Getting number from string?

[Former Member]

HI.

I have a question.

I'd like to get just number from string that consist of numbers and char.

for example. String A1313.33BBC

In this string, i want to get 1313.3. Is there any method for this?

thanks.

[Former Member]

Re: How to find number from text string??

Posted: Oct 10, 2008 1:21 PM in response to: Mrunal Mhaskar Reply

You can do like this.

The below mentioned is code separate material

DO 18 TIMES.

LV_C = LV_C + 1.

LV_C1 = LV_C - 1.

IF GW_MATNR-MATNR+LV_C1(1) CN '0,1,2,3,4,5,6,7,8,9' .

IF ( ( LV_MAT EQ SPACE ) ) ." and ( lv_prefix eq space ) ) .

LV_PREFIX = 'X'.

CONCATENATE GW_TEMP-PREFIX GW_MATNR-MATNR+LV_C1(1) INTO

GW_TEMP-PREFIX.

ELSE.

LV_SUFFIX = 'X'.

CONCATENATE GW_TEMP-SUFFIX GW_MATNR-MATNR+LV_C1(1) INTO

GW_TEMP-SUFFIX.

ENDIF.

ELSEIF ( ( LV_SUFFIX EQ SPACE ) ) .

LV_MAT = 'X'.

CONCATENATE GW_TEMP-MATNR1 GW_MATNR-MATNR+LV_C1(1) INTO

GW_TEMP-MATNR1.

ELSEIF ( ( LV_MAT EQ 'X' ) ) .

CONCATENATE GW_TEMP-SUFFIX GW_MATNR-MATNR+LV_C1(1) INTO

GW_TEMP-SUFFIX.

ENDIF.

ENDDO.

[Former Member]

Hello

Try:

data: str1(20),
      str2(60).

str1 = 'A1313.33BBC'.
str2 = 'A B C D E F G H I J K L M N O P Q R S T U V W X Y Z'.
TRANSLATE str1 USING str2.

[Former Member]

Hi,

Check the below code.

DATA: v_str TYPE string VALUE 'A1313.33BBC',

v_str1 TYPE string,

v_str2 TYPE string.

SPLIT v_str AT '.' INTO v_str1 v_str2.

REPLACE ALL OCCURRENCES OF REGEX '\D' IN v_str1 WITH ''.

REPLACE ALL OCCURRENCES OF REGEX '\D' IN v_str2 WITH ''.

IF sy-subrc NE 0.

REPLACE ALL OCCURRENCES OF REGEX '\D' IN v_str WITH ''.

ELSE.

CONCATENATE v_str1 v_str2 INTO v_str SEPARATED BY '.'.

ENDIF.

WRITE:/ v_str.

Regards,

Kumar Bandanadham

[Former Member]

Hi,

Use the below code;

Data : lv_string type string.
Data : lv_num type string.
lv_string = 'A1313.33BBC'.
lv_num = lv_string.
REPLACE ALL OCCURRENCES OF '[[:alpha:]]' IN lv_num WITH ''.

This will remove all other characters other than numbers and decimal points, now lv_num will have the numeric value.

Check and revert back.

Regards

Karthik D

[Former Member]

You can also do as;

Data : lv_string TYPE string, lv_num type string.
lv_string = 'A1313.33BBC'.
FIND REGEX '(\d*[.]\d*)' IN lv_string SUBMATCHES lv_num.

This will also give the same result.

Regards

Karthik D

[thelazyhangman]

The proper syntax is:

REPLACE ALL OCCURRENCES OF REGEX '([[:alpha:]])'

[Former Member]

Hi Kim,

just go through the fiollowing code... this is working exactly the way you want....

i hope u will get the solution : )

REPORT ZTEST1.

data: f1 type string,

result type string,

count type i,

cnt type i,

final type string.

cnt = 0.

f1 = 'AAA1246VGB,23123'.

count = strlen( f1 ).

if count is not initial.

do count times.

result = f1+cnt(1).

if result CA 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'.

clear result.

else.

concatenate final result into final.

endif.

cnt = cnt + 1.

enddo.

endif.

write:/ final.

OUTPUT is 1246,23123

Thanks & Regards

Ashu Singh

[viquar_iqbal]

Hi

DATA: v_str TYPE string VALUE 'AA1313.33BBC',
vstr1(6) TYPE n,
vstr2 TYPE p .
vstr1 = v_str.
vstr2 = vstr1.
WRITE  vstr2 NO-GROUPING ROUND 2 DECIMALS 1 .

This would give you the required output

Thanks

Viquar Iqbal