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After using grep command, i will give you a small example
grep test
test this is first line : 1
test this is second line : 2
test this is third line : 3
test this is fourth line : 4
How to filter the last line after grep command executed
finally i need the result 4
If I don't misunderstand you:
$grep test | tail -1
>test this is fourth line : 4
$grep test | tail -1 | awk '{print $7}'
>4
$7 is 'the seventh column' in awk.
Related
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Suppose I have a file with 3 lines:
output.txt:
Maruti
Zen
Suzuki
I used the command wc -l output.txt to get no. of lines
i got output as 3
Based on the above output I have to execute a command
echo CREATE FROM (sed -n 1p OUTPUT.txt)
echo CREATE FROM (sed -n 2p OUTPUT.txt)
echo CREATE FROM (sed -n 3p OUTPUT.txt)
:
:
echo CREATE FROM (sed -n np OUTPUT.txt)
Can you please suggest a command to replace 1 2 3 .....n in the above command based on the output i get (i.e., no. of lines in my file)
I just gave a sample explanation of my use case. Please suggest a command to execute n no. of times.
You just need one command.
sed 's/^/CREATE FROM /' output.txt
See also Counting lines or enumerating line numbers so I can loop over them - why is this an anti-pattern?
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I am having a text file containing a list of date and time just like the sample below -
posted_at"
2012-06-09 11:48:31"
2012-08-09 12:40:02"
2012-04-09 13:10:00"
2012-03-09 13:40:00"
2012-10-09 14:30:01"
2012-12-09 15:30:00"
2012-11-09 16:20:00"
I want to extract the month from each line.
P.S - grep should not be used at any point of the code
Thanks in advance!
First select date pattern :
egrep '[0-9]{4}-[0-9]{2}-[0-9]{2} ' content_file
Second, extract the month :
awk -F '-' '{print $2}'
Third redirect to desired file :
>> desired_file
So mix of all this with | to final solution :
egrep '[0-9]{4}-[0-9]{2}-[0-9]{2} ' content_file| awk -F '-' '{print $2}'>> desired_file
Voilà
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I have a file name test,
it contains a String James Bond 007,
and i want to print only James Bond.
I tried the following commands:
$ strings -n 2 test
$ sed -n '/James/,/Bond/p' test
$ awk '{print substr($1,10)}' test
To print the first two words, you can use awk:
awk '{print $1, $2}' test
To print the first ten characters, you can put the file contents in a variable, then use the bash substring operation:
contents=$(cat test)
echo "${contents:0:10}"
Or in awk:
awk '{print substr($0, 1, 10)}' test
Notice that $0 means the whole line, and you have to give both a starting index and length to substr(). Indexes in awk start at 1 rather than 0.
In sed, /James/,/Bond/ is a line range expression; it processes all the lines starting from a line containing James until a line containing Bond. It doesn't process just part of the lines.
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My requirement is to play an alert (Play Sound) in putty if any Exception occurs in the Application , so for this purpose i have tried the below way
tail -f flex.log | grep "Exception" --color paplay alert.wav
But even though the word Exception occurs in flex Log File during tail , but it is not playing the sound .
Please let me know if there is any mistake in the above command .
I am using centOS 8 as OS and script is bash .
This will find all words with Exception and replace it with the bell character, your terminal should beep/flash/whatever you've set up to happen during a terminal bell.
tail -f flex.log | grep "Exception" | sed -e $'s/Exception/Exception\a/'
To see ALL lines of flex.log but bell only on "Exception":
tail -f flex.log | sed -e $'s/Exception/Exception\a/'
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I have comma deliminated file that basically has the structure of:
1, 2, 3, 4, 5 ,,,, 6, etc
I have to count the number of unique 6th columns. Pleasseee help
(btw this is an intro to unix/linux class so this should be able to be done with basic commands)
cut -d "," -f 6 myFile |sort |uniq -c |wc -l
Looking into my crystal ball, I see that your class is discussing awk. Try
awk -F, '!a[$6]++{c++} END{ print c }' input-file