How to filter text, removing lines where a specific position is empty

Question

# lshw -c network -businfo
Bus info          Device      Class          Description
========================================================
pci@0000:03:00.0  enp3s0      network        I210 Gigabit Network Connection
pci@0000:04:00.0  mec_test_1  network        I350 Gigabit Network Connection
pci@0000:04:00.1  mec_test_2  network        I350 Gigabit Network Connection
pci@0000:04:00.2  mec_test_3  network        I350 Gigabit Network Connection
pci@0000:04:00.3  mec_test_4  network        I350 Gigabit Network Connection
pci@0000:b5:00.0              network        Intel Corporation
pci@0000:b5:00.1              network        Intel Corporation
pci@0000:b5:00.2  mec_test_5  network        Intel Corporation
pci@0000:b5:00.3  enp181s0f3  network        Intel Corporation

If I want to pass Device is empty, how to do that? like

Bus info          Device      Class          Description
========================================================
pci@0000:03:00.0  enp3s0      network        I210 Gigabit Network Connection
pci@0000:04:00.0  mec_test_1  network        I350 Gigabit Network Connection
pci@0000:04:00.1  mec_test_2  network        I350 Gigabit Network Connection
pci@0000:04:00.2  mec_test_3  network        I350 Gigabit Network Connection
pci@0000:04:00.3  mec_test_4  network        I350 Gigabit Network Connection
pci@0000:b5:00.2  mec_test_5  network        Intel Corporation
pci@0000:b5:00.3  enp181s0f3  network        Intel Corporation

Answer 1

One simple way would be to ask grep to look for a non-space character in the 19th column:

lshw -c network -businfo | grep -E '^.{18}[^ ]'

The grep command is looking for pattern that's anchored to the beginning of the line; there follows 18 of any character, then something that's not a space.

Alternatively, if there's a TAB there instead of space, use the [:space:] class:

lshw -c network -businfo | grep -E '^.{18}[^[:space:]]'