Separate a text file with sed - linux

I have the following sample file:
evtlog.161202.002609.debugevtlog.161201.162408.debugevtlog.161202.011046.debugevtlog.161202.002809.debugevtlog.161201.160035.debugevtlog.161201.155140.debugevtlog.161201.232156.debugevtlog.161201.145017.debugevtlog.161201.154816.debug
I want to separate the string and add a newline after matching "debug" like this:
evtlog.161202.002609.debug
evtlog.161201.162408.debug
So far I tried almost everything with sed, but it doesn't seem to do what I want.
sed 's/debug/{G}' latest_evtlogs.out
sed '/debug/i "SAD"' latest_evtlogs.out
etc...
sed 's/debug/\n/g' latest_evtlogs.out doesn't work when I add it as a pipe in the script , but it does when I run it manually.
Here's how I generate the file:
printf $(ls -l $EVTLOG_PATH/evtlog|tail -n 10|awk '{printf $8 , "%s\n\n"}'|sed 's/debug/\n/g') >> latest_evtlogs.out
Initially I wanted to just add newline with awk, but it doesn't work either.
Any ideas why I can't separate the string with a newline ?
I'm using :
Distributor ID: Debian
Description: Debian GNU/Linux 5.0.10 (lenny)
Release: 5.0.10
Codename: lenny

Just add a new line after debug:
sed 's/debug/&\n/g' file
Note & prints back the matched text, so it is a way to print "debug" back.
This returns:
evtlog.161202.002609.debug
evtlog.161201.162408.debug
evtlog.161202.011046.debug
evtlog.161202.002809.debug
evtlog.161201.160035.debug
evtlog.161201.155140.debug
evtlog.161201.232156.debug
evtlog.161201.145017.debug
evtlog.161201.154816.debug

The problem is, that you are using the output of sed in a command expansion. In this context your shell will replace all newlines with spaces. The spaces are then used to do the word splitting, so that printf sees each line as a separate argument, interpreting the first line as the format argument and ignoring the rest as there are printf-placeholders in the format.
It should work if you drop the outer printf $() from your command and just redirect the output from your pipeline to your file:
ls -l $EVTLOG_PATH/evtlog|tail -n 10|awk '{printf $8 , "%s\n\n"}'|sed 's/debug/\n/g' >> latest_evtlogs.out

Maybe Perl is "happier" than sed on your system:
perl -pe 's/debug/&\n/g' < YourLogFile

Get will append what is in the hold buffer unto the pattern space (Usually just the current line read from the input file) So this cannot be used.
insert will print the specified text to standard output. So this cannot be used.
What you you want to to replace all debug with debug^J, where ^J is a newline, dependent on the sed version, you can either do:
sed 's/debug/&\n/g' input_file
But \n is - afaik - not strictly specified in POSIX sed. One can however use c strings:
sed 's/debug/&'$'\n''/g' input_file
Or a multi line string:
sed 's/debug/&\
/g' input_file

Thank you all for the answers.I finally did it like this :
echo $(ls -l $EVTLOG_PATH/evtlog|tail -n 10|awk '{printf $8 , "%s\n\n"}'|sed 's/debug/&\n/g') > temp.out
sed 's/ /\n/g' /share/sqa/dumps/5314577631/checks/temp.out > latest_evtlogs.out
It's not at all elegant, but it finally works.

Related

How to use grep and sed in order to replace the substring after searching some specific string?

I want to know how to use two 'grep' and 'sed' utilities or something else in order to replace the substring. I will explain what I want to do below.
We have the file 'test.txt' with the following string:
A1='AA1', A2='AA2', A3='AA3', A4='AA4', A5{ATTR}='AA5', A6='keyword_A'
After searching 'keyword_A' using grep, I want to replace the value of A5 with other string, for example, "NEW".
A1='AA1', A2='AA2', A3='AA3', A4='AA4', A5{ATTR}='NEW', A6='keyword_A'
I tried to use two commands like
grep keyword_A test.txt | sed -e 's/blabla/blabla/'
After trying all I know, I gave up at all.
Please let me know the right solution.
First, you never need grep and sed. Sed has a full regular-expression search engine, so it is a superset of grep. This command will read test.txt, change the lines that you've indicated, and print the entire result on standard output:
sed "/keyword_A/s/A5{ATTR}='[A-Z0-9]*'/A5{ATTR}='NEW'/g" < test.txt
If you want to store the results back into the file test.txt, use the -i (in-place editing) switch to sed:
sed "/keyword_A/s/A5{ATTR}='[A-Z0-9]*'/A5{ATTR}='NEW'/g" -i.bak test.txt
If you want to select only the indicated lines, modify those, and print only those lines to standard out, use a combination of the p (print) command and the -n (no output) switch.
sed "/keyword_A/s/A5{ATTR}='[A-Z0-9]*'/A5{ATTR}='NEW'/gp" -n test.txt
Using grep+sed is always the wrong approach. Here's one way to do it with GNU awk:
$ awk '/keyword_A/{ $0=gensub(/(A5({[^}]+})?=\047)[^\047]+/,"\\1NEW",1) } 1' file
A1='AA1', A2='AA2', A3='AA3', A4='AA4', A5{ATTR}='NEW', A6='keyword_A'
Using a couple variables you could define the keyword and replacement ( if they change at all ):
q="keyword_A"
r="NEW"
Then with sed:
sed -r "s/^(.+\{.+\}=')(.+)('.+"${q}".+)$/\1"${r}"\3/" file
Result:
A1='AA1', A2='AA2', A3='AA3', A4='AA4', A5{ATTR}='NEW', A6='keyword_A'
A5="NEW"
A6="keyword_A"
# with sed
sed "s/='[^']*\(',[[:blank:]]*A6='${A6}'\)/='${A5}\1/" YourFile
# with awk
awk -F "'" -v A5="${A5}" -v A6="${A6}" '
BEGIN { OFS="\047" }
$12 == A6 { $10 = A5; $0 = $0 }
7
' YourFile
Change by the end of the string, for sed and using ' as field separator in awk instead of traditional space.
assuming there is no ' in value (or need to treat the escaping method) for awk version
We can just directly replace the fifth column when the sting keyword_A is found as shown below:
awk -F, 'BEGIN{OFS=",";}/keyword_A/{$5="A5{ATTR}='"'"NEW"'"'"}1' filename
Couple of slight alternatives:
sed -r "/keyword_A/s/(A5[^']*')[^']*/\1NEW/"
awk -F"'" '/keyword_A/{$10 = "NEW"}1' OFS="'"
Of course the negative with awk is afterwards you would have to rename the new file.

