如何打印BASH中搜索到的字符后出现的n个字符？

Question

I have a text file like this:

Once upon a time, there lived a rabbit who lived in the forest.
One day, rabbit found bear.
Bear said, "Hello!"

Given a letter, eg "e", I need to output all searches for e + the next 2 characters , eg, this will print the following:

e u
e, 
ere
e l
ed 
ed 
est
e d
ear
ear
ell

• Any character, including punctuation or symbols can be printed, but not a newline.
• This is case sensitive. If "e" is searched, it does not look for "E".

How can I print a searched for character plus the next 2 characters each time the searched character appears in a file in BASH?

Answer 1

Can you use perl ?

perl -lne 'print "e",$_ for $_ =~ /(?<=e)(..)/g' file

$ perl -lne 'print "e",$_ for $_ =~ /(?<=e)(..)/g' file
e u
e, 
ere
e l
ed 
ed 
e f
est
e d
ear
ear
ell

Answer 2

To include overlapping strings, you can use this:

while IFS= read -r line
do
    while [ "${#line}" -gt 2 ]
    do
        if [ "${line:0:1}" = 'e' ]
        then
            echo "${line:0:3}"
        fi
        line="${line:1}"
    done
done

Answer 3

如果您的grep支持-o，那么很简单

grep -o e..

Answer 4

The hard way

echo $(cat file.txt ) > temp
grep -o . temp | nl | grep e | cut -f1 | xargs -i -n1 echo "{}; {} + 2;" | bc | xargs -n2 echo | sed 's/ /-/' |  xargs -i{} -n 1 cut -c {} temp