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[Former Member]

Give the code for removing redundency of character's in a word.

[Former Member]

hI..

Data:

w_string(30) type c, " Copy Input String.

w_length type i, " Length of string.

w_char type c. " Character to check.

w_length = strlen( p_string ).

w_string = p_string.

write : /'Given string is:'(001),w_string.

write : /'Distinct Chars :'(002).

condense w_string no-gaps.

while w_length gt 0.

w_char = w_string+0(1).

write: w_char.

replace all occurrences of w_char in w_string with space.

condense w_string no-gaps.

w_length = strlen( w_string ).

endwhile.

Message was edited by:

Rammohan Nagam

[Former Member]

hi,

*"Selection screen elements............................................

PARAMETER

p_string(20) TYPE c. " String

*" Data declarations...................................................

"----

---

• Work variables *

"----

---

DATA:

w_index TYPE i , " Index

w_index1 TYPE i VALUE 1, " Index

w_strlen TYPE i. " String length

IF P_STRING CO 'ABCDEFGHIJKLMNOPQRSTUVWXYZ '.

w_strlen = STRLEN( p_string ).

*finding all distinct characters....................................

DO w_strlen TIMES.

IF w_index NE 0.

w_index1 = w_index - 1.

DO w_index TIMES.

IF p_stringw_index(1) EQ p_stringw_index1(1).

EXIT.

ENDIF. " ENDIF P_STRING..

SUBTRACT 1 FROM w_index1.

ENDDO. " ENDDO W_INDEX..

IF w_index1 EQ -1.

WRITE p_string+w_index(1).

ENDIF. " ENDIF W_INDEX1..

ADD 1 TO w_index.

ELSE.

WRITE p_string+w_index(1).

ADD 1 TO w_index.

ENDIF. " ENDIF W_INDEX..

ENDDO. " ENDDO W_STRLEN..

ELSE.

WRITE 'ENTER CHARACTERS ONLY'(001).

ENDIF.

regards,

ananth