How Grep Conditional matches works? - grep

Here I am practicing the grep commands .Am not clear with the following grep condition how it works?
a_file:
boot
record
boots
process
broken
commands
I had tried the following commands :-
1. grep -A0 "boo" a_file
result:
boot
--
boots
2.grep -A1 "boo" a_file
result:
boot
record
boots
process
3.grep -A2 "boo" a_file
result:
boot
record
boots
process
broken
4.grep -A3 "boo" a_file
result:
boot
record
boots
process
broken
commands
Note:I had studied this grep command from terminal man grep.
My Query:
1.What is the purpose of switch -A?
2.How the context lines are ordered for every numeric values (i.e 1,2,3)?

For your first question, -A or -B is really useful while you dubug a long and complex log. You are allowed to use -A and -B option to see more details behind and before the searched pattern in the log also speed up the debug efficiency.
For your second question, from the src of grep, there's no specific limitation for option -A. The argument of it ,i.e. out_after, is defined as an long int in the src.
static intmax_t out_after; /* Lines of trailing context. */
I assume that the trailing lines to be printed until EOF or next matched pattern is hit (the count of trailing line would be reset).

Related

Grep with RegEx Inside a Docker Container?

Dipping my toes into Bash coding for the first time (not the most experienced person with Linux either) and I'm trying to read the version from the version.php inside a container at:
/config/www/nextcloud/version.php
To do so, I run:
docker exec -it 1c8c05daba19 grep -eo "(0|[1-9]\d*)\.(0|[1-9]\d*)\.(0|[1-9]\d*)(?:-((?:0|[1-9]\d*|\d*[a-zA-Z-][0-9a-zA-Z-]*)(?:\.(?:0|[1-9]\d*|\d*[a-zA-Z-][0-9a-zA-Z-]*))*))?(?:\+([0-9a-zA-Z-]+(?:\.[0-9a-zA-Z-]+)*))?" /config/www/nextcloud/version.php
This uses a semantic versioning RegEx pattern (I know, a bit overkill, but it works for now) to read and extract the version from the line:
$OC_VersionString = '20.0.1';
However, when I run the command it tells me No such file or directory, (I've confirmed it does exist at that path inside the container) and then proceeds to spit out the entire contents of the file it just said doesn't exist?
grep: (0|[1-9]\d*).(0|[1-9]\d*).(0|[1-9]\d*)(?:-((?:0|[1-9]\d*|\d*[a-zA-Z-][0-9a-zA-Z-])(?:.(?:0|[1-9]\d|\d*[a-zA-Z-][0-9a-zA-Z-]))))?(?:+([0-9a-zA-Z-]+(?:.[0-9a-zA-Z-]+)*))?: No such file or directory
/config/www/nextcloud/version.php:$OC_Version = array(20,0,1,1);
/config/www/nextcloud/version.php:$OC_VersionString = '20.0.1';
/config/www/nextcloud/version.php:$OC_Edition = '';
/config/www/nextcloud/version.php:$OC_VersionCanBeUpgradedFrom = array (
/config/www/nextcloud/version.php: 'nextcloud' =>
/config/www/nextcloud/version.php: 'owncloud' =>
/config/www/nextcloud/version.php:$vendor = 'nextcloud';
Anyone able to spot the problem?
Update 1:
For the sake of clarity, I'm trying to run this from a bash script. I just want to fetch the version number from that file, to use it in other areas of the script.
Update 2:
Responding to the comments, I tried to login to the container first, and then run the grep, and still get the same result. Then I cat that file and it shows it's contents no problem.
Many containers don't have the GNU versions of Unix tools and their various extensions. It's popular to base containers on Alpine Linux, which in turn uses a very lightweight single-binary tool called BusyBox to provide the base tools. Those tend to have the set of options required in the POSIX specs, and no more.
POSIX grep(1) in particular doesn't have an -o option. So the command you're running is
grep \
-eo \ # specify "o" as the regexp to match
"(regexps are write-only)" \ # a filename
/config/www/nextcloud/version.php # a second filename
Notice that the grep output in the interactive shell only contains lines with the letter "o", but not for example the line just containing array.
POSIX grep doesn't have an equivalent for GNU grep's -o option
Print only the matched (non-empty) parts of matching lines, with each such part on a separate output line. Output lines use the same delimiters as input....
but it's easy to do that with sed(1) instead. Ask it to match some stuff, the regexp in question, and some stuff, and replace it with the matched group.
sed -e 's/.*\(any regexp here\).*/\1/' input-file
(POSIX sed only accepts basic regular expressions, so you'll have to escape more of the parentheses.)
Well, for any potential future readers, I had no luck getting grep to do it, I'm sure it was my fault somehow and not grep's, but thanks to the help in this post I was able to use awk instead of grep, like so:
docker exec -it 1c8c05daba19 awk '/^\$OC_VersionString/ && match($0,/\047[0-9]+\.[0-9]+\.[0-9]+\047/){print substr($0,RSTART+1,RLENGTH-2)}' /config/www/nextcloud/version.php
That ended up doing exactly what I needed:
It logs into a docker container.
Scans and returns just the version number from the line I am looking for at: /config/www/nextcloud/version.php inside the container.
Exits stage left from the container with just the info I needed.
I can get right back to eating my Hot Cheetos.

