I have this in my trim_csv.bat
(I intend to trim trailing spaces of every entry in the csv).
FOR /F "delims=;" %i IN (csv_exports\account.csv) DO @echo %i
(I have also tried enclosing the path with quotes, no difference.)
Executing it through cmd results to:
i was unexpected at this time.
What do I misunderstand? I tried following syntax coming from HELP FOR
but I feel like that is something simple that I just can't see.
PS: It's quite new in batch scripting as you can imagine.
If the code is going to be a part of a
bat
file, percents should be escaped as%i
->%%i
: