How delete first two characters ?

[Former Member]

Hi all,

i have problem like this i want to delete first two (~~) characters . i am written concatenate statement in the loop so v_string taking spaces

so i want to remove this ~~ .

plz any tell how to do this.

this is code in the loop.

CONCATENATE v_string ','

itab-arktx

itab-vrkme

total

total1

itab-vat_rate

itab-matkl

total2

INTO v_string SEPARATED BY '~~'.

present ouput:

~_00500000123~ 2~_0003000963~~~test6

~_00500000123~ 2~_0003000963~~~test6

act output :

00500000123~~ 2~_0003000963~~~test6

00500000123~~ 2~_0003000963~~~test6

thaks for advance

regards

sai

[Former Member]

SHIFT str by 2 places

[Former Member]

add this

shift v_string by 2 places left deleting '~~'.

[Former Member]

hi sai,

you better use the SHIFT command.

i.e., Shift v_string by 2 places

see the below code.

CONCATENATE v_string ','

itab-arktx

itab-vrkme

total

total1

itab-vat_rate

itab-matkl

total2

INTO v_string SEPARATED BY '~~'.

SHIFT str by 2 places left deleting '~~

i didn't see the syntax.but check the syntax also.

if it is useful Reward me.

[former_member156446]

Hi Sai

CONCATENATE v_string ','

itab-arktx

itab-vrkme

total

total1

itab-vat_rate

itab-matkl

total2

INTO v_string SEPARATED BY '~~'.

lv_len = strln (v_string).

v_string = v_string+2(lv_len-2).

[Former Member]

Hi,

Sai, Will TRIM work in your case..

Regards,

Sai

[Former Member]

Sai

CONCATENATE v_string ','   "--> Guess at this point v_string is EMPTY, try to avoid this and check the result
   itab-arktx
   itab-vrkme
   total
   total1
   itab-vat_rate
   itab-matkl
   total2
   INTO v_string SEPARATED BY '~~'.

Regards

Eswar

[Former Member]

hi,

u can do like this:

DATA: str(10) TYPE c VALUE '1234567890'.

DATA: str1(10) TYPE c.

str1 = str+2(8).

This means that take character from third position, and 8 characters from the point. You can specify it as your field's length minus 2.

Where str can be ur string.

Regards,

Renjith Michael.

Edited by: Renjith Michael on Jan 18, 2008 11:00 AM

[Former Member]

Thank u for all