I have a log file (application.log) which might contain the following string of normal & special characters on multiple lines:
*^%Q&$*&^@$&*!^@$*&^&^*&^&
I want to search for the line number(s) which contains this special character string.
grep '*^%Q&$*&^@$&*!^@$*&^&^*&^&' application.log
The above command doesn't return any results.
What would be the correct syntax to get the line numbers?
Tell grep to treat your input as fixed string using -F option.
grep -F '*^%Q&$*&^@$&*!^@$*&^&^*&^&' application.log
Option -n is required to get the line number,
grep -Fn '*^%Q&$*&^@$&*!^@$*&^&^*&^&' application.log
^, $, ., * and \ . or you can try fgrep, too.fgrep is same as grep -F.
grep -rinF '-static' prints Invalid option «t». The @Mani's answer worked, i.e. grep -rin "\-static".
grep -Fn '#!/usr/bin/python' --include=*.py, no any output shown and seems stucked in console.The one that worked for me is:
grep -e '->'
The -e means that the next argument is the pattern, and won't be interpreted as an argument.
From: http://www.linuxquestions.org/questions/programming-9/how-to-grep-for-string-769460/
-, see also stackoverflow.com/questions/2427913/…To grep for carriage return, namely the \r character, or 0x0d, we can do this:
grep -F $'\r' application.log
Alternatively, use printf, or echo, for POSIX compatibility
grep -F "$(printf '\r')" application.log
And we can use hexdump, or less to see the result:
$ printf "a\rb" | grep -F $'\r' | hexdump -c
0000000 a \r b \n
Regarding the use of $'\r' and other supported characters, see Bash Manual > ANSI-C Quoting:
Words of the form $'string' are treated specially. The word expands to string, with backslash-escaped characters replaced as specified by the ANSI C standard
grep -n "\*\^\%\Q\&\$\&\^\@\$\&\!\^\@\$\&\^\&\^\&\^\&" test.log
1:*^%Q&$&^@$&!^@$&^&^&^&
8:*^%Q&$&^@$&!^@$&^&^&^&
14:*^%Q&$&^@$&!^@$&^&^&^&
You could try removing any alphanumeric characters and space. And then use -n will give you the line number. Try following:
grep -vn "^[a-zA-Z0-9 ]*$" application.log
Try vi with the -b option, this will show special end of line characters (I typically use it to see windows line endings in a txt file on a unix OS)
But if you want a scripted solution obviously vi wont work so you can try the -f or -e options with grep and pipe the result into sed or awk. From grep man page:
Matcher Selection -E, --extended-regexp Interpret PATTERN as an extended regular expression (ERE, see below). (-E is specified by POSIX.)
-F, --fixed-strings
Interpret PATTERN as a list of fixed strings, separated by newlines, any of which is to be matched. (-F is specified
by POSIX.)
^@: means^followed by@or nul character?