So far, I used grep for these kind of questions, but here, I think it can't be use. Indeed, I want a match in the first section of each line, and then print the whole line.
I wrote something like that:
cat file.txt | cut -d " " -f1 | grep root
But of course, this command does not print the whole line, but only the first section of each line that contains "root". I've heard of awk command, but even with the manual, I do not understand how to reach my goal.
Thank you in advance for your answer
Simple awk should do it:
awk '$1 ~ /root/' file.txt
Related
I have a file with many lines, one is:
COMPOSER_HOME=/home/glen/.composer
I want to extract the string /home/glen/.composer from this file in my shell script. How can I?
I can get the whole line with grep but not sure how to remove the first part.
Here:
grep 'COMPOSER_HOME=' file| cut -d= -f2
cut cut's by delimiter = and the 2nd portion would be whatever is After the = e.g.: /home/glen/.composer , with -f1 you would get COMPOSER_HOME
Since you tagged linux, you have GNU grep which includes PCRE
grep -oP 'COMPOSER_HOME=\K.+' file
The \K means match what comes before, then throw it out and operate on the rest of the line.
You can also use awk
awk -F "=" '$1 == "COMPOSER_HOME" {print $2}' file
Maybe this is enough
sed -nE 's/COMPOSER_HOME=(.*)/\1/p' your_file
It does not print any line unless you explicitly request it (-n), it matches the line starting with COMPOSER_HOME= and captures what follows (.*) (using () instead of \(\), thanks to -E), and puts in the replacement only what is captured. Then requests the printing of the line with the p flag of the substitution command.
I am trying to capture the string between 2 specific letters/words using sed/awk. This is what I am trying to do:
The input is a file test.log containing
Owner: CN=abc.samplecerrt.com,o=IN,DC=com
Owner: CN=abc1.samplecerrt.com,o=IN,DC=com
I want to extract only "CN=abc.samplecerrt.com"
I tried
sed 's/.*CN=\(.*\),.*/\1/p' test.log >> result.log
But this returns "abc.samplecerrt.com,o=IN,DC=com"
How do I go about this?
test file:
$ cat logs.txt
CN=abc.samplecerrt.com,o=IN,DC=com Owner: CN=abc1.samplecerrt.com,o=IN,DC=com
command and output:
$ grep -oP 'CN=(?:(?!CN=).)*?.com' logs.txt
CN=abc.samplecerrt.com
CN=abc1.samplecerrt.com
This might work for you (GNU sed):
sed -n 's/.*\(CN=[^,]*\).*/\1/p' file
Or:
sed 's/.*\(CN=[^,]*\).*/\1/p;d' file
The first turns off implicit printing -n so as to act like grep.
Matches and captures the string CN= followed by zero or more non-comma characters and prints the captured group \1 if a match is made.
The second solution is much the same except it deletes all lines and only prints the captured group as above.
With awk you can get the field where is the string you need. For it, you can set FS=:|, Now if you run
awk -v FS=":|," '{print $2}' file
CN=abc.samplecerrt.com
CN=abc1.samplecerrt.com
you get the field. But you only want one, so
awk -v FS=":|," '$2 !~ /abc1/ {print $2}' file
CN=abc.samplecerrt.com
I am trying to number the lines of a txt file and hide the empty ones . I use this code :
cat -n file.txt | grep . file.txt
But it doesnt work . It ignores the cat command . I want to display all the non-empty lines and number them ( the txt file is not a static one , like a list that a user can type in ).
edit : Given the great solutions below , i would also add that grep . file.txt | cat -n also worked .
I assume you want to number the lines that remain after the empty lines are removed.
Solution #1
Use sed '/^$/d' to delete the empty lines then pipe its output to cat -n to number them:
sed '/^$/d' file.txt | cat -n
The sed program contains only one command: d (delete the line). The sed commands can be prefixed by zero, one or two addresses that tell what lines the command applies to.
In this case there is only one address /^$/. It is a regex (enclosed in /) that selects the empty lines; the lines where start of the line (^) is followed by the end of the line ($).
Solution #2
You can also use grep -v '^$' to filter out the empty lines:
grep -v '^$' file.txt | cat -n
Again, ^$ is a regular expression that matches the empty lines. -v reverses the condition and tells grep to display the lines that do not match the regex.
The commands above do not modify the file. They read the content of file.txt, process it and display the result on screen.
Update
As #robc suggests in a comment, nl is even better than cat -n to number the lines. Thank you #robc, I didn't know about nl until now (I didn't know about cat -n either). It is never too late to learn new things.
This could be easily done with awk. This will print line with line numbers and ignore empty lines.
awk 'NF{print FNR,$0}' file.txt
Explanation: Adding detailed explanation for above code.
awk ' ##Starting awk program from here.
NF{ ##Checking condition if NF(number of fields) is NOT NULL in current line then do following.
print FNR,$0 ##Printing current line number by mentioning FNR and then current line value.
}
' file.txt ##Mentioning Input_file name which we are passing to awk program here.
Please, accept my apologies, if this question was asked before. I am new and do not know how to do it. I have a file containing the data like this:
name=1|surname=2|phone=3|email=4
phone=5|surname=6|name=7|email=8
surname=9|phone=10|email=11|name=12
phone=13|email=14|name=15|surname=6
I would like to have a file like this:
name=1
name=7
name=12
name=15
Thanks in advance!
Say names.txt is your file, then use something like :
cat names.txt | tr "|" "\n" | grep "^name="
tr transforms | to newlines
grep filters for the lines with name
And here is a one command solution with GNU awk:
awk -v RS="[|\n]" '/^name=/' names.txt
the -v RS="[|\n]' set the record separatro to|` or newline
the /^name=/ filters for records starting with name= (and implicitly prints them)
I would go for the solution of #Lars, but I wanted to test this with "lookbehind".
With grep you can get the matches only with grep -o, but the following line will also find surname:
grep -o "name=[0-9]*" names.txt
You can fix this a little by looking for the character before name (start of line with ^ or |).
grep -o "(^|\|)name=[0-9]*" names.txt
What a fix! Now you get the right names, but sometimes with an extra |.
With \K (and grep option -P) you can tell grep to use something for the matching but skip it during output.
grep -oP "(^|\|)\Kname=[0-9]*" names.txt
I have a file: users.txt that looks like the below:
Name
Jack
Name
Jackson
Name
Peter
I need to remove EVERYTHING ABOVE the last match found Name so that the file looks like:
Name
Peter
I have tried searching the net, but only find awk command's that can achieve this.
I need it to be sed in order to use the -i option to save. I have come across where a file is sometimes empty when redirecting > standard output using awk.
Your time and assistance is greatly appreciated.
You can use tac twice with awk:
tac file | awk '{print} /Name/{exit}' | tac
Name
Peter
Through sed,
tac file | sed -n '0,/Name/p' | tac
Using sed without tac:
sed -n '/Name/h; //!H; ${g;p}' file
When Name is seen, replace the hold space with the current line
When Name is NOT seen, append the current line to the hold space
At the last line, replace the pattern space with the hold space and print.
Since we don't have sed in a pipeline, you can use -i to replace the file.
All in awk
awk 'x{x=x"\n"$0}/Name/{x=$0}END{print x}' file
This might work for you (GNU sed):
sed 'H;/Name/h;$!d;x' file
Hold matches in the hold space and at the end of file print the hold space with the last match.