000031:22:3:000031:19:3-000030:20:3-000031:21:3~length

3₁002431:19:3-002430:20:3-002431:21:3-002431:22:3~length

Required output

CUDC_RECORDDATA~LENGTH

1₁(0022)32:2:1-23:2:3-46:2:2-(0023)06:2:3-14:6:8-21:6:8-46:6:6-(0024)33:7:3-33:8:3-33:9:3-13:10:10-35:10:3-31:19:3-30:20:3-31:21:3-31:22:3~102

2₁(0022)32:2:1-23:2:3-46:2:2-(0023)06:2:3-14:6:8-21:6:8-46:6:6-(0024)33:7:3-33:8:3-33:9:3-13:10:10-35:10:3-31:19:3-30:20:3-31:21:3-31:22:3-31:19:3-(0000)30:20:3-31:21:3-31:22:3-31:19:3-30:20:3-31:21:3~length

3₁(0024)31:19:3-30:20:3-31:21:3-21:22:3~length

the length at end of each row is for calculating the length of the data

The first six digits is for time hhmmss,

I need to club (hhmm)ss-ss... and so on and club the records based on CUDC as well.

the digits after colon( is not to be analysed.

thanks and regards.

[Former Member]

You can use the code I wrote above.

temp1-temp2-temp3 contains data that you want. If you don't want to concatenate data after ':' then you can take the part before it:.(ie concatenate tempx itab-DATA+6 into tempx)

Kind Regards,

Kerim.

[former_member777515]

you need to first breakdown data table into another internal table, before merging them back.

for identification sake, i named the table as below:

- CUDC_DATA, your original table

- CUDC_BREAKDOWN, the breakdown data table

- CUDC_RESULT, result table

1) Concatenate the DATA for each set of CUDC from CUDC_DATA table into a string variable (e.g. tmpstring)

2) Process each CUDC's tmpstring into CUDC_BREAKDOWN where you should

- get every single group of hhmmss:xx:xx, the delimiter should be '-'. Use SPLIT function

- break the group to 2 different fields, where the 1st 4 char(HHMM) is field #1, and the remaining field #2

- append the current CUDC value, field #1 and field #2 into CUDC_BREAKDOWN

3) Repeat the process until all CUDC's DATA has been breakdown into CUDC

Now, you should merge back the data back to CUDC_RESULT, from CUDC_BREAKDOWN table

4) For every CUDC, it would indicate a start of a new record

5) For each new set of field #1 encountered, concatenate the HHMM value to the data field in CUDC_RESULT

6) Then concatenate every field #2 to the data field in CUDC_RESULT

7) Do until all CUDC are completed.