string

[sushant_singh]

hi ,

in my program the requirement is to find out the no. of times a character is present in a string. how can i do this.

plz send the code.

thanks .

[Former Member]

same thread----

data:  var1(30)  type c value 'ghghj#ghjgjgh#ghjghjg#ghjg#'.
 
data : totcnt type i,
       cnt type i, 
       v type c,
       n type i.
 
cnt = strlen( var1 ).
 
 
do cnt times.
move var1+n(1) to v.
if v eq '#'.
totcnt = totcnt + 1.
endif.
n = n + 1.
if n = cnt .
exit.
endif.
enddo.
 
write:/ 'No of #s', totcnt .

kishan negi

[Former Member]

try like this

data : len type i,

count type i,

pos type i ,

val.

compute len = strlen( str1 ).

do len times.

val = str1+pos(1).

if val = <your required char>.

count = count + 1.

endif.

pos = pos + 1.

enddo.

regards

shiba dutta

[Former Member]

DATA: L_NAME TYPE STRING.

DATA: TEMP TYPE STRING.

DATA: CNT TYPE I.

DATA: FL(1).

CLEAR.

WHILE FL <> 'X'.

SEARCH L_NAME FOR 'A'.

IF SY-SUBRC = 0.

SPLIT FILE AT 'A' INTO PATH FILE.

CNT = CNT + 1.

ELSE.

FL = 'X'.

ENDIF.

ENDWHILE.

WRITE: CNT.

[Former Member]

Hi,

Check this example

DATA: v_string TYPE string.

DATA: v_index TYPE sy-index.

DATA: v_count TYPE sy-index.

DATA: v_char.

DATA: v_len TYPE int4.

• String

v_string = 'AadfdAAA'.

• Get the length

v_len = strlen( v_string ).

• Calculate the times.

DO v_len TIMES.

v_index = sy-index - 1.

v_char = v_string+v_index(1).

IF v_char = 'A'.

v_count = v_count + 1.

ENDIF.

ENDDO.

WRITE: / 'Number of times A ',v_count.

**I am checking how many times A has been present in the string.

Thanks,

Naren

[Former Member]

DATA: L_NAME TYPE STRING.

DATA: TEMP TYPE STRING.

DATA: FL(1).

DATA: CNT TYPE I.

TEMP = L_NAME.

WHILE FL <> 'X'.

SEARCH TEMP FOR 'A'.

IF SY-SUBRC = 0.

SPLIT FILE AT 'A' INTO L_NAME TEMP.

CNT = CNT + 1.

ELSE.

FL = 'X'.

ENDIF.

ENDWHILE.

WRITE: CNT.

[Former Member]

use this demo code -

PARAMETERS : P_CHAR TYPE C LOWER CASE.

data str type string.

dATA: LEN TYPE I,

COUNT TYPE I,

PASS TYPE I VALUE '0'.

str = 'amita'.

CONDENSE STR.

LEN = STRLEN( STR ).

DO LEN TIMES.

IF P_CHAR = STR+pass(1).

COUNT = COUNT + 1.

ENDIF.

pass = pass + 1.

ENDDO.

[Former Member]

hi try this

data : str(20) type c value 'ABABCASDAFERA',

cnt type i,

i type i ,

len type i,

val type c.

len = strlen( str ).

do len times.

val = str+i(1).

if val eq 'A'.

cnt = cnt + 1.

endif.

i = i + 1.

enddo.

write cnt.