I have this file named korad and I need to print all lines of this file that start with "d" and end with number, This is what I tried:
grep "^d*[0-9]$" korad
What about this:
grep "^d" korad | grep "[0-9]$"
This first filters the lines, starting with letter "d" and afterwards filters those results with the lines, ending with a number. Like that, you don't need to worry about anything being present between the first and the last character.
In case you don't understand the vertical bar, it's called a pipe, which is (amongst others) explained here.
grep -E '^d(?:.*)\d$' korad
To get all lines starting with d and ending with a digit
Regex101 Demo
Suggesting:
grep "^d.*[[:digit:]]$" korad
Related
I have a file that contains:
/usr/bin/alias, /usr/bin/clear, /usr/bin/echo, /usr/bin/cat, /usr/bin/netstat,
/usr/sbin/shutdown, /usr/bin/less
and I need to move the last words to another file
alias
clear
echo
cat
netstat
shutdown
less
I have tried awk, grep, sed, cut all kinds of combinations but can't seem to get the right result
Thank you in advance for your help.
I used this:
grep -Po '(?<=/)[^/]+$' filename
-P means to use Perl style regex
-o means to output just the matched text
(?<=/) is a zero-width lookbehind to match the leading slash
[^/]+$ matches any text except a slash to the end of the line
I want to find the number of 8 letter words that do not contain the letter "e" in a number of text files (*.txt). In the process I ran into two issues: my lack of understanding in quantifiers and how to exclude characters.
I'm quite new to the Unix terminal, but this is what I have tried:
cat *.txt | grep -Eo "\w+" | grep -i ".*[^e].*"
I need to include the cat command because it otherwise includes the names of the text files in the pipe. The second pipe is to have all the words in a list, and it works, but the last pipe was meant to find all the words that do not have the letter "e" in them, but doesn't seem to work. (I thought "." for no or any number of any character, followed by a character that is not an "e", and followed by another "." for no or any number of any character.)
cat *.txt | grep -Eo "\w+" | grep -wi "[a-z][a-z][a-z][a-z][a-z][a-z][a-z][a-z]"
This command works to find the words that contain 8 characters, but it is quite ineffective, because I have to repeat "[a-z]" 8 times. I thought it could also be "[a-z]{8}", but that doesn't seem to work.
cat *.txt | grep -Eo "\w+" | grep -wi "[a-z][a-z][a-z][a-z][a-z][a-z][a-z][a-z]" | grep -i ".*[^e].*"
So finally, this would be my best guess, however, the third pipe is ineffective and the last pipe doesn't work.
You may use this grep:
grep -hEiwo '[a-df-z]{8}' *.txt
Here:
[a-df-z]{8}: Matches all letters except e
-h: Don't print filename in output
-i: Ignore case search
-o: Print matches only
-w: Match complete words
In case you are ok with GNU awk and assuming that you want to print only the exact words and could be multiple matches in a line if this is the case one could try following.
awk -v IGNORECASE="1" '{for(i=1;i<=NF;i++){if($i~/^[a-df-z]{8}$/){print $i}}}' *.txt
OR without the use of IGNORCASE one could try:
awk '{for(i=1;i<=NF;i++){if(tolower($i)~/^[a-df-z]{8}$/){print $i}}}' *.txt
NOTE: Considering that you want exact matches of 8 letters only in lines. 8 letter words followed by a punctuation mark will be excluded.
Here is a crazy thought with GNU awk:
awk 'BEGIN{FPAT="\\<\\w{8}\\>"}{c+=NF}END{print c}' file
Or if you want to make it work only on a select set of characters:
awk 'BEGIN{FPAT="\\<[a-df-z]{8}\\>"}{c+=NF}END{print c}' file
What this does is, it defines the fields, to be a set of 8 characters (\w as a word-constituent or [a-df-z] as a selected set) which is enclosed by word-boundaries (\< and \>). This is done with FPAT (note the Gory details about escaping).
