add charecter

[Former Member]

Hi Guys:

I have to add one charecter inthe existing filed.

example:

data : begin of itab,

string_a(10) type c,

end of itab.

string_a = '1234567890'.

now i want to add X after 12345 it mean string_a = 12345X67890.I need exactly in 6th postion X and then modify the internal table.

Advanced Thanks

Suresh Babu Karanam

[Former Member]

Hi Guys:

I have to add one charecter inthe existing filed.

example:

data : begin of itab,

string_a(10) type c,

end of itab.

string_a = '1234567890'.

now i want to add X after 12345 it mean string_a = 12345X67890.I need exactly in 6th postion X and then modify the internal table.

Advanced Thanks

Suresh Babu Karanam

[Former Member]

but here string_a must be at least 11 characters.

data:

zi type i

zi = strlen( string_a ).

right = zi - 7.

string_a7(right) = string_a7.

string_a+6(1) = 'X'.

-

---

if it helps please give points.

Message was edited by: Fuat Ulugay

[Former Member]

now out is like string_a = 12345X7890. I am loosing actuval 6 number in the string.

[Former Member]

I modified it try it again with new code.

and tested it.

The code below works.

DATA:

zi TYPE i,

right TYPE i.

string_a = '1234567890'.

zi = strlen( string_a ).

right = zi - 6.

string_a7 = string_a6(right) .

string_a+6(1) = 'X'.

Message was edited by: Fuat Ulugay

[Former Member]

Thanks Fuat Ulugay. You are right i have increased string length.

loop at itab.

REPLACE FIRST OCCURRENCE OF '6' IN itab-string_a WITH 'X6'.

modify

endloop.

[Former Member]

Yes this is a better solution

shortest is the best.

REPLACE '6' into string_a WITH 'X6'.

[Former Member]

Hi Suresh,

Check the below code. It will give the exact result you need,

REPORT zc1test_string .

DATA : old_str(10) TYPE c,

new_str(11) TYPE c.

old_str = '1234567890'.

• This moves 5 characters to new_str

MOVE old_str TO new_str PERCENTAGE 50.

*now removing the first 5 chars in old_str

SHIFT old_str BY 5 PLACES LEFT.

*joining them together

CONCATENATE new_str 'X' old_str into new_str.

write:/ new_str.

In your program make sure the final field should be of size 11 to hold the extra value.

Thanks and Regards,

Kathirvel.

[Former Member]

Hi Suresh,

You can use the coding below.It is an example for both string and char variables :

DATA : lv_str TYPE string VALUE '1234567890',

lv_char TYPE char11 VALUE '1234567890'.

CONCATENATE lv_str+0(5)

'X'

lv_str+5(5)

INTO lv_str.

WRITE / lv_str.

CONCATENATE lv_char+0(5)

'X'

lv_char+5(5)

INTO lv_char.

WRITE / lv_char.