Hello I have a fairly simple question I want to search for only the last '1+1' before the = sign from this statement
1+1+1+1+1+1=6
I have tried the grep statement
`grep '.+' 'text.txt'`
it returns
1+1+1+1+1+
as it highlights all of the + symbols. I believe I need to create a min range such as:
grep '.+{4,}' 'text.txt' This does not work, can anyone explain why? A solution with explanation would be appreciated.
Thanks
To get the last +-expression you could use grep -o and pipe that into tail -1:
grep -o '[0-9]+[0-9]' <<< '1+1+1+1+2+1=6' | tail -1
This regex matches a number (between 0 and 9) [0-9], followed by a literal + sign, followed by a number again. That will match more than one part of your expression string. The -o option prints each match on a separate line, and the tail selects the last one of these lines.
If you have the expression in a file text.txt, just use:
grep -o '[0-9]+[0-9]' text.txt | tail -1
+ is a metacharacter that means "1 or more of the previous match". When you say .+ you're saying "any character, and then 1 or more of any character."
If you want to search for a literal metacharacter, you must escape it with a backslash.
grep '.\+' text.txt
+ is a normal character for basic regular expressions, unless it is escaped to become a metacharacter = grep == grep -G 'a+' finds literal a+ but grep 'a\+' finds one or more a in a sequence; while it's opposite for egrep == grep -E 'a+' the + is already a metacharacter, so grep -E 'a+' finds one or more a but grep -E 'a\+' escapes the meta + and matches literal a+ — based on man 1 grep for "BSD General Commands" on OS X. An online page for man 1 grep on Linux tells me it's the same, but I haven't tested.