This is an example of a filename.csv file
"Sort Order","Common Name","Formal Name","Type","Sub Type"
"1","Afghanistan","Islamic State of Afghanistan","Independent State"
"2","Albania","Republic of Albania","Independent State"
"3","Algeria","People's Democratic Republic of Algeria","Independent State"
"4","Andorra","Principality of Andorra","Independent State"
"5","Angola","Republic of Angola","Independent State"
So what is the grep command to search for angola in common name and print it like this:
"5","Angola","Republic of Angola","Independent State"
I know that we can use:
grep '"Algeria"' filename.csv
However, what if I am being more specific. Let's say Algeria might exist in other column however, we only need to print the one in Common Name column.
I tried
grep '"Common Name | Algeria"' filename.csv
Seems not to work.
You could try the below grep command to print the lines which contains the string Angola in the second column.
grep -E '^"[^"]*","Angola"' file
This could be easily done through awk,
awk -F, '$2=="\"Angola\""' file
try awk
awk -F"," '$2~/Algeria/' file
Use cut command,
grep Algeria filename.csv|cut -f2 -d','|awk -F '[""]' '{print $2}'
Output:
Algeria
Here is the explanation of all command I have added with your grep and discarded some unnecessary quotes you used in it.
cut command:
-f It will extract 2nd column depending on delimiter
-d It will choose delimiter i.e.: which is here ,
awk command:
It will extract value between two "" (block quotes) and print it.
Related
I need to grep some pattern and further i need to print some output within that. Currently I am using the below command which is working fine. But I like to eliminate using multiple pipe and want to use single awk command to achieve the same output. Is there a way to do it using awk?
root#Server1 # cat file
Jenny:Mon,Tue,Wed:Morning
David:Thu,Fri,Sat:Evening
root#Server1 # awk '/Jenny/ {print $0}' file | awk -F ":" '{ print $2 }' | awk -F "," '{ print $1 }'
Mon
I want to get this output using single awk command. Any help?
You can try something like:
awk -F: '/Jenny/ {split($2,a,","); print a[1]}' file
Try this
awk -F'[:,]+' '/Jenny/{print $2}' file.txt
It is using muliple -F value inside the [ ]
The + means one or more since it is treated as a regex.
For this particular job, I find grep to be slightly more robust.
Unless your company has a policy not to hire people named Eve.
(Try it out if you don't understand.)
grep -oP '^[^:]*Jenny[^:]*:\K[^,:]+' file
Or to do a whole-word match:
grep -oP '^[^:]*\bJenny\b[^:]*:\K[^,:]+' file
Or when you are confident that "Jenny" is the full name:
grep -oP '^Jenny:\K[^,:]+' file
Output:
Mon
Explanation:
The stuff up until \K speaks for itself: it selects the line(s) with the desired name.
[^,:]+ captures the day of week (in this case Mon).
\K cuts off everything preceding Mon.
-o cuts off anything following Mon.
I understand that grep can extract the specific content from a file line by line.
Just wondering how can add another column before or after each line as an index.
For example:
grep "aaa" text.txt > tmp.txt
In the tmp.txt file, we can see the content as follows,
aaawekjre
qejrteraaa
wrgeaaaere
However, I would like to add a specific index as an extra column.
Therefore, the tmp.txt might look like this:
John aaawekjre
John qejrteraaa
John wrgeaaaere
You can use awk:
awk '/aaa/{print "John", $0}' text.txt > tmp.txt
$ sed -n '/aaa/ s/^/John /p' text.txt
John aaawekjre
John qejrteraaa
John wrgeaaaere
How it works
-n
This tells sed not to print anything unless we explicitly ask it to.
/aaa/ s/^/John /p
This selects lines that contain aaa. For those lines, we do a substitution (s/^/John /) to put John at the beginning of the line and we print the line (p).
In this way, lines that do not contain aaa are never printed. Thus, there is no need for a separate grep process.
try this
grep "aaa" text.txt | awk '{print "John " $0}' > tmp.txt
This image is the /etc/passwd fileLinux: What single sed command can be used to output matches from /etc/passwd that have Smith or Jones in their description (5th field) to a file called smith_jones.txt?
I wouldn't use sed, but it looks like you're referencing a standard /etc/passwd file, so something that may do what you're looking for is this:
cat /etc/passwd | awk -F ":" '{if ($5 ~ /Smith/ || $5 ~ /Jones/) print}'
So awk '{print $5}' is commonly used to print the 5th column of something piped to it, in this case the /etc/passwd file. However, as it's not tabular data, I've supplied -F argument with the delimiter ":" as that's what splits our values.
It's then a fairly easy if statement essentially saying, if this string contains Smith OR Jones in it somewhere, print it.
I have a text file like mentioned below, and along with that I will pass an input for which I want a corresponding output.
Input file: test.txt
abc:abc_1
abcd:abcd_1
1_abcd:1_abcd_bkp
xyz:xyz_2
so if I use abc with the above test.txt file, I want abc_1; and if I pass abcd, I need abcd_1 as output.
I tried cat text.txt | grep abc | cut -d":" -f2,2, but I am getting the output
abc_1
abcd_1
1_abcd_bkp
when I want only abc_1.
With GNU grep:
grep -Po "^abc:\K.*" file
Output:
abc_1
\K keeps the text matched so far out of the overall regex match.
You want to use a regular expression with the -e switch.
In particular, regular expressions allow you to use caret (^) to express the start of a line.
Since you only care about abc when it's at the start of a line and it's followed by :, you want:
cat test.txt | grep -e "^abc:" | cut -d":" -f2,2
Output:
abc_1
awk to the rescue!
awk -F: -v key="abc" '$1==key{print $2}'
using : as the delimiter do the look up for key on field 1 to return field 2.
Or, by moving the key in the script
awk -F: '$1=="abc"{print $2}'
you can try the exclude -v:
cat text.txt | grep abc | grep -vi abc[a-z]
not sure if that would work exactly, try something with that kind of idea
Without specifying second field to be printed the whole line will be or in other cases lines.
awk -F: '/abc_/{print $2}' file
abc_1
awk -F: 'NR==1,/abc/{print $2}' file
abc_1
I am having following syntax for one of my file.Could you please anyone explain me what is this command doing
path = /document/values.txt
where we have different username specified e.g username1 = john,username2=marry
cat ${path} | grep -e username1 | cut -d'=' -f2`
my question here is cat command is reading from the file value of username1 but why why we need to use cut command?
Cat is printing the file. The file has username1=something in one of the lines. The cut command splits this and prints out the second argument.
your command was not written well. the cat is useless.
you can do:
grep -e pattern "$path"|cut ...
you can of course do it with single process with awk if you like. anyway the line in your question smells not good.
awk example:
awk -F'=' '/pattern/{print $2}' inputFile
cut -d'=' -f2`
This cut uses -d'=' that means you use '=' as 'field delimiter' and -f2 will take only de second field.
So in this case you want only the value after the "=" .