grep for special characters in Unix
Tell grep to treat your input as fixed string using -F option.
grep -F '*^%Q&$*&^@$&*!^@$*&^&^*&^&' application.logOption -n is required to get the line number,
grep -Fn '*^%Q&$*&^@$&*!^@$*&^&^*&^&' application.log
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/
A related note
To grep for carriage return, namely the \r character, or 0x0d, we can do this:
grep -F $'\r' application.logAlternatively, use printf, or echo, for POSIX compatibility
grep -F "$(printf '\r')" application.logAnd we can use hexdump, or less to see the result:
$ printf "a\rb" | grep -F $'\r' | hexdump -c0000000 a \r b \nRegarding 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