How to remove line breaks generated by sed

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.

Shell Linux : grep exact sentence with NULL character

I have a file like
key\0value\n
akey\0value\n
key2\0value\n
I have to create a script that take as argument a word. I have to return every lines having a key exactly the same than the argument.
I tried
grep -aF "$key\x0"
but grep seems to do not understand the \x0 (\0 same result). Futhermore, I have to check that the line begins with "$key\0"
I only can use sed grep and tr and other no maching commands
To have the \0 taken into account try :
grep -Pa "^key\x0"
it works for me.
Using sed
sed will work:
$ sed -n '/^key1\x00/p' file
key1value
The use of \x00 to represent a hex character is a GNU extension to sed. Since this question is tagged linux, that is not a problem.
Since the null character does not display well, one might (or might not) want to improve the display with something like this:
$ sed -n 's/^\(akey\)\x00/\1-->/p' file
akey-->value
Using sed with keys that contain special characters
If the key itself can contain sed or shell active characters, then we must escape them first and then run sed against the input file:
#!/bin/bash
printf -v script '/^%s\\x00/p' "$(sed 's:[]\[^$.*/]:\\&:g' <<<"$1")"
sed -n "$script" file
To use this script, simply supply the key as the first argument on the command line, enclosed in single-quotes, of course, to prevent shell processing.
To see how it works, let's look at the pieces in turn:
sed 's:[]\[^$.*/]:\\&:g' <<<"$1"
This puts a backslash escape in front of all sed-active characters.
printf -v script '/^%s\\x00/p' "$(sed 's:[]\[^$.*/]:\\&:g' <<<"$1")"
This creates a sed command using the escaped key and stores it in the shell variable script.
sed -n "$script" file
This runs sed using the shell variable script as the sed command.
Using awk
The question states that awk is not an acceptable tool. For completeness, though, here is an awk solution:
$ awk -F'\x00' -v k=key1 '$1 == k' file
key1value
Explanation:
-F'\x00'
awk divides the input up into records (lines) and divides the records up into fields. Here, we set the field separator to the null character. Consequently, the first field, denoted $1, is the key.
-v k=key1
This creates an awk variable, called k, and sets it to the key that we are looking for.
$1 == k
This statement looks for records (lines) for which the first field matches our specified key. If a match is found, the line is printed.

Append text to file without line breaking

On a Linux machine, I have list of IPs as follows:
107.6.38.55
108.171.207.62
108.171.244.138
108.171.246.87
I want to use some function to add the word "or" at the end of each line without breaking each line, like this:
107.6.38.55 or
108.171.207.62 or
108.171.244.138 or
108.171.246.87 or
Every implementation I have experimented with in sed or awk has given me incorrect results as it keeps trying to line break or add input in strange spots. What is the easiest way to achieve this goal?
With awk '$0=$0" or"' and the sed suggestions I've tried thus far I get the following formatting:
107.6.38.55
or
108.171.207.62
or
108.171.244.138
or
108.171.246.87
or
Not sure what you have been trying but the following works for me on Ubuntu 12.04
awk '{print $0" or"}'
Or as fedorqui suggests
awk '$0=$0" or"'
Or as glenn jackman suggests
awk '{print $0, "or"}'
[EDIT]
It turns out the OP's file had CRLF line breaks so dos2unix had to be run first to address the format issue
The following two worked for me:
sed 's/.*/& or/'
sed 's/$/ or/'
Or use ed, the standard text editor:
With bash you can use the lovely here-strings together with ANSI-C quotings
ed -s filename <<< $',s/.$/& or/\nwq'
or a pipe with printf
printf "%s\n" ',s/.$/& or/' 'wq' | ed -s filename
or if you like echo better
{ echo ',s/.$/& or/'; echo "wq"; } | ed -s filename
or interactively (if you love question marks):
$ ed filename
,s/.$/& or/
wq
Remark. I'm using the substitution s/.$/& or/ and not s/$/ or/ just so as not to append or in an empty line.

Insert newline before first line

I am trying to insert a newline before the first line of text in a file. The only solution i have found so far is this:
sed -e '1 i
')
I do not like to have an actual newline in my shell script. Can this be solved any other way using the standard (GNU) UNIX utilities?
For variety:
echo | cat - file
Here's a pure sed solution with no specific shell requirements:
sed -e '1 s|^|\n|'
EDIT:
Please note that there has to be at least one line of input for this (and anything else using a line address) to work.
A $ before a single-quoted string will cause bash to interpret escape sequences within it.
sed -e '1 i'$'\n'
You could use awk:
$ awk 'FNR==1{print ""} 1' file
Which will work with any number of files.

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