I have a requirement where i'd like to omit the 1st line from the output of ls -latr "some path" Since I need to remove total 136 from the below output
So I wrote ls -latr /home/kjatin1/DT_901_linux//autoInclude/system | tail -q which excluded the 1st line, but when the folder is empty it does not omit it. Please tell me how to omit 1st line in any linux command output
The tail program can do this:
ls -lart | tail -n +2
The -n +2 means “start passing through on the second line of output”.
Pipe it to awk:
awk '{if(NR>1)print}'
or sed
sed -n '1!p'
ls -lart | tail -n +2 #argument means starting with line 2
This is a quick hacky way: ls -lart | grep -v ^total.
Basically, remove any lines that start with "total", which in ls output should only be the first line.
A more general way (for anything):
ls -lart | sed "1 d"
sed "1 d" means only print everything but first line.
You can use awk command:
For command output use pipe: | awk 'NR>1'
For output of file: awk 'NR>1' file.csv
Related
I'm connecting to an exadata and want to get information about "ORACLE_HOME" variable inside them. So i'm using this command:
ls -l /proc/<pid>/cwd
this is the output:
2 oracle oinstall 0 Jan 23 21:20 /proc/<pid>/cwd -> /u01/app/database/11.2.0/dbs/
i need the get the last part :
/u01/app/database/11.2.0 (i dont want the "/dbs/" there)
i will be using this command several times in different machines. So how can i get this substring from whole output?
Awk and grep are good for these types of issues.
New:
ls -l /proc/<pid>/cwd | awk '{print ($NF) }' | sed 's#/dbs/##'
Old:
ls -l /proc/<pid>/cwd | awk '{print ($NF) }' | egrep -o '^.+[.0-9]'
Awk prints the last column of the input which is your ls command and then grep grabs the beginning of that string up the last occurrence of numbers and dots. This is a situational solution and perhaps not the best.
Parsing the output of ls is generally considered sub-optimal. I would use something more like this instead:
dirname $(readlink -f /proc/<pid>/cwd)
I have a number of log files in a directory. I am trying to write a script to search all the log files for a string and echo the name of the files and the line number that the string is found.
I figure I will probably have to use 2 grep's - piping the output of one into the other since the -l option only returns the name of the file and nothing about the line numbers. Any insight in how I can successfully achieve this would be much appreciated.
Many thanks,
Alex
$ grep -Hn root /etc/passwd
/etc/passwd:1:root:x:0:0:root:/root:/bin/bash
combining -H and -n does what you expect.
If you want to echo the required informations without the string :
$ grep -Hn root /etc/passwd | cut -d: -f1,2
/etc/passwd:1
or with awk :
$ awk -F: '/root/{print "file=" ARGV[1] "\nline=" NR}' /etc/passwd
file=/etc/passwd
line=1
if you want to create shell variables :
$ awk -F: '/root/{print "file=" ARGV[1] "\nline=" NR}' /etc/passwd | bash
$ echo $line
1
$ echo $file
/etc/passwd
Use -H. If you are using a grep that does not have -H, specify two filenames. For example:
grep -n pattern file /dev/null
My version of grep kept returning text from the matching line, which I wasn't sure if you were after... You can also pipe the output to an awk command to have it ONLY print the file name and line number
grep -Hn "text" . | awk -F: '{print $1 ":" $2}'
I am trying to find the occurance of tab in a file some_file and print those line with leading line number.
grep -nP "\t" some_file works well for me but I want sed or awk equivalent command for the same.
To emulate: grep -nP "\t" file.txt
Here's one way using GNU awk:
awk '/\t/ { print NR ":" $0 }' file.txt
Here's one way using GNU sed:
< file.txt sed -n '/\t/{ =;p }' | sed '{ N;s/\n/:/ }'
Well, you can always do it in sed:
cat -n test.txt | sed -n "/\t/p"
Unfortunately, sed can only print line numbers to stdout with a new line, so in any case, more than one command is necessary. A more lengthy (unnecessary so) version of the above, but one only using sed, would be:
sed = test.txt | sed -n "N;s/\n/ /;/\t/p"
but I like the one with cat more. CATS ARE NICE.
In a very large file I need to find the position (line number) of a string, then extract the 2 lines above and below that string.
To do this right now - I launch vi, find the string, note it's line number, exit vi, then use sed to extract the lines surrounding that string.
Is there a way to streamline this process... ideally without having to run vi at all.
Maybe using grep like this:
grep -n -2 your_searched_for_string your_large_text_file
Will give you almost what you expect
-n : tells grep to print the line number
-2 : print 2 additional lines (and the wanted string, of course)
You can do
grep -C 2 yourSearch yourFile
To send it in a file, do
grep -C 2 yourSearch yourFile > result.txt
Use grep -n string file to find the line number without opening the file.
you can use cat -n to display the line numbers and then use awk to get the line number after a grep in order to extract line number:
cat -n FILE | grep WORD | awk '{print $1;}'
although grep already does what you mention if you give -C 2 (above/below 2 lines):
grep -C 2 WORD FILE
You can do it with grep -A and -B options, like this:
grep -B 2 -A 2 "searchstring" | sed 3d
grep will find the line and show two lines of context before and after, later remove the third one with sed.
If you want to automate this, simple you can do a Shell Script. You may try the following:
#!/bin/bash
VAL="your_search_keyword"
NUM1=`grep -n "$VAL" file.txt | cut -f1 -d ':'`
echo $NUM1 #show the line number of the matched keyword
MYNUMUP=$["NUM1"-1] #get above keyword
MYNUMDOWN=$["NUM1"+1] #get below keyword
sed -n "$MYNUMUP"p file.txt #display above keyword
sed -n "$MYNUMDOWN"p file.txt #display below keyword
The plus point of the script is you can change the keyword in VAL variable as you like and execute to get the needed output.
I'm using this command to sort and remove duplicate lines from a file.
sort file2.txt | uniq > file2_uniq.txt
After performing the command, I find the last line with this value: \n which cause me problems. What can I do to avoid it ?
You could also let sort take care of uniquing the output, omitting the first line would avoid empty lines:
sort -u file2.txt | tail -n +2
Edit
If you also wanted to remove all empty lines I would suggest using:
grep -v '^$' | sort -u file2.txt
Just filter out what you don't want:
sort file2.txt | egrep -v "^$" | uniq > file2_uniq.txt
The problem solved by removing the last line using:
sed '$d' infile > outfile