CODEWITHSUNDEEP
unixsearchgrepspecial-characters
Avinash Sonee
Avinash Sonee

Reputation: 1811

grep for special characters in Unix

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?

Upvotes: 147

Views: 476930

Answers (6)

user7648830
user7648830

Reputation: 1

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.)

Upvotes: -4

R. Kumar
R. Kumar

Reputation: 109

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

Upvotes: 3

ryenus
ryenus

Reputation: 17381

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

Upvotes: 9

Mani
Mani

Reputation: 965

grep -n "\*\^\%\Q\&\$\&\^\@\$\&\!\^\@\$\&\^\&\^\&\^\&" test.log
1:*^%Q&$&^@$&!^@$&^&^&^&
8:*^%Q&$&^@$&!^@$&^&^&^&
14:*^%Q&$&^@$&!^@$&^&^&^&

Upvotes: 3

Student
Student

Reputation: 28345

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/

Upvotes: 132

Prince John Wesley
Prince John Wesley

Reputation: 63688

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

Upvotes: 223

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