如何从文件的“ xxTHISISMYSTRING”开头grep包含查询字符串的字符串，该查询字符串包含任意2个字符？

Question

I have a multi-lined file in the format:

hhhhhhhhhhhhhhhhhhh hhaaaa hhhhhhhhhhhhhh hhhhhhhhhhhhhhhhhhh oaaaaa hhhhhhhhhhhhhh hhhhhhhhhhhh hbaaaa hhhhhhhhhhhhhhhhhhhhh hhhhhhhhhhhhhhhhhhhhh fbaaaa hhhhhhhhhhhh

I want to find all strings that contain the "aaaa" motif as well as the two letters preceding it.

How would I grep out the strings: hhaaaa, oaaaaa, hbaaaa, fbaaaa ? With " aaaa " as my input.

Answer 1

To match any character in a regex, use . :

$ grep -o ..aaaa file
hhaaaa
hoaaaa
hbaaaa
fbaaaa

The -o option tells grep to print only the matches, not the context for the matches.

To restrict the match to alphabetic characters, use the alphabetic class:

$ grep -Eo '[[:alpha:]]{2}aaaa' file
hhaaaa
hoaaaa
hbaaaa
fbaaaa

[[:alpha:]] matches any alphabetic character. Unlike AZ , this is unicode-safe. The {2} indicates two such characters. To avoid backslashes, we have added the -E flag to turn on extended regex.

Answer 2

grep -oh "..aaaa" file.txt

will do.

-h, --no-filename
Suppress the prefixing of file names on output. This is the default
when there is only one file (or only standard input) to search.
-o, --only-matching
Print only the matched (non-empty) parts of a matching line,
with each such part on a separate output line.

Answer 3

grep -o '..aaaa' file

should do it. Had the objective been to count the total matches, then do:

grep -o '..aaaa' file | wc -l

GREP manpage says :

-o, --only-matching
Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.

WC manpage says :

-l, --lines
print the newline counts