Suppose I have a file named as test.txt having content .
I want to find the line containing the words starting with "r" character and ending with "i" character?
That would be something like:
grep '\b[Rr][A-Za-z]*[Ii]\b' test.txt
That's case insensitive so, if you want to ensure specific capitalisation, you would adjust the individual character classes in the expression.
Related
I'm trying to isolate lines that contain the following: '[Homosapiens]' from a file.
My file looks something like this:
br
blabla
>blabldi[Homosapiens]
>skadlfjkl[Musmusculus]
I only want to isolate the third line.
I have tried the following:
grep -F '\*[Homosapiens]' mytext.txt
and
fgrep '\*[Homosapiens]' mytext.txt
but both are not working.
Can anyone solve this problem?
What about the following:
fgrep "[Homosapiens]" mytext.txt
Or:
grep "\[Homosapiens\]" mytext.txt
Two remarks:
grep (or whatever of its family members fgrep, egrep, ...) search for an entry inside a line of text, so there is no need to try to fit the whole line inside your grep expression.
The square brackets have a special meaning (grep [a-e] means a search for all letters from 'a' to 'e'). Using a backslash in front of a square bracket disables that feature and gives you the opportunity to look for a square bracket.
The txt file is :
bar
quux
kabe
Ass
sBo
CcdD
FGH
I would like to grep the words with only one capital letter in this example, but when I use "grep [A-Z]", it shows me all words with capital letters.
Could anyone find the "grep" solution here? My expected output is
Ass
sBo
grep '\<[a-z]*[A-Z][a-z]*\>' my.txt
will match lines in the ASCII text file my.txt if they contain at least one word consisting entirely of ASCII letters, exactly one of which is upper case.
You seem to have a text file with each word on its own line.
You may use
grep '^[[:lower:]]*[[:upper:]][[:lower:]]*$' file
See the grep online demo.
The ^ matches the start of string (here, line since grep operates on a line by lin basis by default), then [[:lower:]]* matches 0 or more lowercase letters, then an [[:upper:]] pattern matches any uppercase letter, and then [[:lower:]]* matches 0+ lowercase letters and $ asserts the position at the end of string.
If you need to match a whole line with exactly one uppercase letter you may use
grep '^[^[:upper:]]*[[:upper:]][^[:upper:]]*$' file
The only difference from the pattern above is the [^[:upper:]] bracket expression that matches any char but an uppercase letter. See another grep online demo.
To extract words with a single capital letter inside them you may use word boundaries, as shown in mathguy's answer. With GNU grep, you may also use
grep -o '\b[^[:upper:]]*[[:upper:]][^[:upper:]]*\b' file
grep -o '\b[[:lower:]]*[[:upper:]][[:lower:]]*\b' file
See yet another grep online demo.
I want to search a file in which there any words which contain alphanumeric words (i.e. words that have both combination of alpha and numeral)
I have tried using different grep combinations but not able to find the exact result I want to achieve
for example if I have a file that contains multiple lines
asbcd acblk54 lkasdfn
098213 102938 091283
aalk adsf adf
lkjas 0098324 0980 assdf
alkj30lkl 093adflkj 0lkdsf094
since lines 1 and 5 contain words which are alphanumeric only two lines should be filtered. how can I achieve this using grep.(line 2 contains numerals only, line 3 contains alpha only, line 4 contains words that are either alpha or numeral but not combination of both)
What you are interested in is a grep that matches full words. So you need the -w option:
-w, --word-regexp: Select only those lines containing matches that form whole words. The test is that the matching substring must either be at the beginning of the line, or preceded by a non-word constituent character. Similarly, it must be either at the end of the line or followed by a non-word constituent character. Word-constituent characters are letters, digits, and the underscore. This option has no effect if -x is also specified.
source: man grep
The regex you search for uses indeed [[:alnum:]] but you have to ensure that it has both a [[:alpha:]] and a [[:digit:]]. A word containing both must thus have a sequence [[:alpha:]][[:digit:]] or [[:digit:]][[:alpha:]]. The regex you are after is thus: [[:alnum:]]*([[:alpha:]][[:digit:]]|[[:digit:]][[:alpha:]])[[:alnum:]]*
The following grep will do the matches:
$ grep -w -E '[[:alnum:]]*([[:alpha:]][[:digit:]]|[[:digit:]][[:alpha:]])[[:alnum:]]*' file
I've been trying to find a way to filter a specific letter within a word using a regular expression. For exemple, filtering the letter "a" in the word "latin". Filtering only a letter would be simple using something like :
grep "\ba\b"
but I can't find a way to get the "a" only in a certain word.
Thanks for your help!
You can pipe to another grep, like this:
grep "\ba\b" /path/to/input/file | grep -o "a"
The latter part of the pipe uses the o flag which only outputs the matched part. Alternatively grep -o "a" should return all a's.
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.