I need to grep a string from a file, which contains '`' character. How to achieve this? Wrapping string around " and ' doesn't work.
To grep from a file which has a backtick in its name you can put a backspace in front of it. The same works for finding a string that holds the backtick. e.g.:
grep \` stupid\`filename
Related
Im trying to figure out how to search a string in Linux i hope someone can help me out.
grep "Test\|Account" test.txt
The above command works if i only want to search for one word.
But when i try to search "Create Test 'account'" not sure how to use grep since im a newbie in Linux.
With a GNU grep, you can use
grep "Create Test 'account'\|Create Test \`account\`" test.txt
Here, the backticks are escaped since they are used inside a double quoted string where they are evaluated. The | regex alternation operator is escaped because it is considered a literal pipe char otherwise.
Details:
Create Test 'account' - a literal text
\| - or
Create Test `account` - a literal text
I have found on this answer the regex to find a string between two characters. In my case I want to find every pattern between ‘ and ’. Here's the regex :
(?<=‘)(.*?)(?=’)
Indeed, it works when I try it on https://regex101.com/.
The thing is I want to use it with grep but it doesn't work :
grep -E '(?<=‘)(.*?)(?=’)' file
Is there anything missing ?
Those are positive look-ahead and look behind assertions. You need to enable it using PCRE(Perl Compatible Regex) and perhaps its better to get only matching part using -o option in GNU grep:
grep -oP '(?<=‘)(.*?)(?=’)' file
In some book I have seen a grep command example as
$grep '^no(fork\|group)' /etc/group
I need explanation for "why to use single quotes for the patteren and \ before the characters ( | )".
The advantage of using single quotes with grep, is that you do not need to escape double quotes when you need to grep for them. For example, if you wanted to search for "findthis" (including searching for the quotes) with grep, using single quotes, it would look like this:
grep '"findthis"' yourfile.txt
If you were using double quotes you would need to escape the quotes with a \, so it would look like this:
grep "\"findthis\"" yourfile.txt
The reason a backslash is needed to search for certain characters is that grep assumes that those characters have special meanings. For example grep uses " to find out the beginning and end of what you are searching for (among other things). But that means that you cannot ever search for " unless there is some way around this. The solution is to place a \ before the " like so: \". If you do that, then grep knows that you actually want to search for " rather than end the string.
quoting arguments for a command is always recommended. single quote won't expand variable. in your example, it makes no different to use single/double quotes.
take an example:
kent$ cat f
foo
bar
ooo
without quote:
kent$ grep foo|bar f
zsh: correct 'bar' to 'bzr' [nyae]? n
zsh: command not found: bar
you see, my zsh thought you want to pipe output to a command "bar"
now why escape |:
Assume your grep is not an alias. grep use BRE by default, in BRE you need to escape some char to give them special meaning, | is one of them.
You can however let grep work in ERE or PCRE mode, with -E, -P option. then you don't need escape those char any longer:
kent$ grep -E 'foo|bar' f
foo
bar
in ERE or PCRE, you escape some char, to take the special meaning away.
I am downlading the info.0.json file from xkcd and trying to parse just the alt text. I don't care if there are quotes around it or not. The problem it that the info.0.json file is all one line, and the alt text is in quotes after the word "alt=". Trying cat info.0.json | grep alt just returns the whole file (because it's all one line). What is the grep or sed code that will get me the alt text?
Use the -o switch:
-o, --only-matching show only the part of a line matching PATTERN
Example:
grep -o 'alt="[^"]*"'
I have lines in a file which look like the following
....... DisplayName="john" ..........
where .... represents variable number of other fields.
Using the following grep command, I am able to extract all the lines which have a valid 'DisplayName' field:
grep DisplayName="[0-9A-Za-z[:space:]]*" e:\test
However, I wish to extract just the name (ie "john") from each line instead of the whole line returned by grep. I tried piping the output into the cut command but it does not accept string delimiters.
This works for me:
awk -F "=" '/DisplayName/ {print $2}'
which returns "john". To remove the quotes for john use:
awk -F "=" '/DisplayName/ {gsub("\"","");print $2}'
Specifically:
sed 's/.*DisplayName="\(.*\)".*/\1/'
Should do, sed semantics is s/subsitutethis/forthis/ where "/" is delimiter. The escaped parentheses in combination with escaped 1 are used to keep the part of the pattern designated by parentheses. This expression keeps everything inside the parentheses after displayname and throws away the rest.
This can also work without first using grep, if you use:
sed -n 's/.*DisplayName="\(.*\)".*/\1/p'
The -n option and p flag tells sed to print just the changed lines.
More in: http://www.grymoire.com/Unix/Sed.html