grep printout, whitespaces consolidated

My own bash script searches compilation logs (higher-level logs generated by Bitbake cooker while cross-building a Linux software package) for compilation command line strings. Those command strings in searched logs are huge in length, more than 2300 columns of ascii text (due several -I's and -W's).
Log file is searched for compilation command line strings using following regex
grep -E -e "^${TARGETCANONICAL}${COMPCMDPATT}"
where
TARGETCANONICAL='arm-oe-linux-gnueabi-'
COMPCMDPATT="gcc[[:space:]].+[[:space:]]-c[[:space:]].+\w+\.c$"
Used grep command is constructed to print out each line matching given pattern, with no manipulation of matching input line printed to grep output. Ubuntu is build machine.
Each compilation command string in log-file includes at few points sequences of multiple subsequent spaces (blanks). For instance:
-ftree-vectorize -Wno-error=maybe-uninitialized
In lines printed out by grep (regex matches) all those original sequences of blanks are consolidated to one single blank, whitespace.
That consolidation of whitespaces is not what was expected here.
What might be the possible reason?
Update:
grep output goes to command substitution (bash expansion) whose output is assigned to variable. Then it is printed out using echo or used in compound conditional command - which is the level where myself conducts observations/debugging. Possibly rootcause of observed phenomenon is located even in subsequent command expansion or in assignment.

Remove a certain amount of files with a sequence of characters in a directory

I stopped a process to troubleshoot something. Now, I would like to start the process where it left off in CentOS 6.4.
This script I stopped runs a perl script in a loop to process all of the files in /dev/shm/split/. These files were split into many parts from a larger file. An example of how they are named are as follows:
file.txt.aa
file.txt.ab
file.txt.ac
...and so on.
I have identified that the script left off at file.txt.fy. So, I would like to remove all of the files in /dev/shm/split/ that are from file.txt.aa through file.txt.fy.
I tried to create a whitelist for the rm command by doing:
ls /dev/shm/split/ > whitelist
cat whitelist | egrep -v 'file.txt.[aa-fz]' | tee whitelist.tmp
This did not do what I had intended.
Please help me! Thank you!
The problem with your command is that you cannot match two characters with the square bracket pattern in bash. You should use something like that instead:
ls file.txt.[a-e]? file.txt.f[a-y]
Basically decompose your range into two ranges, the first will match .aa to .ez, and the second .fa to .fy (included).
Note that I have used the ls command here. I always find it a good idea to first echo or ls the commands/files you're going to run when the operations you do are potentially destructive. When you're sure it produces the right output, go on and use rm instead of ls.

How to grep for two words existing on the same line? [duplicate]

