grep for special characters in Unix

Question

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

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.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