I have a log file that continuously gets updated. I can view the file with tail -f and see all the changes as they come in. However, there is an issue - the text contains the characters \n instead of newlines. I would like to change the characters \n into actual newlines. How do I go about doing that?
Found a solution that works for me:
tail -f logfile | sed 's/\\n/\
/g'
I'm trying to parse MAC address with sed but I faced something which I cannot explain.
Here there are 2 examples - good and wrong:
good one:
# echo "01:23:45:67:89:ab" | sed -r 's/^([^:]{2}):([^:]{2}):([^:]{2}):([^:]{2}):([^:]{2}):([^:]{2})$/\1_\2_\3_\4_\5_\6/'
01_23_45_67_89_ab
wrong one:
# echo "tadam" | sed -r 's/^([^:]{2}):([^:]{2}):([^:]{2}):([^:]{2}):([^:]{2}):([^:]{2})$/\1_\2_\3_\4_\5_\6/'
tadam
The first example is working fine - it returns parsed properly string. But how to explain the second one? Why sed is returning any string (tadam) while this string doesn't match te regexp?
Anyone can explain it? I'd like to achieve the situation where sed is returning nothing when the regexp is not matching the string. How to do it?
sed's default setting is to print the complete pattern space.
Add option -n (suppress printing of pattern space) and add command p (print current pattern space):
echo "..." | sed -r -n 's/.../.../p'
Now sed only prints the pattern space if something could be replaced.
Im trying to append a line in a file using following sed command
sed -i 's/id="1asda"/id="1asda"\n peer="120"/g' a.xml.
Now the problem is that even if I'm giving a wrong id, sed is not popping an error.How to get an error if the pattern is not found.
sed is used to replace text. Having nothing to replace is not an error.
Use something like grep to check whether there's something to replace or not. For example:
if grep -q 'id="1asda"' a.xml
then
sed -i 's/id="1asda"/id="1asda"\n peer="120"/g' a.xml
else
echo 'nothing to do' >& 2
fi
I would suggest to use awk:
awk 'BEGIN{e=1}gsub(/foo1/, "bar",$0){e=0};1;END{exit e}' file
The above command will give you a return value of 1 if no substitution has been performed, otherwise 0.
I have a file called sms:
gsm versi jadul
29 sender: +62896666666
date: 15/02/14,03:55:12
reboot router
when I type in:
sed -n '6p' sms > /tmp/result
The /tmp/result always looks like this:
Notice the line break there, I want to get rid of the line break on the second line, so the final result will be like this:
How do I do that?
You could trim it off with tr like this:
sed -n '6p' sms | tr -d '\n' > /tmp/result
You can use awk instead of sed:
awk 'NR==6 {printf $0}' sms > result
NR==6 specifies line number
printf $0 prints that line without any \n
There's nothing wrong with your sed command, your input file contains trailing control-Ms. Remove them with dos2unix or similar before running sed.
A correct implementation of the POSIX sed command does not add such a blank line. 6p should print the sixth line. I cannot reproduce the issue on, for example, Ubuntu 12 Linux. You have some line ending problem or some such issue.
Trying to debug an issue with a server and my only log file is a 20GB log file (with no timestamps even! Why do people use System.out.println() as logging? In production?!)
Using grep, I've found an area of the file that I'd like to take a look at, line 347340107.
Other than doing something like
head -<$LINENUM + 10> filename | tail -20
... which would require head to read through the first 347 million lines of the log file, is there a quick and easy command that would dump lines 347340100 - 347340200 (for example) to the console?
update I totally forgot that grep can print the context around a match ... this works well. Thanks!
I found two other solutions if you know the line number but nothing else (no grep possible):
Assuming you need lines 20 to 40,
sed -n '20,40p;41q' file_name
or
awk 'FNR>=20 && FNR<=40' file_name
When using sed it is more efficient to quit processing after having printed the last line than continue processing until the end of the file. This is especially important in the case of large files and printing lines at the beginning. In order to do so, the sed command above introduces the instruction 41q in order to stop processing after line 41 because in the example we are interested in lines 20-40 only. You will need to change the 41 to whatever the last line you are interested in is, plus one.
# print line number 52
sed -n '52p' # method 1
sed '52!d' # method 2
sed '52q;d' # method 3, efficient on large files
method 3 efficient on large files
fastest way to display specific lines
with GNU-grep you could just say
grep --context=10 ...
No there isn't, files are not line-addressable.
There is no constant-time way to find the start of line n in a text file. You must stream through the file and count newlines.
Use the simplest/fastest tool you have to do the job. To me, using head makes much more sense than grep, since the latter is way more complicated. I'm not saying "grep is slow", it really isn't, but I would be surprised if it's faster than head for this case. That'd be a bug in head, basically.
What about:
tail -n +347340107 filename | head -n 100
I didn't test it, but I think that would work.
I prefer just going into less and
typing 50% to goto halfway the file,
43210G to go to line 43210
:43210 to do the same
and stuff like that.
Even better: hit v to start editing (in vim, of course!), at that location. Now, note that vim has the same key bindings!
