How to split the string?

[Former Member]

Hi Experts,

Can anyone explain me with code how to separate the single character from the given input string?

For Example: Input String str = 'RAGHU'

Output should be:

R

RA

RAGH

RAGHU

RAGH

RAG

RA

R

Can anyone help me with code for above required output?

Thanks in Advance,

Regards,

Raghu.

[Former Member]

TRY THIS

DATA : TEXT(10) VALUE 'RAGHU',
       LEN TYPE I,
       POS TYPE I,
       LNG TYPE I,
       VCH(10).
LEN = STRLEN( TEXT ).
LNG = LEN * 2.
LNG = LNG - 1.
DO LNG TIMES.
IF SY-INDEX LE LEN.
POS = POS + 1.
ELSE.
POS = POS - 1.
ENDIF.
VCH = TEXT+0(POS).
WRITE : / VCH.
ENDDO.

REGARDS

SHIBA DUTTA

[Former Member]

Hi,

Really the code you given works fine.And I felt that the code given by you is good compared to others.

Yesterday I tried that code and understand myself.Its really coooooooooooool.

Once again thanks for answering me with optimized code.

[Former Member]

DATA: s TYPE string VALUE 'RAGHU',

int TYPE i ,

count type i.

int = STRLEN( s ) .

DO int TIMES.

count = sy-index.

WRITE:/ s(count).

ENDDO.

DO int TIMES.

count = int - sy-index.

WRITE:/ s(count).

ENDDO.

Rewards if useful...........

Minal

[Former Member]

Hi,

Thanks for replying me.Can you explain me the logic inside the Do Loop?

Can you explain me theoretically for below part of the code?What SY-IDNEX will returns?

IF SY-INDEX LE LEN.

POS = POS + 1.

ELSE.

POS = POS - 1.

ENDIF.

VCH = TEXT+0(POS).

[Former Member]

Hii

data str type string.

data: len type i,

I TYPE I.

str = 'RAGHU'.

len = strlen( str ).

I = 1.

do len times.

write :/1(I) str.

I = I + 1.

enddo.

I = LEN.

do len times.

write :/1(I) str.

I = I - 1.

enddo.

[Former Member]

HI,

sy-index will give u the iteration number.

first of all in this program u are counting the length of the string and it was in variable LEN.

we are always printing the string from 0 index(1st character) to POS number of characters.

length is 5 so LNG = 5 * 2 = 10 and LNG = 10 - 1 = 9.and first sy-index = 1.so,1 <= 5 so POS = 0 + 1 = 1. it is printing the first char

second sy-index = 2.so,2 <= 5 so POS = 1 + 1 = 2. it is printing the first two chars

third sy-index = 3.so,3 <= 5 so POS = 2 + 1 = 3. it is printing the first three chars

forth sy-index = 4.so,4 <= 5 so POS = 3 + 1 = 4. it is printing the first four chars

fifth sy-index = 5.so,5 <= 5 so POS = 4 + 1 = 5. it is printing the first five chars

sixth sy-index = 6.so,6 > 5 so POS = 5 - 1 = 4. it is printing the first four chars

seventh sy-index = 7.so,7 > 5 so POS = 4 - 1 = 3. it is printing the first three chars

eighth sy-index = 8.so,8 > 5 so POS = 3 - 1 = 2. it is printing the first two chars

ninth sy-index = 9.so,9 > 5 so POS = 2 - 1 = 1. it is printing the first char

rgds,

bharat.

[Former Member]

in do loop sy-index is giving the counter of the loop . i.e. how much time it processes the loop.

so from first to the last of the string it should increase 1 character printing in each loop pass.

But when it reaches last of the string it should decrease 1 character from the printing.

so by i am adding one upto when the loop pass is reaching to the last. Then

I am subtracting from the total length.

regards

shiba dutta

[Former Member]

suppose in ur selection screen the field is :

S_name:raghu

code :

data : v_char type string.

v_char = s_name+0(1).Then v_char will have R

v_char = s_name+0(2).Then v_char will have RA

v_charg = s_name+0(3).Then v_char will have RAG.

LIKE THIS YOU HAVE TO PROCEED.

this CODING HELPS YOU. Give marks if it helps