How do I use grep to extract a single word from a string using positive lookbehind and lookahead?-CodePudding

I have a string literal stored in a variable.

fullString="Line 1 Line 2 captcha_decode userEnteredCaptcha poting Line 3"

I only want to extract the word poting from this string. If I echo the string and pipe the result through a grep as follows:

echo $fullString | grep "captcha_decode userEnteredCaptcha"

I am returned the entire line again in the stdout.

Line 1 Line 2 captcha_decode userEnteredCaptcha poting Line 3

If I use -o option as following

echo $fullString | grep -o "captcha_decode userEnteredCaptcha"

I am returned only the part I had supplied as argument to the -o option

captcha_decode userEnteredCaptcha

I want to extract only the word poting which comes after captcha_decode userEnteredCaptcha and is succeeded by other characters Line 3. How do I use grep to achieve a positive lookbehind and lookahead and extract only poting?

CodePudding user response：

You may use this sed solution:

fullString="Line 1 Line 2 captcha_decode userEnteredCaptcha poting Line 3"
sed -E 's/.* captcha_decode userEnteredCaptcha ([^ ] ).*/\1/' <<< "$fullString"

poting

Or this gnu-grep solution would also work:

grep -oP 'captcha_decode userEnteredCaptcha \K\S ' <<< "$fullString"

poting

CodePudding user response：

1st solution: With your shown samples, please try following awk code. Written and tested in GNU awk.

echo "$fullString" | 
awk -v RS='captcha_decode userEnteredCaptcha [^ ]*' '
RT{
  num=split(RT,arr,FS)
  print arr[num]
}'

2nd solution: Using awk's match function in GNU awk's array option for matches found in regex for capturing group.

echo "$fullString" | 
awk 'match($0,/captcha_decode userEnteredCaptcha ([^ ]*)/,arr){print arr[1]}'

CodePudding user response：

mawk   NF OFS= FS='^.*captcha_decode userEnteredCaptcha | .*$'
gawk '$_ = $2' FS='^.*captcha_decode userEnteredCaptcha | .*$'

poting