I want to match a string in a file, print the line which matches and the line 10 lines before the match occurs. I am trying using awk, sed and grep, but cant get the right thing. Can anyone help?
Try this command:
grep -B 10 PATTERN file.txt
The -A and -B options to GNU grep allow you to specify how much context should be displayed after and before the matching line.
Related
I'm quite new to bash scripting. I have a script where I want to extract part of the value of a particular line in a separate config file and then use that value as a variable in the script.
For example:
Line 75 in a file named config.cfg
"ssl_cert_location=/etc/ssl/certs/thecert.cer"
I want just the value at the end of "thecert.cer" to then use in the script. I've tried awk and various uses of grep but I can't quite get just the name of the certificate.
Any help would be appreciated. Thanks
These are some examples of the commands I ran:
awk -F "/" '{print $4}' config.cfg
grep -o *.cer config.cfg
Is this possible to extract the value on that line and then edit the output so it just contains the name of the certificate file?
This is a pure Bash version of the basic functionality of basename:
cert=${line##*/}
which removes everything up to and including the final slash. It presupposes that you've already read the line.
Or, using sed:
cert=$(sed -n '75s/^.*\///p' filename)
or
cert=$(sed -n '/^ssl_cert_location=/s/^.*\///p' filename)
This gets the specified line based on the line number or the setting name and replaces everything up to and including the final slash with nothing. It ignores all other lines in the file (unless the setting is repeated in the case of the text match version). The text match version is better because it works no matter what line number the setting is on.
grep uses regular expressions (as does sed). The grep command in your command appears to have a glob expression which won't work. One way to use grep (GNU grep) is to use the PCRE feature (Perl Compatible Regular Expressions):
cert=$(grep -Po '^ssl_cert_location=.*/\K.*' filename)
This works similarly to the sed command.
I have anchored the regular expressions to the beginning of the line. If there may be leading white spaces (the line may be indented), change the regex so it looks something like this:
^[[:space:]]*ssl_cert_location=
which works for both indented and unindented lines.
There are many variants, but a simple one that comes to mind with grep is first getting the line, then matching only non-slashes at the end of the line:
<config.cfg grep '^ssl_cert_location=' | grep -o '[^/]*$'
Why didn't your grep command (grep -o *.cer config.cfg) work? Becasue *.cer is a shell glob pattern and will be expanded by the shell to matching file names, even before the grep process is even started. If there are no matching files, it will be passed verbatim, but * in regular expressions is a quantifier which needs a preceeding expression. . in regex is "match any single character". So what you wanted is probably grep -o '.*\.cer', but .* matches anything, including slashes.
An awk solution would look like the following:
awk -F/ '/^ssl_cert_location=/{print $NF}' config.cfg
It uses "/" as separator, finds only lines starting with "ssl_cert_location" and then prints the last (NF) field in from this line.
Or an equivalent sed solution, which matches the same line and then deletes everything including the last slash:
sed -n '/^ssl_cert_location=/s#^.*/##p' config.cfg
To store the output of any command in a variable, use command substitution:
var="$(command with arguments)"
So I am trying to grep for a specific pattern and then print everything above and below that pattern up to a specific indicator. I don't know if this is possible with grep or if I should you some other tool like awk, sed, or generate some shell script. So if I have the following:
---------------
.....
process: failed
......
----------------
and
----------------
.....
process: frozen
.....
----------------
I want to grep for 'process: frozen' and want everything between the dashed lines when 'process: frozen' is found. However the number of lines between the dashed lines may vary for different 'process: frozen' messages, so I can't count the number of lines above or below and use the -A and -B option of grep. Thank you in advance.
I would use GNU awk and set the record separator to a string which contains 16 hypens:
awk '/process: (failed|frozen)/' RS='-{16}' input.file
You can simply delete the line containing the process: frozen or process: failed message in sed.
sed "/process: frozen/d"
In looking for a command to delete a line (or lines) from a text file that contain a certain string.
For example
I have a text file as follows
Sat 21-12-2014,10.21,78%
Sat 21-12-2014,11.21,60%
Sun 22-12-2014,09.09,21%
I want to delete all lines that have "21-12-2014" in them.
I'm not able to find a solution that works.
According to #twalberg there is more three alternate solution for this question, which I'm explaining is as follows for future reader of this question for more versatile solutions:
With grep command
grep -v 21-12-2014 filename.txt
explanations:
-v is used to find non-matching lines
With awk command
awk '! /21-12-2014/' filename.txt
explanations:
! is denoting it will print all other lines that contain match of the string. It is not operator signify ignorance.
With sed command
sed -e '/21-12-2014/d' < filename.txt
explanations:
-e is signify scripted regex to be executed
d is denoting delete any match
< is redirecting the input file content to command
Try doing this :
sed -i.bak '/21-12-2014/d' *
A bit of explanations :
sed : the main command line, put the mouse pointer on sed
-i.bak : replace the file in place and make a backup in a .bak file
// is the regex
d means: delete
I need to read a file line by line in Linux, find a substring in each line, remove it and place it at the end of that line.
Example:
Line in the original file:
a,b,c,substring,d,e,f
Line in the output file:
a,b,c,d,e,f,substring
How do I do it with the Linux command? Thanks!
sed '/substring/{ s///; s/$/substring/;} '
will handle a fixed substring. Note that if substring begins with a ,, this handles your example case well. If the substring is not fixed but may be a general regular expression:
sed 's/\(substring\)\(.*\)/\2\1'
If you are looking for general csv parsing, you should rephrase the question. (It will be difficult to apply this solution to find a fixed string at the start of a line if you are thinking of the input as comma separated fields.)
I always prefer to use perl's command line to do such regex tasks - perl is powerful enough to cover awk and sed in most of my usages, and both available in windows and linux, it is just easy and handy to me, so the solution in perl would be like:
perl -ne "s/^(.*?)(?:(?<comma>,)(?<substr>substring)|(?<substr>substring)(?<comma>,))(?<right>.*)$/$1$+{right}$+{comma}$+{substr}/; print" input.txt > output.txt
or a simpler one:
perl -lpe "if(s/(,substring|substring,)//){ s/$/,substring/ }" input.txt > output.txt
input.txt
substring,a,b,c,d,e,f
a,b,c,substring,d,e,f
a,b,c,d,e,f,substring
substring,a
a,substring
substring
a
output.txt
a,b,c,d,e,f,substring
a,b,c,d,e,f,substring
a,b,c,d,e,f,substring
a,substring
a,substring
substring
a
You can edit based on your actual input:
If there are any space between words and commas
If you are using tab as separator
Some explanation of the command line:
use perl's -n -e options: -n means process the input line by line in a loop; -e means one line program in the command line
use perl's -l -p options: -l means process multilines; -p means always print
The one line program is just a regex replacement and a print
(?:pattern) means group but don't capture the match
(?<comma>) is a named group, you then need to use $+{comma} hash to access it
I am trying to extract no.s from a file, so I created a script, but grep is giving error:grep: line too long. Can anyone tell me where am I wrong. command is:
echo $(cat filename|grep '\<[0-9]*\>')
Thanks in advance
grep is line-oriented; it will print matching lines to output. Probably you have a huge line in your file, and the resulting line cannot be converted into a string value by shell, as $(...) requires.
First of all, try just cat filename | grep '\<[0-9]*\>' > results and see what is in the results file. Maybe it's enough.
But if you have multiple numbers in a line and you want to extract them all, use -o: grep -o '\<[0-9]*\>'. This will print only matching parts, every match on a new line, even if original matches are on the same line. If you need line numbers, too, add -n.