Sometimes you might also have words which contain diatrics, so you have to expand. Then this might be the best solution:
awk 'BEGIN{FPAT="\\<\\w{8}\\>"}{for(i=1;i<=NF;++i) if($i !~ /e/) c++}END{print c}' file
I have a file "test.txt" with the lines below and also lot bunch of extra stuff after the "version"
soainfra_metrics{metric_group="sca_composite",partition="test",is_active="true",state="on",is_default="true",composite="test123"} map:stats version:1.0
soainfra_metrics{metric_group="sca_composite",partition="gello",is_active="true",state="on",is_default="true",composite="test234"} map:stats version:1.8
soainfra_metrics{metric_group="sca_composite",partition="bolo",is_active="true",state="on",is_default="true",composite="3415"} map:stats version:3.1
soainfra_metrics{metric_group="sca_composite",partition="solo",is_active="true",state="on",is_default="true",composite="hji"} map:stats version:1.1
I tried:
egrep -r 'partition|is_active|state|is_default|composite' test.txt
It's displaying every line, but I need only specific mentioned fields like this below,ignoring rest of the data/stuff or lines
in a nut shell, i want to display only these fields from a line not the rest
partition="test",is_active="true",state="on",is_default="true",composite="test123"
partition="gello",is_active="true",state="on",is_default="true",composite="test234"
partition="bolo",is_active="true",state="on",is_default="true",composite="3415"
partition="solo",is_active="true",state="on",is_default="true",composite="hji"
If your version of grep supports Perl-style regular expressions, then I'd use this:
grep -oP '.*?,\K[^}]+' file
It removes everything up to the first comma (\K kills any previous output) and prints everything up to the }.
Alternatively, using awk:
awk -F'}' '{ sub(/[^,]+,/, ""); print $1 }' file
This sets the field separator to } so the part you're interested in is the first field. It then uses sub to remove the part up to the first comma.
For completeness, you could also use sed:
sed 's/[^,]*,\([^}]*\).*/\1/' file
This captures the part after the first , up to the } and replaces the content of the line with it.
After the grep to pick out the lines you want, use sed to edit the lines:
sed 's/.*\(partition[^}]*\)} map.*/\1/'
This means: "whenever you see anything .*, followed by partition and
any number of non-}, then } map and anything else, grab the part
from partition up to but not including the brace \(...\) as group 1.
The replacement text is just group 1 \1.
Use a pipe | to connect the output of egrep to the input of sed:
egrep ... | sed ...
As far as i understood your file might have more lines you don't want to see, so i would use:
sed -n 's/.*\(partition.*\)}.*/\1/p' file
we use -n p to show only lines where we made substitution. The substitution part just gets the part of the line you need substituting the whole line with the pattern.
This might work for you (GNU sed):
sed -r 's/(partition|is_active|state|is_default|composite)="[^"]*"/\n&\n/g;s/[^\n]*\n([^\n]*)\n[^\n]*/\1,/g;s/,$//' file
Treat the problem as if it were a "decomposed club sandwich". Identify the fillings, remove the bread and tidy up.
I want to last few lines when a pattern is matched in a file using sed.
if file has following entries:
This is the first line.
This is the second line.
This is the third line.
This is the forth line.
This is the Last line.
so, search for pattern, "Last" and print last few lines ..
Find 'Last' using sed and pipe it to tail command which print last n lines of the file -n specifies the no. of lines to be read from end of the file, here I am reading last 2 lines of the file.
sed '/Last/ p' yourfile.txt|tail -n 2
For more info on tail use man tail.
Also, | symbol here is known as a pipe(unnamed pipe), which helps in inter-process communication. So, in simple words sed feeds data to tail command using pipe.
I assume you mean "find the pattern and also print the previous few lines". grep is your friend: to print the previous 3 lines:
$ grep -B 3 "Last" file
This is the second line.
This is the third line.
This is the forth line.
This is the Last line.
-B n for "before". There's also -A n ("after"), and -C n ("context", both before and after).
This might work for you (GNU sed):
sed ':a;$!{N;s/\n/&/2;Ta};/Last/P;D' file
This will print the line containing Last and the two previous lines.
N.B. This will only print the lines before the match once. Also more lines can by shown by changing the 2 to however many lines you want.
Echo "Hello everybody!"
I need to check whether the input argument of a linux script does comply with my security needs. It should contain only a-z characters, 0-9 digits, some spaces and the "+" sign. Eg.: "for 3 minutes do r51+r11"
This didn't worked for me:
if grep -v '[0123456789abcdefghijklmnopqrstuvwxyz+ ]' /tmp/input; then
echo "THIS DOES NOT COMPLY!";
fi
Any clues?
You are telling grep:
Show me every line that does not contain [0123456789abcdefghijklmnopqrstuvwxyz+ ]
Which would only show you lines that contains neither of the characters above. So a line only containing other characters, like () would match, but asdf() would not match.
Try instead to have grep showing you every line that contains charachter not in the list above:
if grep '[^0-9A-Za-z+ ]' file; then
If you find something that's not a number or a letter or a plus, then.
You want to test the entire row (assuming there is only one row in /tmp/input), not just whether a single character anywhere matches, so you need to anchor it to the start end end of the row. Try this regexp:
^[0123456789abcdefghijklmnopqrstuvwxyz+ ]*$
Note that you can shorten this using ranges:
^[0-9a-z+ ]*$