Extract file name from path in awk program

Several options:

awk '
  function basename(file) {
    sub(".*/", "", file)
    return file
  }
  {print FILENAME, basename(FILENAME)}' /path/to/file

Or:

awk '
  function basename(file, a, n) {
    n = split(file, a, "/")
    return a[n]
  }
  {print FILENAME, basename(FILENAME)}' /path/to/file

Note that those implementations of basename should work for the common cases, but not in corner cases like basename /path/to/x/// where they return the empty string instead of x or / where they return the empty string instead of /, though for regular files, that should not happen.

The first one will not work properly if the file paths (up to the last /) contain sequences of bytes that don't form valid characters in the current locale (typically this kind of thing happens in UTF-8 locales with filenames encoded in some 8 bit single byte character set). You can work around that by fixing the locale to C where every sequence of byte form valid characters.

Try this awk one-liner,

$ awk 'END{ var=FILENAME; split (var,a,/\//); print a[5]}' /home/abc/imp/asgd.csv
asgd.csv

On the systems where basename command is available, one could use awk's system() function or expression | getline var structure to call external basename command. This can help accounting for corner cases mentioned in Stephane's answer.

$ awk '{cmd=sprintf("basename %s",FILENAME);cmd | getline out; print FILENAME,out; exit}' /etc///passwd
/etc///passwd passwd