That looks to be fixed width columns, 25 characters per column. You could load each line into a variable and use substring operations to get the values.
@echo off
setlocal enableDelayedExpansion
for /f delims^=^ eol^= %%A in (test.txt) do (
set "ln=%%A"
echo col1=[!ln:~0,25!]
echo col2=[!ln:~25,25!]
echo col3=[!ln:~50!]
echo(
)
But that leaves you with the problem of removing trailing spaces from the end of each value. It is doable, but not easy with batch. My output from the above script encloses the values within square brackets so you can easily see the trailing space issue.
Instead of messing with the issue of removing trailing spaces, I would use my REPL.BAT utility to transform the data from fixed width to delimited format. REPL.BAT is a hybrid JScript/batch utility that performs a regular expression search/replace operation on stdin and writes the result to stdout. It is pure script that will run on any modern Windows machine from XP onward without needing any 3rd party executables. Full documentation is embedded within the script.
I would use REPL.BAT to insert a delimiter after the 25th and 50th characters. I would then use another REPL.BAT to strip out any spaces that precede the delimiter, and then a normal FOR /F can safely parse the values. I chose to use a pipe (|
) as the delimiter.
@echo off
for /f "eol=| delims=| tokens=1-3" %%A in (
'type test.txt ^| repl "^(.{25})(.{25})" "$1|$2|" ^| repl " *\|" "|"'
) do (
echo col1=[%%A]
echo col2=[%%B]
echo col3=[%%C]
echo(
)
If you know that no value contains consecutive spaces, and all values are separated by at least 2 spaces, then you can get by with a single REPL that replaces 2 or more spaces with a delimiter
@echo off
for /f "eol=| delims=| tokens=1-3" %%A in (
'type test.txt ^| repl " {2,}" "|"'
) do (
echo col1=[%%A]
echo col2=[%%B]
echo col3=[%%C]
echo(
)