topic Re: Problem with the Read Statement in Application Development and Automation Discussions

Problem with the Read Statement

Former Member — Tue, 11 Sep 2007 05:49:06 GMT

Hi All,

I have one issue while using the read table .Iam using Read table with binary search .but even though there is matching key Sy-subrc is returning 4.Since I used the binary search there is no need of using sort table before read.

Can some one help me to solve this issue .

Thanks in Advance

-Padma

Re: Problem with the Read Statement

Former Member — Tue, 11 Sep 2007 05:50:48 GMT

before binary search sorting should be done.

Re: Problem with the Read Statement

Former Member — Tue, 11 Sep 2007 05:51:04 GMT

Hi,

u need to sort the table before using BINARY SEARCH in the read statement.

cud u pls give us the code....

here is a sample:

sort i_final by BUKRS .

LOOP AT i_final INTO w_final.

READ TABLE i_output INTO w_output WITH KEY

bukrs = w_final-bukrs

BINARY SEARCH.

ENDLOOP.

Regards,

Anjali

Re: Problem with the Read Statement

Former Member — Tue, 11 Sep 2007 05:51:36 GMT

HI,

You must sort the table with key which u r using in Read statement before using Binary search.

Regards,

Balavardhan.K

Re: Problem with the Read Statement

Former Member — Tue, 11 Sep 2007 05:51:37 GMT

example

FETCHING NAME1

SELECT werks

name1

FROM t001w

INTO TABLE t_name1

FOR ALL ENTRIES IN t_itab1

WHERE werks = t_itab1-werks.

SORT t_itab1 BY werks.

SORT t_name1 BY werks.

IF sy-subrc = 0.

LOOP AT t_itab1 INTO wa_itab1.

hold_tabix = sy-tabix.

READ TABLE t_name1 INTO wa_name1 WITH KEY werks = wa_itab1-werks

BINARY SEARCH.

IF sy-subrc = 0.

wa_itab1-name1 = wa_name1-name1.

MODIFY t_itab1 INDEX hold_tabix FROM wa_itab1.

ENDIF.

Re: Problem with the Read Statement

Former Member — Tue, 11 Sep 2007 05:52:20 GMT

HI,

SORTING IS A PRE REQUISIT FOR BINARY SEARCH. BEFORE REAT TABLE ST SORT UR INTERNAL TABLE WITH A KEY FIELD AND TRY.

WITH REGARDS,

SURESH ALURI.

Re: Problem with the Read Statement

Former Member — Tue, 11 Sep 2007 05:52:50 GMT

can you paste ur code

Re: Problem with the Read Statement

Former Member — Tue, 11 Sep 2007 06:01:58 GMT

Thanks all !!

The problem has been resolved!

Re: Problem with the Read Statement

Former Member — Tue, 11 Sep 2007 06:04:45 GMT

Hi,

<u><b>Binary Search in Standard Tables</b></u>

If you read entries from standard tables using a key other than the default key, you

can use a binary search instead of the normal linear search. To do this, include the addition BINARY SEARCH in the corresponding READ statements.

READ TABLE <itab> WITH KEY = <f> <result> BINARY SEARCH.

and

READ TABLE <itab> WITH KEY <k1> = <f1> ... <kn> = <fn> <result>

BINARY SEARCH.

The standard table must be sorted in ascending order by the specified search key. The BINARY SEARCH addition means that you can access an entry in a standard table by its key as quickly as you would be able to in a sorted table.

Example

DATA: BEGIN OF LINE,

COL1 TYPE I,

COL2 TYPE I,

END OF LINE.

DATA ITAB LIKE STANDARD TABLE OF LINE.

DO 4 TIMES.

LINE-COL1 = SY-INDEX.

LINE-COL2 = SY-INDEX ** 2.

APPEND LINE TO ITAB.

ENDDO.

SORT ITAB BY COL2.

READ TABLE ITAB WITH KEY COL2 = 16 INTO LINE BINARY SEARCH.

WRITE: 'SY-SUBRC =', SY-SUBRC.

The output is:

SY-SUBRC = 0

The program fills a standard table with a list of square numbers and sorts them into

ascending order by field COL2. The READ statement uses a binary search to look

for and find the line in the table where COL2 has the value 16.

Regards,

Bhaskar