‎2009 Sep 24 1:11 PM
Hi,
Following goes for dump:
data:path TYPE FILE_INFO.
DATA:l_path TYPE string,
l_len TYPE i.
l_len = STRLEN( path ).
l_path = path+9(l_len).
Can you suggest?
‎2009 Sep 24 1:24 PM
Hi, Ginger
Please Check the following Sample Code hope will solve out your problem,
DATA: path LIKE file_info-filename.
DATA: l_path TYPE string,
l_len TYPE i.
path = '123456789ABC'.
l_len = STRLEN( path ).
l_path = path+9(l_len).
WRITE: l_path.Faisal
‎2009 Sep 24 1:16 PM
Hi,
Path must be character-like data Object.
‎2009 Sep 24 1:16 PM
Hi,
Path should be a character type data object. Here ur using file_info which is a structure.
This structure contains date time integer fields which are not suitable for string operations
‎2009 Sep 24 1:16 PM
Hi Dear,
FILE_INFO is a structure and not a field.
"PATH" must be a character-type data object (data type C, N, D, T or String)
Regards,
Vijay
‎2009 Sep 24 1:17 PM
strlen function can't be used on structure ' FILE_INFO'
‎2009 Sep 24 1:18 PM
Hello Ginger,
The code does nto even syntax check, how can you get a dump? (May be you did a "activate anyways").
BR,
Suhas
‎2009 Sep 24 1:20 PM
Now suggest!!!please...
data:path TYPE FILE_INFO.
DATA:l_path TYPE string,
l_len TYPE i.
l_len = STRLEN( path-name ).
l_path = path-name+9(l_len).
Edited by: Ginger on Sep 24, 2009 8:20 AM
‎2009 Sep 24 1:23 PM
this will work!!!!
data:path TYPE FILE_INFO.
DATA:l_path TYPE string,
l_len TYPE i.
l_len = STRLEN( path-FILENAME ).
l_path = path+9(l_len).
‎2009 Sep 24 1:25 PM
Hi Ginger again,
TRy this.
DATA:path TYPE file_info.
DATA:l_path TYPE string,
l_len TYPE i.
l_len = STRLEN( path-filename ).
l_path = path-filename+9(l_len).
Regards,
VIjay
‎2009 Sep 24 1:38 PM
data:path LIKE RLGRAP-FILENAME.
DATA:l_path LIKE RLGRAP-FILENAME.
path = 'C:\file.txt'.
data : len TYPE i.
translate path to upper case.
len = strlen( path ).
l_path = path+9(len).
‎2009 Sep 24 1:20 PM
Hi,
Declare path as file_info-filename
data:path TYPE FILE_INFO-filename.
DATA:l_path TYPE string,
l_len TYPE i.
l_len = STRLEN( path ).
l_path = path+9(l_len).
Regards,
Vikranth
‎2009 Sep 24 1:24 PM
Hi, Ginger
Please Check the following Sample Code hope will solve out your problem,
DATA: path LIKE file_info-filename.
DATA: l_path TYPE string,
l_len TYPE i.
path = '123456789ABC'.
l_len = STRLEN( path ).
l_path = path+9(l_len).
WRITE: l_path.Faisal