I want to extract a specific part out of the filenames to work with them.
Example:
ls -1
REZ-Name1,Surname1-02-04-2012.png
REZ-Name2,Surname2-07-08-2013.png
....
So I want to get only the part with the name.
How can this be achieved ?
There are several ways to do this. Here's a loop:
for file in REZ-*-??-??-????.png
do
name=${file#*-}
name=${name%-??-??-????.png}
echo "($name)"
done
Given a variety of filenames with all sorts of edge cases from spacing, additional hyphens and line feeds:
REZ-Anna-Maria,de-la-Cruz-12-32-2015.png
REZ-Bjørn,Dæhlie-01-01-2015.png
REZ-First,Last-12-32-2015.png
REZ-John Quincy,Adams-11-12-2014.png
REZ-Ridiculous example # this is one filename
is ridiculous,but fun-22-11-2000.png # spanning two lines
it outputs:
(Anna-Maria,de-la-Cruz)
(Bjørn,Dæhlie)
(First,Last)
(John Quincy,Adams)
(Ridiculous example
is ridiculous,but fun)
If you're less concerned with correctness, you can simplify it further:
$ ls | grep -o '[^-]*,[^-]*'
Maria,de
Bjørn,Dæhlie
First,Last
John Quincy,Adams
is ridiculous,but fun
In this case, cut makes more sense than grep:
ls -l | cut -f2 -d-
cut the second field from the input, using '-' as the field delimiter. That other guy's answer will correctly handle some cases mine will not, but for one off uses, I generally find the semantics of cut to be much easier to remember.
Related
Say I have two files:
1.json
{"foo":"bar"}\n
2.json
{"foo":"bar"}
when using checksum routines, is there a way to ignore the trailing whitespace?
maybe something like this:
md5sum < <(cat file | trim_somehow)
You can use sed or xargs.
xargs is much simpler, but be careful with it. I'm not sure if it is safe to use in this context. Read the comments below this answer https://stackoverflow.com/a/12973694/4330274. (There are a lot of answers to your question in that post).
md5sum < <(cat file | xargs) will delete trailing/leading whitespaces (Also, as stated by dave_thompson_085 on the comments down below, it will compress each whiltespace sequence to one whitespace and it will remove quotation marks and backslashes) from the file before passing it to the md5sum utility.
Note: xargs appends a new line to the end of the input.
I recommend using sed for this purpose. It is much safer. Read this answer https://stackoverflow.com/a/3232433/4330274
I'm trying to reduce a .sm file1 - around 10 GB by filtering it using a fair long set of words (around 180.108 items) listed in a text file file2.
File1 is structured as follows:
word <http://internet.address.com> 1
i.e. one word followed by a blank space, an internet address, and a number.
File2 is a simple .txt file, a list of words, one on each line.
My aim is to create a third file File3 containing only those lines in file1 whose first word matches with the word-list of file2, and disregard the rest.
My attempt is the following:
grep -w -F -f file2.txt file1.sm > file3.sm
I've also attempted something along this line:
gawk 'FNR==NR {a[$1]; next } !($2 in a)' file2.txt file1.sm > file3.sm
but with no success. I understand /^ and \b might play a part here, but I don't know how to fit them in the syntax. I've looked around extensively but no solution seems to fit.
My problem is that here grep reads the entire file1's line, and it can happen that the matching word lies in the webpage address, which I'm not interested in finding out.
sed 's/^/^/' file2.txt | grep -f - file1.sm
join is the best tool for this, not grep/awk:
join -t' ' <(sort file1.sm) <(sort file2.txt) >file3.sm
I am trying to grep the output of a command that outputs unknown text and a directory per line. Below is an example of what I mean:
.MHuj.5.. /var/log/messages
The text and directory may be different from time to time or system to system. All I want to do though is be able to grep the directory out and send it to a variable.
I have looked around but cannot figure out how to grep to the end of a word. I know I can start the search phrase looking for a "/", but I don't know how to tell grep to stop at the end of the word, or if it will consider the next "/" a new word or not. The directories listed could change, so I can't assume the same amount of directories will be listed each time. In some cases, there will be multiple lines listed and each will have a directory list in it's output. Thanks for any help you can provide!
If your directory paths does not have spaces then you can do:
$ echo '.MHuj.5.. /var/log/messages' | awk '{print $NF}'
/var/log/messages
It's not clear from a single example whether we can generalize that e.g. the first occurrence of a slash marks the beginning of the data you want to extract. If that holds, try
grep -o '/.*' file
To fetch everything after the last space, try
grep -o '[^ ]*$' file
For more advanced pattern matching and extraction, maybe look at sed, or Awk or Perl or Python.
Your line can be described as:
^\S+\s+(\S+)$
That's assuming whitespace is your delimiter between the random text and the directory. It simply separates the whitespace from the non-whitespace and captures the second part.
Or you might want to look into the word boundary character class: \b.
I know you said to use grep, but I can't help to mention that this is trivially done using awk:
awk '{ print $NF }' input.txt
This is assuming that a whitespace is the delimiter and that the path does not contain any whitespaces.
Is there any way to do the opposite of showing only the matching part of strings in grep (the -o flag), that is, show everything except the part that matches the regex?
That is, the -v flag is not the answer, since that would not show files containing the match at all, but I want to show these lines, but not the part of the line that matches.
EDIT: I wanted to use grep over sed, since it can do "only-matching" matches on multi-line, with:
cat file.xml|grep -Pzo "<starttag>.*?(\n.*?)+.*?</starttag>"
This is a rather unusual requirement, I don't think grep would alternate the strings like that. You can achieve this with sed, though:
sed -n 's/$PATTERN//gp' file
EDIT in response to OP's edit:
You can do multiline matching with sed, too, if the file is small enough to load it all into memory:
sed -rn ':r;$!{N;br};s/<starttag>.*?(\n.*?)+.*?<\/starttag>//gp' file.xml
You can do that with a little help from sed:
grep "pattern" input_file | sed 's/pattern//g'
I don't think there is a way in grep.
If you use ack, you could output Perl's special variables $` and $' variables to show everything before and after the match, respectively:
ack string --output="\$`\$'"
Similarly if you wanted to output what did match along with other text, you could use $& which contains the matched string;
ack string --output="Matched: $&"
I'm using the operating systems dictionary file to scan. I'm creating a java program to allow a user to enter any concoction of letters to find words that contain those letters. How would I do this using grep commands?
To find words that contain only the given letters:
grep -v '[^aeiou]' wordlist
The above filters out the lines in wordlist that don't contain any characters except for those listed. It's sort of using a double negative to get what you want. Another way to do this would be:
grep '^[aeiou]+$' wordlist
which searches the whole line for a sequence of one or more of the selected letters.
To find words that contain all of the given letters is a bit more lengthy, because there may be other letters in between the ones we want:
cat wordlist | grep a | grep e | grep i | grep o | grep u
(Yes, there is a useless use of cat above, but the symmetry is better this way.)
You can use a single grep to solve the last problem in Greg's answer, provided your grep supports PCRE. (Based on this excellent answer, boiled down a bit)
grep -P "(?=.*a)(?=.*e)(?=.*i)(?=.*o)(?=.*u)" wordlist
The positive lookahead means it will match anything with an "a" anywhere, and an "e" anywhere, and.... etc etc.