You can use the ex command, a standard Unix editor (part of Vim now), e.g.
display a single line (e.g. 2nd one):
ex +2p -scq file.txt
corresponding sed syntax: sed -n '2p' file.txt
range of lines (e.g. 2-5 lines):
ex +2,5p -scq file.txt
sed syntax: sed -n '2,5p' file.txt
from the given line till the end (e.g. 5th to the end of the file):
ex +5,p -scq file.txt
sed syntax: sed -n '2,$p' file.txt
multiple line ranges (e.g. 2-4 and 6-8 lines):
ex +2,4p +6,8p -scq file.txt
sed syntax: sed -n '2,4p;6,8p' file.txt
Above commands can be tested with the following test file:
seq 1 20 > file.txt
Explanation:
+ or -c followed by the command - execute the (vi/vim) command after file has been read,
-s - silent mode, also uses current terminal as a default output,
q followed by -c is the command to quit editor (add ! to do force quit, e.g. -scq!).
I'd first split the file into few smaller ones like this
$ split --lines=50000 /path/to/large/file /path/to/output/file/prefix
and then grep on the resulting files.
If your line number is 100 to read
head -100 filename | tail -1
Get ack
Ubuntu/Debian install:
$ sudo apt-get install ack-grep
Then run:
$ ack --lines=$START-$END filename
Example:
$ ack --lines=10-20 filename
From $ man ack:
--lines=NUM
Only print line NUM of each file. Multiple lines can be given with multiple --lines options or as a comma separated list (--lines=3,5,7). --lines=4-7 also works.
The lines are always output in ascending order, no matter the order given on the command line.
sed will need to read the data too to count the lines.
The only way a shortcut would be possible would there to be context/order in the file to operate on. For example if there were log lines prepended with a fixed width time/date etc.
you could use the look unix utility to binary search through the files for particular dates/times
Use
x=`cat -n <file> | grep <match> | awk '{print $1}'`
Here you will get the line number where the match occurred.
Now you can use the following command to print 100 lines
awk -v var="$x" 'NR>=var && NR<=var+100{print}' <file>
or you can use "sed" as well
sed -n "${x},${x+100}p" <file>
With sed -e '1,N d; M q' you'll print lines N+1 through M. This is probably a bit better then grep -C as it doesn't try to match lines to a pattern.
Building on Sklivvz' answer, here's a nice function one can put in a .bash_aliases file. It is efficient on huge files when printing stuff from the front of the file.
function middle()
{
startidx=$1
len=$2
endidx=$(($startidx+$len))
filename=$3
awk "FNR>=${startidx} && FNR<=${endidx} { print NR\" \"\$0 }; FNR>${endidx} { print \"END HERE\"; exit }" $filename
}
To display a line from a <textfile> by its <line#>, just do this:
perl -wne 'print if $. == <line#>' <textfile>
If you want a more powerful way to show a range of lines with regular expressions -- I won't say why grep is a bad idea for doing this, it should be fairly obvious -- this simple expression will show you your range in a single pass which is what you want when dealing with ~20GB text files:
perl -wne 'print if m/<regex1>/ .. m/<regex2>/' <filename>
(tip: if your regex has / in it, use something like m!<regex>! instead)
This would print out <filename> starting with the line that matches <regex1> up until (and including) the line that matches <regex2>.
It doesn't take a wizard to see how a few tweaks can make it even more powerful.
Last thing: perl, since it is a mature language, has many hidden enhancements to favor speed and performance. With this in mind, it makes it the obvious choice for such an operation since it was originally developed for handling large log files, text, databases, etc.
print line 5
sed -n '5p' file.txt
sed '5q' file.txt
print everything else than line 5
`sed '5d' file.txt
and my creation using google
#!/bin/bash
#removeline.sh
#remove deleting it comes move line xD
usage() { # Function: Print a help message.
echo "Usage: $0 -l LINENUMBER -i INPUTFILE [ -o OUTPUTFILE ]"
echo "line is removed from INPUTFILE"
echo "line is appended to OUTPUTFILE"
}
exit_abnormal() { # Function: Exit with error.
usage
exit 1
}
while getopts l:i:o:b flag
do
case "${flag}" in
l) line=${OPTARG};;
i) input=${OPTARG};;
o) output=${OPTARG};;
esac
done
if [ -f tmp ]; then
echo "Temp file:tmp exist. delete it yourself :)"
exit
fi
if [ -f "$input" ]; then
re_isanum='^[0-9]+$'
if ! [[ $line =~ $re_isanum ]] ; then
echo "Error: LINENUMBER must be a positive, whole number."
exit 1
elif [ $line -eq "0" ]; then
echo "Error: LINENUMBER must be greater than zero."
exit_abnormal
fi
if [ ! -z $output ]; then
sed -n "${line}p" $input >> $output
fi
if [ ! -z $input ]; then
# remove this sed command and this comes move line to other file
sed "${line}d" $input > tmp && cp tmp $input
fi
fi
if [ -f tmp ]; then
rm tmp
fi
You could try this command:
egrep -n "*" <filename> | egrep "<line number>"
Easy with perl! If you want to get line 1, 3 and 5 from a file, say /etc/passwd:
perl -e 'while(<>){if(++$l~~[1,3,5]){print}}' < /etc/passwd
I am surprised only one other answer (by Ramana Reddy) suggested to add line numbers to the output. The following searches for the required line number and colours the output.
file=FILE
lineno=LINENO
wb="107"; bf="30;1"; rb="101"; yb="103"
cat -n ${file} | { GREP_COLORS="se=${wb};${bf}:cx=${wb};${bf}:ms=${rb};${bf}:sl=${yb};${bf}" grep --color -C 10 "^[[:space:]]\\+${lineno}[[:space:]]"; }