This question already has an answer here:
How can I retrieve an entry from /etc/passwd for a given username?
(1 answer)
Closed 5 years ago.
I have this line
Username:x:120:101:somethingsomething
and I need to get the '101' part after the third ':', how can I do that?
do I use grep or sed?
cut -d':' -f4 /etc/passwd
awk, only with string:
mstr="Username:x:120:101:somethingsomething"; awk -F: '{print $4}' <<< "$mstr"
Related
This question already has answers here:
How do I get the last word in each line with bash
(7 answers)
Closed 3 years ago.
for a file file.txt, the contents are:
Had repulsive dashwoods suspicion sincerity but advantage now him.
Remark easily garret nor nay.
Civil those mrs enjoy shy fat merry. You greatest jointure saw horrible.
He private he on be imagine suppose.
Fertile beloved evident through no service elderly is.
Now I want to cut such a way that I only get the last words of the lines
I tried
cut -d" " -f1- file.txt
but that just gives all from start to end.
cut -d" " -f-1 file.txt
This just gives the first word.
You can do that easily with awk:
echo "Hi there Zeeshan." | awk '{print $NF}'
would print Zeeshan.
For your file:
awk '{print $NF}' file.txt
This question already has answers here:
How to read the last line of a text file into a variable using Bash? [closed]
(2 answers)
Print the last line of a file, from the CLI
(5 answers)
Closed 4 years ago.
I am working on a tool project. I need to grab the last line from a file & assign into a variable. This is what I have tried:
line=$(head -n $NF input_file)
echo $line
Maybe I could read the file in reverse then use
line=$(head -n $1 input_file)
echo $line
Any ideas are welcome.
Use tail ;)
line=$(tail -n 1 input_file)
echo $line
Combination of tac and awk here. Benefit in this approach could be we need NOT to read complete Input_file in it.
tac Input_file | awk '{print;exit}'
With sed or awk :
sed -n '$p' file
sed '$!d' file
awk 'END{print}' file
However, tail is still the right tool to do the job.
This question already has answers here:
Extract filename and extension in Bash
(38 answers)
Closed 7 years ago.
I have files with "multiple" extension , for better manipulation I would like to create new folder for each last extension but first I need to retrieve the last extension.
Just for example lets assume i have file called info.tar.tbz2 how could I get "tbz2" ?
One way that comes to my mind is using cut -d "." but in this case I would need to specify -f parameter of the last column, which I don't know how to achieve.
What is the fastest way to do it?
You may use awk,
awk -F. '{print $NF}' file
or
sed,
$ echo 'info.tar.tbz2' | awk -F. '{print $NF}'
tbz2
$ echo 'info.tar.tbz2' | sed 's/.*\.//'
tbz2
This question already has answers here:
How to split a string into an array in Bash?
(24 answers)
Closed 7 years ago.
Suppose I have a file with a name ABC_DE_FGHI_10_JK_LMN.csv. I want to extract the ID from the file-name i.e. 10 with the help of ID position and file-name separator. I have following two inputs
File-name_ID_Position=4; [since 10 is at fourth position in file-name]
File-name_Delimiter="_";
Here ID can be numeric or alpha-numeric. So how extract the 10 from above file with the help of above two inputs. How to achieve this in bash?
Instead of writing a regex in bash, I would do it with awk:
echo 'ABC_DE_FGHI_10_JK_LMN.csv' | awk -F_ -v pos=4 '{print $pos}'
or if you want the dot to also be a delimiter (requires GNU awk):
echo 'ABC_DE_FGHI_10_JK_LMN.csv' | awk -F'[_.]' -v pos=4 '{print $pos}'
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Using awk with variables
The following command is wrong, the point is I want to use $curLineNumber in awk, how can I do it? Any solution?
curLineNumber = 3
curTime=`ls -l | awk 'NR==$curLineNumber {print $NF}'`
Thanks
curTime=$(ls -l | awk -v line=$curLineNumber 'NR == line { print $NF }'
The -v option is used to specify variables initialized on the command line. I chose the name line for the awk variable.