String comparision

[Former Member]

Hi,

I need a progarm on string comparision.

>if the string are not identical we need to count the number of alphabets which are same in both the strings

>we need to write the alphabet with its respective count number

[Former Member]

hi check it once

REPORT Z_SAMPLE.

data : string1 type string value 'Ram uh Rag',

string2 type string value 'Raghu'.

data: len type i,

len1 type i,

n type i,

m type i,

count type i.

if string1 eq string2.

write 'strings r identical'.

else.

len = strlen( string1 ).

len1 = strlen( string2 ).

n = 0.

count = 0.

while n < len.

m = 0.

while m < len1.

if string1n(1) = string2m(1).

count = count + 1.

write 😕 count,string1+n(1).

endif.

m = m + 1.

endwhile.

n = n + 1.

endwhile.

endif.

[Former Member]

Hi suneetha

Here are string operations

Here are those

String Operations

Concatenate

Split

Shift

Replace

Translate

Offset

String length

String comparision

Concatenate

data : a(10),b(10),c(10),d(40).

A = ‘Apple’. B = ‘Orange’. C = ‘Banana’.

Concatenate A B C into D.

Write:/ d.

Concatenate A B C into D separated by ‘/’.

Write:/ d.

Split

data : a(10),b(10),c(10),d(40).

D = ‘Apple/Orange/Banana’.

Split d at ‘/’ into a b c.

Write:/ a

/ b,

/ c.

Shift

Data : a(6) value ‘ABCDEF’.

Shift a.(by default shifts to left by one place)

write: / a. BCDEF

A = ‘ABCDEF’.

Shift a by 2 places.

Write:/ a. CDEF

Data : a(6) value ‘ABCDEF’.

Shift a right.

Write:/ a. ABCDE

A = ‘ABCDEF’.

Shift a right by 2 places.

Write:/ a. ABCD

Data : a(6) value ‘ABCDEF’.

Shift a up to ‘C’. Write:/ a. CDEF

a = ‘ABCDEF’.

Shift a circular. Write:/ a. BCDEFA

Replace

Data p(6) value ‘ABCABC’.

Replace ‘ABC’ with ‘DEF’ into p.

Write:/ p. DEFABC

(Replaces first occurrence only)

Translate

Data p(11) ‘ABC ABC ABC’.

Translate p using ‘ADBECF’.

WRITE:/ P. DEF DEF DEF

Offset

Data p(6) value ‘ABCDEF’, q(3).

q = p+2(3).

Write 😕 q. CDE

q = p+0(1).

Write:/ q. A

String Length.

Data: a(50) value ‘PQRPQRPQRXYZ’,

b type i.

b = strlen( a ). 15

String Comparison

Contains any ‘ca’

Contains only ‘co’

Contains string ‘cs’ and

Contains pattern ‘cp’.

Contains any (ca) [ case sensitive ]

If ‘SAP’ ca ‘ABAP/4’.

Write:/ ‘True’.

Else.

Write:/ ‘False’.

Endif.

True

Contains only (co) [ case sensitive ]

If ‘SAP’ co ‘ABAP/4’.

Write:/ ‘True’.

Else.

Write:/ ‘False’.

Endif.

False

Contains string (cs) [ not case sensitive ]

If ‘ABAP/4’ cs ‘Ab’.

Write:/ ‘True’.

Else.

Write:/ ‘False’.

Endif.

True

Contains pattern (cp) [ not case sensitive ]

If ‘ABAP/4’ cp ‘*aP++’.

Write:/ ‘True’.

Else.

Write:/ ‘False’.

Endif.

True

check this sample code

DATA: V_CHAR(4).

DATA : v_string TYPE string.
DATA:BEGIN OF itab OCCURS 0,
matnr LIKE mara-matnr,
meins LIKE mara-meins,
mhdhb LIKE mara-mhdhb,
END OF itab.

V_CHAR = ITAB-MHDHB.

CONCATENATE itab-matnr v_maktx itab-meins V_CHAR
INTO v_string SEPARATED BY '|'.

<b>2nd program</b>

data: l_char(10) type c value '23.54',
      l_pchar(8) type p decimals 2.
 
l_pchar = l_char.
if l_pchar gt 1.
  write:/ 'Success'.
endif.

Reward all helpfull answers

Regards

Pavan

[Former Member]

data: a type char20 value 'abcde',

b type char10 value 'as',

g type char1.

data: c type i,

d type i,

e TYPE i,

f type i.

c = STRLEN( a ).

d = STRLEN( b ).

do d times.

e = e + 1.

g = b+f(e).

f = f + 1.

do c TIMES.

if g ca a.

write / g.

else.

write 😕 'fail' .

endif.

enddo.

enddo.

try this code

regards,

venkatesh

[Former Member]

Hi,

Take 2 Strings and put the values in those strings then

IF STRING1 CO STRING2.

Write:/ 'Both are same'.

ELSE.

Write:/ 'Both are not same'.

ENDIF.

<f1> CO <f2>

is true if <f1> contains only characters from <f2>. The comparison is case-sensitive. Trailing blanks are included. If the comparison is true, the system field SY-FDPOS contains the length of <f1>. If it is false, SY-FDPOS contains the offset of the first character of <f1> that does not occur in <f2> .

Regards

Sudheer

[Clemenss]

Hi Sudheer,

IF 'ABC' CO 'CBA'.

Write:/ 'Both are same'.

Sorry, I don't believe.

Please read more about striing comparisons:

<a href="http://help.sap.com/saphelp_erp2005vp/helpdata/en/fc/eb3516358411d1829f0000e829fbfe/content.htm">Comparisons Between Character Strings and Byte Strings</a>

Regards,

Clemens

[Former Member]

TRY THIS

DATA : BEGIN OF ICHAR OCCURS 0,
       ALPH,
       COUNT TYPE I,
       END OF ICHAR.
DATA : TEXT1(30) VALUE 'MATCHING STRING',
       TEXT2(30) VALUE 'SEARCH STRING',
       V_CHAR,
       LEN TYPE I,
       POS TYPE I.
CONDENSE TEXT1 NO-GAPS.
LEN = STRLEN( TEXT1 ).

DO LEN TIMES.
V_CHAR = TEXT1+POS(1).

IF TEXT2 CA V_CHAR.
READ TABLE ICHAR WITH KEY ALPH = V_CHAR.

ICHAR-COUNT = ICHAR-COUNT + 1.

IF SY-SUBRC = 0.
MODIFY ICHAR INDEX SY-TABIX.
ELSE.
ICHAR-ALPH = V_CHAR.
APPEND ICHAR.
ENDIF.
ENDIF.
POS = POS + 1.

ENDDO.
SORT ICHAR BY ALPH.
LOOP AT ICHAR.
WRITE : / ICHAR-ALPH, ICHAR-COUNT.
ENDLOOP.

REGARDS

SHIBA DUTTA