This question already has answers here:
Match two strings in one line with grep
(23 answers)
Closed 3 years ago.
How do I grep for lines that contain two input words on the line? I'm looking for lines that contain both words, how do I do that? I tried pipe like this:
grep -c "word1" | grep -r "word2" logs
It just stucks after the first pipe command.
Why?
Why do you pass -c? That will just show the number of matches. Similarly, there is no reason to use -r. I suggest you read man grep.
To grep for 2 words existing on the same line, simply do:
grep "word1" FILE | grep "word2"
grep "word1" FILE will print all lines that have word1 in them from FILE, and then grep "word2" will print the lines that have word2 in them. Hence, if you combine these using a pipe, it will show lines containing both word1 and word2.
If you just want a count of how many lines had the 2 words on the same line, do:
grep "word1" FILE | grep -c "word2"
Also, to address your question why does it get stuck : in grep -c "word1", you did not specify a file. Therefore, grep expects input from stdin, which is why it seems to hang. You can press Ctrl+D to send an EOF (end-of-file) so that it quits.
Prescription
One simple rewrite of the command in the question is:
grep "word1" logs | grep "word2"
The first grep finds lines with 'word1' from the file 'logs' and then feeds those into the second grep which looks for lines containing 'word2'.
However, it isn't necessary to use two commands like that. You could use extended grep (grep -E or egrep):
grep -E 'word1.*word2|word2.*word1' logs
If you know that 'word1' will precede 'word2' on the line, you don't even need the alternatives and regular grep would do:
grep 'word1.*word2' logs
The 'one command' variants have the advantage that there is only one process running, and so the lines containing 'word1' do not have to be passed via a pipe to the second process. How much this matters depends on how big the data file is and how many lines match 'word1'. If the file is small, performance isn't likely to be an issue and running two commands is fine. If the file is big but only a few lines contain 'word1', there isn't going to be much data passed on the pipe and using two command is fine. However, if the file is huge and 'word1' occurs frequently, then you may be passing significant data down the pipe where a single command avoids that overhead. Against that, the regex is more complex; you might need to benchmark it to find out what's best — but only if performance really matters. If you run two commands, you should aim to select the less frequently occurring word in the first grep to minimize the amount of data processed by the second.
Diagnosis
The initial script is:
grep -c "word1" | grep -r "word2" logs
This is an odd command sequence. The first grep is going to count the number of occurrences of 'word1' on its standard input, and print that number on its standard output. Until you indicate EOF (e.g. by typing Control-D), it will sit there, waiting for you to type something. The second grep does a recursive search for 'word2' in the files underneath directory logs (or, if it is a file, in the file logs). Or, in my case, it will fail since there's neither a file nor a directory called logs where I'm running the pipeline. Note that the second grep doesn't read its standard input at all, so the pipe is superfluous.
With Bash, the parent shell waits until all the processes in the pipeline have exited, so it sits around waiting for the grep -c to finish, which it won't do until you indicate EOF. Hence, your code seems to get stuck. With Heirloom Shell, the second grep completes and exits, and the shell prompts again. Now you have two processes running, the first grep and the shell, and they are both trying to read from the keyboard, and it is not determinate which one gets any given line of input (or any given EOF indication).
Note that even if you typed data as input to the first grep, you would only get any lines that contain 'word2' shown on the output.
Footnote:
At one time, the answer used:
grep -E 'word1.*word2|word2.*word1' "$#"
grep 'word1.*word2' "$#"
This triggered the comments below.
you could use awk. like this...
cat <yourFile> | awk '/word1/ && /word2/'
Order is not important. So if you have a file and...
a file named , file1 contains:
word1 is in this file as well as word2
word2 is in this file as well as word1
word4 is in this file as well as word1
word5 is in this file as well as word2
then,
/tmp$ cat file1| awk '/word1/ && /word2/'
will result in,
word1 is in this file as well as word2
word2 is in this file as well as word1
yes, awk is slower.
The main issue is that you haven't supplied the first grep with any input. You will need to reorder your command something like
grep "word1" logs | grep "word2"
If you want to count the occurences, then put a '-c' on the second grep.
git grep
Here is the syntax using git grep combining multiple patterns using Boolean expressions:
git grep -e pattern1 --and -e pattern2 --and -e pattern3
The above command will print lines matching all the patterns at once.
If the files aren't under version control, add --no-index param.
Search files in the current directory that is not managed by Git.
Check man git-grep for help.
See also:
How to use grep to match string1 AND string2?
Check if all of multiple strings or regexes exist in a file.
How to run grep with multiple AND patterns?
For multiple patterns stored in the file, see: Match all patterns from file at once.
You cat try with below command
cat log|grep -e word1 -e word2
Use grep:
grep -wE "string1|String2|...." file_name
Or you can use:
echo string | grep -wE "string1|String2|...."

To understand the practical use of Grep's option -H in different situations

This question is based on this answer.
Why do you get the same output from the both commands?
Command A
$sudo grep muel * /tmp
masi:muel
Command B
$sudo grep -H muel * /tmp
masi:muel
Rob's comment suggests me that Command A should not give me masi:, but only muel.
In short, what is the practical purpose of -H?
Grep will list the filenames by default if more than one filename is given. The -H option makes it do that even if only one filename is given. In both your examples, more than one filename is given.
Here's a better example:
$ grep Richie notes.txt
Richie wears glasses.
$ grep -H Richie notes.txt
notes.txt:Richie wears glasses.
It's more useful when you're giving it a wildcard for an unknown number of files, and you always want the filenames printed even if the wildcard only matches one file.
If you grep a single file, -H makes a difference:
$ grep muel mesi
muel
$ grep -H muel mesi
masi:muel
This could be significant in various scripting contexts. For example, a script (or a non-trivial piped series of commands) might not be aware of how many files it's actually dealing with: one, or many.
When you grep from multiple files, by default it shows the name of the file where the match was found. If you specify -H, the file name will always be shown, even if you grep from a single file. You can specify -h to never show the file name.
Emacs has grep interface (M-x grep, M-x lgrep, M-x rgrep). If you ask Emacs to search for foo in the current directory, then Emacs calls grep and process the grep output and then present you with results with clickable links. Clickable links, just like Google.
What Emacs does is that it passes two options to grep: -n (show line number) and -H (show filenames even if only one file. the point is consistency) and then turn the output into clickable links.
In general, consistency is good for being a good API, but consistency conflicts with DWIM.
When you directly use grep, you want DWIM, so you don't pass -H.

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