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To match a character that is special to grep –E, put a backslash ( \ ) in front of the character. It is usually simpler to use grep –F when you don't need special pattern matching.
To find a character string, type / followed by the string you want to search for, and then press Return. vi positions the cursor at the next occurrence of the string. For example, to find the string “meta,” type /meta followed by Return.
grep is very often used as a "filter" with other commands. It allows you to filter out useless information from the output of commands. To use grep as a filter, you must pipe the output of the command through grep . The symbol for pipe is " | ".
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/
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
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