This question already has answers here:
Using grep to search for a string that has a dot in it
(9 answers)
Closed 5 years ago.
It seemed to me that grep -v displays the files that don't contain the following string.
How comes the file named highscore.txt doesn't appear when using grep -v ".c" ?
$ ls -1
a.out
easy.txt
hard.txt
highscores.txt
main.c
main.txt
util.c
$ ls -1 | grep -v ".c"
a.out
easy.txt
hard.txt
medium.txt
The ".c" in your grep command is a regular expression, and . means "any character".
To fix this, you can
Escape the period:
grep -v '\.c$'
I've added the "end of string" anchor $ to exclude false positives for files like something.cpp.
Use the -F option for "fixed strings":
grep -vF '.c'
Notice that this would also exclude something.cpp, which probably isn't what you want.
Use extended glob patterns to exclude anything ending in .c:
shopt -s extglob
ls -1 !(*.c)
Here, *.c is not a regular expression, but a glob pattern, where . is a literal period and has no special meaning.
Related
This question already has answers here:
Colorized grep -- viewing the entire file with highlighted matches
(24 answers)
Closed 4 years ago.
How can I modify grep so that it prints full file if its entry matches the grep pattern , instead of printing Just the matching line ?
I tried using(say) grep -C2 to print two lines above and 2 below but this doesn't always works as no. of lines is not fixed ..
I am not Just searching a single file , I am searching an entire directory where some files may contain the given pattern and I want those Files to be completely Printed.
I am also using grep inside grep result without getting printed the first grep output.
Simple grep + cat combination:
grep 'pattern' file && cat file
Use grep's -l option to list the paths of files with matching contents, then print the contents of these files using cat.
grep -lR 'regex' 'directory' | xargs -d '\n' cat
The command from above cannot handle filenames with newlines in them.
To overcome the filename with newlines issue and also allow more sophisticated checks you can use the find command.
The following command prints the content of all regular files in directory.
find 'directory' -type f -exec cat {} +
To print only the content of files whose content matches the regexes regex1 and regex2, use
find 'directory' -type f \
-exec grep -q 'regex1' {} \; -and \
-exec grep -q 'regex2' {} \; \
-exec cat {} +
The linebreaks are only for better readability. Without the \ you can write everything into one line.
Note the -q for grep. That option supresses grep's output. grep's exit status will tell find whether to list a file or not.
Information and Problems
I am learning linux command now, and was simply practicing grep command in a bash.
I want to match every file whose name begins with character "a"...quite a simple requirement...From what I understand the regex should be something like a.*, but it doesn't work as what I thought.
Some of the filenames should be matched doesn't match.
My Command
I typed commands in a Ubuntu Mate 16.04 VirtualBox terminal.
I have created a document called test. In the test document, I have got three files,
a.txt
a1.txt
a2.txt
Here the following is my command using grep.
ls -a | grep -E -e a.*
But the output is simply
a.txt
I think .* should mean any numbers of whatever character. So the a1.txt and a2.txt should match the regex, but it doesn't work.
However if I tried
ls -a | grep -E -e ^a.*
ls -a | grep -E -e a.+
Both of the command work as what I expected, all the filenames matches.
a.txt
a1.txt
a2.txt
I could not figure out what goes wrong?
What I have tried
I have searched through the questions, there exist a question very similar to mine, but the problems is about the extended grep and the basic one, which definitely isn't my situation.
Use more quotes!
With the literal command you ran in your question:
ls -a | grep -E -e a.*
...your shell will replace a.* with a list of filenames in the current directory matching a.* as a glob pattern before grep is started at all. (See also the full bash-hackers page on globbing).
If a.* is placed inside quotes, as in:
ls -a | grep -E 'a.*'
...then this string will no longer be evaluated as a glob. You might also want to anchor the regex with ^, to search only at the beginning:
ls -a | grep -E '^a.*'
That said, ls is not a tool build for programmatic use -- it isn't guaranteed to emit filenames in unmodified literal form, so it's not certain that all possible names will be emitted in such a way that grep or other tools will parse them correctly (indeed, ls can't emit all possible names is literal form, since it uses newline delimiters between names, whereas newline literals are actually possible within names themselves). Consider using find for this kind of processing:
while IFS= read -r -d '' filename; do
printf 'Found file: %q\n' "$filename"
done < <(find . -regex '/^a[^/]*' -print0)
...will work even with files having intentionally difficult-to-process names; consider, for example, mkdir -p $'\n/etc/passwd\n' && touch $'\n/etc/passwd\n/a.txt'.
You are misunderstanding how the shell is parsing your command. When you do this:
ls -a | grep -E -e a.*
The shell globs the command before it is passed to ls or grep. The result of the glob is this:
ls -a | grep -E -e a.txt
Because in globbing, a.* only matches a.txt.
You need to put the regexes in quotes, e.g.
ls -a | grep -E -e 'a.*'
I would like to grep a number of documents by using a set of search terms and to specify the number of characters after match. Here is what I tried
grep -F -o -P "$(<search.txt).{0,4}" foo.txt
but I get the message 'grep: conflicting matchers specified' because -F and '-oP' cannot be combined. It does not work with '-E' either.
-F and -P are conflicting options, simple as that. The first means that the patterns are fixed strings, the second means that the patterns are Perl-compatible regular expressions. Perhaps you meant to use -f instead, which reads patterns from a file or a process substitution.
If you want to match any of the patterns in your file, followed by 4 characters, you could use something like this
grep -oP -f <(awk '{print $0 ".{4}"}' search.txt) file
This dynamically adds the pattern to each line in the file.
Alternatively, a more portable and concise version would be this:
sed 's/$/.{0,4}/' search.txt | grep -f - -oP file
I have a file temp.txt as below.
a.*,super
I want to grep .* to check whether the value is present in the file or not.
Command used:
grep -i ".*" temp.txt
returns nothing
This is because grep considers the pattern as a regular expression.
To make grep interpret it as a literal, use -F.
grep -F ".*" temp.txt
Also, note -i is not needed, because there is no case distinction to take into account (we for example use it to make grep return AB, aB, Ab and ab when doing grep -i "ab").
As man grep says:
-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.)
-i, --ignore-case
Ignore case distinctions in both the PATTERN and the input files. (-i
is specified by POSIX.)
Using awk
awk '/\.\*/' file
or fgrep
fgrep ".*" file
Both ., * have special meaning in regular expression. Escape them to match literally.
$ cat temp.txt
a.*,super
$ grep "\.\*" temp.txt
a.*,super
$ echo $?
0
$ grep "there-is-no-such-string" temp.txt
$ echo $?
1
-i is not need because there's no alphabet in the regular expression.
This question already has answers here:
Grep for literal strings
(6 answers)
Closed 6 years ago.
Is there some way to make grep match an exact string, and not parse it as a regex? Or is there some tool to escape a string properly for grep?
$ version=10.4
$ echo "10.4" | grep $version
10.4
$ echo "1034" | grep $version # shouldn't match
1034
Use grep -F or fgrep.
$ echo "1034" | grep -F $version # shouldn't match
$ echo "10.4" | grep -F $version
10.4
See man page:
-F, --fixed-strings
Interpret PATTERN as a list of fixed strings, separated
by newlines, any of which is to be matched.
I was looking for the term "literal match" or "fixed string".
(See also Using grep with a complex string and How can grep interpret literally a string that contains an asterisk and is fed to grep through a variable?)