String India

[Former Member]

Hi,

Plz tell me how to write the program

1. Accept the String INDIA and Split as I

I N

I N D

I N D I

I N D I A

2. Accept the string and search character and display the number of occurances of that search character in that string.

Thanks

[Former Member]

Hi ,

try this

REPORT yyyvenkateshtesting .

DATA: c(5) TYPE c VALUE 'INDIA'.

DATA: a TYPE i.

DATA: b TYPE i.

a = 0.

b = 0.

DO 5 TIMES.

b = b + 1.

DO b TIMES.

WRITE:/ c+a(b).

exit.

ENDDO.

ENDDO.

reward points if helpful,

thanks & regards,

venkatesh.

[Former Member]

Hi rams,

You can search for using the string operators like CA, CO etc.

This will return the position found in system variable sy-fdpos.

Using that position you can split the string and manipulate as you wish.

Regards,

Yogesh

[varma_narayana]

Hi...

<b>1. Accept the String INDIA and Split as I

I N

I N D

I N D I

I N D I A</b>

<b>Check the Soution</b>

PARAMETERS: PSTR(10) DEFAULT 'INDIA'.

DATA : V_POS TYPE I VALUE 0,

V_LEN TYPE I.

V_LEN = STRLEN( PSTR ).

DO V_LEN TIMES.

WRITE:/ PSTR+V_POS(SY-INDEX).

ENDDO.

<b>2. Accept the string and search character and display the number of occurances of that search character in that string.</b>

Solution:

PARAMETERS: PSTR(10) DEFAULT 'INDIA',

PCHR DEFAULT 'I'.

DATA : V_POS TYPE I VALUE 0,

V_LEN TYPE I,

V_COUNT TYPE I.

V_LEN = STRLEN( PSTR ).

DO V_LEN TIMES.

IF PSTR+V_POS(1) = PCHR.

ADD 1 TO V_COUNT.

ENDIF.

ADD 1 TO V_POS.

ENDDO.

Write:/ 'No of occurrences', V_COUNT.

<b>Reward if Helpful</b>

<b></b>

[Former Member]

hi,

REPORT ZNAZSPLIT .

data : lv_i(5) type c value 'INDIA',

lv type c value 'I'.

data: lv_len type i,

i type i,

j type i,

k type i.

lv_len = strlen( lv_i ).

i = 1.

k = 0.

j = 0.

do lv_len times.

write : / lv_i+0(i).

i = i + 1.

if lv_i+k(1) = lv.

j = j + 1.

endif.

k = K + 1.

enddo.

write : / 'I occurs ' , j , 'times'.

copy and paste the program above.. your result occurs

rewards if useful,

regards,

Nazeer shaik

[Former Member]

Hi Ram,

<b>1. Accept the String INDIA and Split as </b>

PARAMETERS: S_NAME TYPE STRING.

DATA V_LENGTH TYPE I.

DATA V_INDEX TYPE SY-INDEX.

V_LENGTH = STRLEN( S_NAME ).

DO V_LENGTH TIMES.

V_INDEX = SY-INDEX.

DO V_INDEX TIMES.

V_POS = V_INDEX - 1.

WRITE: S_NAME+V_POS(1).

ENDDO.

ENDDO.

<b>2. Accept the string and search character and display the number of occurances of that search character in that string

</b>

PARAMETERS: P_NAME TYPE STRING.

PARAMETERS: P_SC.

DATA V_LENGTH TYPE I.

DATA V_COUNT TYPE I.

V_LENGTH = STRLEN( P_NAME ).

DO V_LENGTH TIMES.

V_INDEX = SY-INDEX - 1.

IF P_NAME+V_INDEX(1) = P_SC.

V_COUNT = V_COUNT + 1.

ENDIF.

ENDDO.

IF V_COUNT IS NOT INITIAL.

WRITE:/ 'The search character found ', V_COUNT, 'times'.

ELSE.

WRITE:/ 'Search character doest not find in the String'.

ENDIF.

Thanks,

Vinay

[Former Member]

Hi Rams,

PARAMETERS: STRING(5) DEFAULT 'INDIA'.

DATA : V_LEN TYPE I,

VAR TYPE I VALUE1.

V_LEN = STRLEN( STRING ).

DO V_LEN TIMES.

IF VAR < V_LEN.

WRITE:/ STRING + 0 ( VAR ).

NEW-LINE.

ENDIF.

VAR = VAR + 1.

ENDDO.

Reward points for all helpful answers,

kiran.M

[satykumar]

try this to split any string

DATA length TYPE i.

PARAMETERS name TYPE string.

length = STRLEN( name ).

DO.

DATA : v_str TYPE string,

v_substr TYPE string,

v_position TYPE i.

v_str = name.

v_position = v_position + 1.

v_substr = v_str+0(v_position).

WRITE 😕 v_substr.

IF v_position = length.

EXIT.

ENDIF.

regards,

Satyendra