awk / sed从模式之间提取字符串

Question

I know there has probably been a few hundred forms of this question asked on stackoverflow, but I can't seem to find a suitable answer to my question.

I'm trying to parse through the /etc/ldap.conf file on a Linux box so that I can specifically pick out the description fields from between (description= and ) :

*-bash-3.2$ grep '^nss_base_passwd' /etc/ldap.conf

nss_base_passwd ou=People,dc=ca,dc=somecompany,dc=com?one?|(description= TD_FI )(description= TD_F6 )(description= TD_F6 )(description= TRI_142 )(description= 14_142 )(description= REX5 )(description= REX5 )(description= 1950 )*

I'm looking to extract these into their own list with no duplicates:

TD_FI
TD_F6
TRI_142
14_142
REX5
1950

(or all on one line with a proper delimiter)

I had played with sed for a few hours but couldn't get it to work - I'm not entirely sure how to use the global option.

Answer 1

You could use grep with -P option,

$ grep '^nss_base_passwd' /etc/ldap.conf | grep -oP '(?<=description\=)[^)]*' | uniq
TD_FI
TD_F6
TRI_142
14_142
REX5
1950

Explanation:

A positive lookbehind is used in grep to print all the characters which was just after to the description= upto the next ) bracket. uniq command is used to remove the duplicates.

Answer 2

Try this:

grep '^nss_base_passwd' /etc/ldap.conf |
grep -oE '[(]description=[^)]*' | sort -u |
cut -f2- -d=

Explanations:

1. With bash , if you end a line with | (or || or && ), the shell knows that the command continues on the next line, so you don't need to use \\ .

2. The second grep uses the -o flag to indicate that the matching expressions should be printed out, one per line. It also uses the -E flag to indicate that the pattern is an "Extended" (ie normal) regular expression.

3. Since -o will print the entire match, we need to extract the part after the prefix, for which we use cut , specifying a delimiter of = . -f2- means "all the fields starting with the second field", which we need in case there is an = in the description.

Answer 3

perl -nE 'say join(",", /description=\K([^)]+)/g) if /^nss_base_passwd/' /etc/ldap.conf

TD_FI,TD_F6,TD_F6,TRI_142,14_142,REX5,REX5,1950

Answer 4

Avinash's answer was very close. Here is my improved version:

grep '^nss_base_passwd' /etc/ldap.conf | grep -Po '\(description=\K[^)]+' | sort -u

There is no need to use lookaround syntax when you can simply use \\K (which is actually a shortcut for a corresponding zero-width assertion).

Also, you said that you want NO duplicates, but uniq will only remove duplicate adjacent lines, it will not remove duplicates if there is something in between. That's why I am using sort -u instead.