Regarding finding the mismatching characters in two strings.

[Former Member]

Hi all,

Please tell me how to find the differing characters between two strings.

For ex str1 = 'asdwe'.

str2 = 'asd'.

Then i should get 'we' as result.

Thanks.

[Former Member]

Probably the safest way would be to process both string character by character, something like this:

DO.
offset = offset + 1.
lv_char1 = str1+offset(1).
lv_char2 = str2+offset(1).

IF lv_char1 <> lv_char2.
***do whatever you want***
ENDIF.
ENDDO.

[Sm1tje]

Probably the safest way would be to process both string character by character, something like this:

DO.
offset = offset + 1.
lv_char1 = str1+offset(1).
lv_char2 = str2+offset(1).

IF lv_char1 <> lv_char2.
***do whatever you want***
ENDIF.
ENDDO.

[GauthamV]

Use comparison operators.

[http://help.sap.com/saphelp_erp2004/helpdata/EN/fc/eb3516358411d1829f0000e829fbfe/frameset.htm]

[Former Member]

Hi,

use the FIND statement to find the string usbset. once found, find the offset at which the subset occurs. The startingt ta offset u can read the remaining characters.

[Former Member]

Hi,

Use the following code,

data: str1(70) VALUE 'asdwedkre[qriomaaaaaawfhupfewpfjsnjiheuwiqfjaaaaa',
      str2(10) VALUE 'a'.


replace ALL OCCURRENCES OF str2 in str1 WITH ''.   

the output is sdwedkre[qriomwfhupfewpfjsnjiheuwiqfj without any blacks or spaces.

Edited by: Rajat Chaturvedi on Mar 16, 2009 11:54 AM

[Former Member]

Hi,

This is very simple, hope the below snippet help you out:

data:String1(5) type c value 'asdwe',
     String2(3) type c value 'asd',
     String3(2) type c.

if String1 CN String2.
String3 = String1+3(2).
endif.

write:/ String3.

PS:if the lenght of the strings are not known then find out the lenght of the 2 strings and use the offset from the fixed value to the length of the strings calculated.

Moreover I am assuming that if String1 contains String2 in the starting position only.

Pooja

[Former Member]

try this:

calculate the length of the 1st string: v_len1

calculate the length of the 2nd string: v_len2

now,

v_len1 = STRLEN( v_text1)

v_len2 = STRLEN( v_text2)

v_next1 = 0.

v_count = 0.

v_next2 = 0.

v_count = 0.

• comparing each letter in 1st string with each letter in 2nd string.

DO v_len1 TIMES.

DO v_len2 TIMES.

if v_textv_next1(v_len1) <> v_text2v_next2(v_len2). v_text3 = v_text+v_next(v_len1).

ENDIF.

v_next2 = v_next2+1.

ENDDO.

v_next1 = v_next1+1.

ENDDO.

v_text3 will hve the difference.

[Former Member]

hi,

do like this,

var1 = asd.

var2 = asdewasdew..

then take strlen(var1) into count.

then count holds value 3.

i =1.

do count times..

var3 = var1+0(i).

i = i + 1.

replace all occurences of var3 in var2 by space..

codense var2.

enddo.

• Here var3 holds value 'a' , in var2 a is been replaced by space and var2 holds sdewsdew..

same way repeat for other two characters also then u r left with ewew..

i think this is ur requireed output..

Rgds.,

subash

[Former Member]

sorry small change in logic

var1 = asd.

var2 = asdewasdew..

then take strlen(var1) into count.

then count holds value 3.

i =1.

a= 0.

do count times..

var3 = var1+a(i)

i = i + 1.

a= a + 1.

replace all occurences of var3 in var2 by space..

codense var2.

enddo.

[Former Member]

check this code.

DATA: st1 TYPE char3 VALUE 'asd', "Important: type length should be equal to value
      st2 TYPE char10 VALUE 'asdwe'.


WRITE: / st2.   "output: asdwe
CONDENSE st1.
REPLACE st1 WITH space INTO st2.
CONDENSE st2.
WRITE: / st2.   "output: we

кu03B1ятu03B9к

[former_member222860]

here's the logic:

DATA: counter TYPE i value '5',
      flag,
      str1 TYPE char12 VALUE 'asdwew',
      str2 TYPE char12 VALUE 'asd',
      str3 TYPE char12,
      len type i.

DO counter TIMES.
  len = strlen( str1 ).
  IF str1(sy-index) = str2(sy-index).
    flag = 'X'.
   ELSE.
    sy-index = sy-index - 1.
    move str1+sy-index(len) to str3.
    write:/ str3.
    EXIT.
  ENDIF.
ENDDO.

thanks\

Mahesh

[Former Member]

Hi,

Try this.

DATA : lv_lenstr1 TYPE i,

lv_lenstr2 TYPE i,

idx_i TYPE i,

idx_j TYPE i,

lv_str3 TYPE string,

lv_char1 TYPE c,

lv_char2 TYPE c.

PARAMETERS : p_str1 TYPE string,

p_str2 TYPE string.

lv_lenstr1 = STRLEN( p_str1 ).

lv_lenstr2 = STRLEN( p_str2 ).

IF ( lv_lenstr1 GT lv_lenstr2 ). "if first string is greater than second

"if second string is greater interchange

"first and second string

CLEAR : idx_i,

idx_j.

DO lv_lenstr1 TIMES.

lv_char1 = p_str1+idx_i(1).

CLEAR : idx_j.

DO lv_lenstr2 TIMES.

IF idx_i LE lv_lenstr2.

lv_char2 = p_str2+idx_j(1).

IF lv_char1 EQ lv_char2.

EXIT.

ELSE.

IF p_str2 CA lv_char1.

EXIT.

ELSE.

CONCATENATE lv_str3 lv_char1 INTO lv_str3.

EXIT.

ENDIF.

ENDIF.

idx_j = idx_j + 1.

ENDIF.

ENDDO.

IF idx_i GT lv_lenstr2.

IF p_str2 CA lv_char1.

idx_i = idx_i + 1.

CONTINUE.

ELSE.

CONCATENATE lv_str3 lv_char1 INTO lv_str3.

ENDIF.

idx_i = idx_i + 1.

ELSE.

idx_i = idx_i + 1.

ENDIF.

ENDDO.

ENDIF.

WRITE lv_str3.

Regards,

Roopa

[Former Member]

declare two internal tables.

data:begin of lt_string1 occurs 0,

line(1) type c,

end of lt_string1.

data: begin of lt_string2 occurs 0,

line(1) type c,

end of lt_string 2.

data: lv_char(10) type c value 'abcd'

data: lv_ch(10) type c value 'ab'.

lt_string1-line = lv_char+0(1)

append lt_string1.

lt_string1-line = lv_char+1(1)

append lt_string1.

and so on....

do the same thing with another internal table .

now the two tables have structures like this

lt_string1 lt_string2

-

---

-

---

a a

b b

c

d

now

loop at lt_string1.

read table lt_string2 with key line = lt_string1-line.

if sy-subrc <> 0.

lt_string3-line = lt_string1-line

append lt_string3.

endif.

lt_string3 will have c and d

reg