sed: strip enclosing square brackets

Question

I have this file:

[default]
region = us-east-1
output = json
[profile automation]
region = us-east-1
output = json
[profile qa]
region = us-east-1
output = json

I need to extract only the profile names like so:

default
automation
qa

My sed so far:

gsed 's?\[[^\ ]*\ \([^\]]*\)?\1?' ~/.aws/config

As I'm testing, I can see that \([^\]]*\) matches only the first character after the space for example: the a in automation, the q in qa etc.. Why it only matches once? Do you have a better way of stripping the []? Thanks.

Answer 1

A backslash inside a bracketed expression is literal. You want [^]]. Your expression [^\]]* would match a non-backslash followed by an optional ].

$ sed -En 's/^\[(.* )?([^]]*)\]$/\2/p' file
default
automation
qa

This allows for optional words terminated by a space as the first substring inside [...], e.g. [group name]. The string after these optional words, up to the ], is captured in the second group.

The default printing of every line is turned off with -n and instead we print explicitly after each successful substitution.

We need -E since the ? modifier is an extended regular expression modifier.

Without -E you could use \{0,1\} in place of the ? (and change the capturing ( and ) into \( and \) respectively).

Answer 2

With GNU grep or compatible built with PCRE support:

grep -Po '^\h*\[\h*(\H+\h+)?\K.*[^]\h](?=\h*\]\s*$)'

We allow leading horizontal whitespace sometimes found for indentation and trailing whitespace, including vertical for the CR character generally found in MS Windows ini files.

On an input like:

[default]
[profile automation]
[profile a  b]
[profile  c d ]

It gives:

default
automation
a  b
c d