Capture word after pattern with slash

user_5954825 0
Viewed 76

I want to extract word1 from:

something /CLIENT_LOGIN:word1 something else

I would like to extract the first word after matching pattern /CLIENT_LOGIN:.

Without the slash, something like this works:

A=something /CLIENT_LOGIN:word1 something else
B=$(echo $A | awk '$1 == "CLIENT_LOGIN" { print $2 }' FS=":")

With the slash though, I can't get it working (I tried putting / and \/ in front of CLIENT_LOGIN). I don't care getting it done with awk, grep, sed, ...

sed grep awk regex
4 Answers

Using sed:

s='=something /CLIENT_LOGIN:word1 something else'
sed -E 's~.* /CLIENT_LOGIN:([^[:blank:]]+).*~\1~' <<< "$s"

word1

Details:

  • We use ~ as regex delimiter in sed
  • /CLIENT_LOGIN:([^[:blank:]]+) matches /CLIENT_LOGIN: followed by 1+ non-whitespace characters that is captured in group #1
  • .* on both sides matches text before and after our match
  • \1 is used in substitution to put 1st group's captured value back in output
user_5482250

1st solution: With your shown samples, please try following GNU grep solution.

grep -oP '^.*? /CLIENT_LOGIN:\K(\S+)' Input_file

Explanation: Simple explanation would be, using GNU grep's o and P options. Which are responsible for printing exact match and enabling PCRE regex. In main program, using regex ^.*? /CLIENT_LOGIN:\K(\S+): which means using lazy match from starting of value to till /CLIENT_LOGIN: to match very first occurrence of string. Then using \K option to forget till now matched values so tat we can print only required values, which is followed by \S+ which means match all NON-Spaces before any space comes.

2nd solution: Using awk's match function along with its split function to print the required value.

awk '
match($0,/\/CLIENT_LOGIN:[^[:space:]]+/){
  split(substr($0,RSTART,RLENGTH),arr,":")
  print arr[2]
}
' Input_file

3rd solution: Using GNU awk's FPAT option please try following solution. Simple explanation would be, setting FPAT to /CLIENT_LOGIN: followed by all non-spaces values. In main program of awk using sub to substitute everything till : with NULL for first field and then printing first field.

awk -v FPAT='/CLIENT_LOGIN:[^[:space:]]+' '{sub(/.*:/,"",$1);print $1}'  Input_file
user_58665800

Performing a regex match and capturing the resulting string in BASH_REMATCH[]:

$ regex='.*/CLIENT_LOGIN:([^[:space:]]*).*'

$ A='something /CLIENT_LOGIN:word1 something else'
$ unset B

$ [[ "${A}" =~ $regex ]] && B="${BASH_REMATCH[1]}"
$ echo "${B}"
word1

Verifying B remains undefined if we don't find our match:

$ A='something without the desired string'
$ unset B

$ [[ "${A}" =~ $regex ]] && B="${BASH_REMATCH[1]}"
$ echo "${B}"
               <<<=== nothing output
user_73661000

Fixing your awk command, you can use

A="/CLIENT_IPADDR:23.4.28.2 /CLIENT_LOGIN:xdfmb1d /MXJ_C"
B=$(echo "$A" | awk 'match($0,/\/CLIENT_LOGIN:[^[:space:]]+/){print substr($0,RSTART+14,RLENGTH-14)}')

See the online demo yielding xdfmb1d. Details:

  • \/CLIENT_LOGIN: - a /CLIENT_LOGIN: string
  • [^[:space:]]+ - one or more non-whitespace chars

The pattern above is what awk searches for, and once matched, the part of this match value after /CLIENT_LOGIN: is "extracted" using substr($0,RSTART+14,RLENGTH-14) (where 14 is the length of the /CLIENT_LOGIN: string).

user_38329700
Related