I am trying to find which words happened in logfiles plus show the logfilename for anything that matches following pattern:
'BA10\|BA20\|BA21\|BA30\|BA31\|BA00'
so if file dummylogfile.log contains BA10002 I would like to get a result such as:
dummylogfile.log:BA10002
it is totally fine if the logfile shows up twice for duplicate matches.
the closest I got is:
for f in $(find . -name '*.err' -exec grep -l 'BA10\|BA20\|BA21\|BA30\|BA31\|BA00' {} \+);do printf $f;printf ':';grep -o 'BA10\|BA20\|BA21\|BA30\|BA31\|BA00' $f;done
but this gives things like:
./register-05-14-11-53-59_24154.err:BA10
BA10
./register_mdw_files_2020-05-14-11-54-32_24429.err:BA10
BA10
./process_tables.2020-05-18-11-18-09_11428.err:BA30
./status_load_2020-05-18-11-35-31_9185.err:BA30
so,
1) there are empty lines with only the second match and
2) the full match (e.g., BA10004) is not shown.
thanks for the help
There are a couple of options you can pass to grep:
-H: This will report the filename and the match
-o: only show the match, not the full line
-w: The match must represent a full word (string build from [A-Za-z0-9_])
If we look at your regex, you use BA01, this will match only BA01 which can appear anywhere in the text, also mid word. If you want the regex to match a full word, it should read BA01[[:alnum:]_]* which adds any sequence of word-constituent characters (equivalent to [A-Za-z0-9_]). You can test this with
$ echo "foo BA01234 barBA012" | grep -Ho "BA01"
(standard input):BA01
(standard input):BA01
$ echo "foo BA01234 barBA012" | grep -How "BA01"
$ echo "foo BA01234 barBA012" | grep -How "BA01[[:alnum:]_]*"
(standard input):BA01234
So your grep should look like
grep -How "\('BA10\|BA20\|BA21\|BA30\|BA31\|BA00'\)[[:alnum:]_]*" *.err
From your example it seems that all files are in one directory. So the following works right away:
grep -l 'BA10\|BA20\|BA21\|BA30\|BA31\|BA00' *.err
If the files are in different directories:
find . -name '*.err' -print | xargs -I {} grep 'BA10\|BA20\|BA21\|BA30\|BA31\|BA00' {} /dev/null
Explanation: the addition of /dev/null to the filename {} forces grep to report the matching filename
Related
Say I have multiple .gz files that I want to search a keyword in them. I can do this by piping zcat result to a grep like this:
zcat some.file.* | grep "keyword_1" | ... | grep "keyword_n"
The output of this command though will be just the matching line and won't have the file name in it. Is there any way I can attach the file name to the zcat output?
Try zgrep instead of zcat:
zgrep -H keyword some.file.*
And if you want to use egrep to get pattern matching:
export GREP=egrep
zgrep -H -e "(keyword1|keyword2)" some.file.*
I have set of files in a directory. In those, few files contain a matching pattern config_dict["backup.moduleDir"] and some characters following them. In few other files the pattern appears exactly at the end of the line (no characters followed after the pattern). Note that, the pattern appears exactly one time in all these files.
Now, I want to find those file names which have some characters following a matching pattern. I use the below code:
find . -type f -name "*.py" -exec grep -El 'config_dict\["backup.moduleDir"].+$' {} \;
Actually I want to avoid the use of regex character '+' and extended pattern option -E of grep. So I tried using the grep -v logic by the following 2 ways, but it did not give me the expected result. What really went wrong in the below 2 methods?
grep -vl 'config_dict\["backup.moduleDir"\]$' `find . -type f -name "*.py" -exec grep -l 'backup.moduleDir' {} \;`
find . -type f -name "*.py" -exec grep -l 'backup.moduleDir' {} \; | xargs grep -vl 'config_dict["backup.moduleDir"]$'
Surprisingly in the above working code, I have to escape only the opening square bracket '[' where as escaping is optional for closing square bracket ']' and for double quotes and for dot character between the strings "backup" and "moduleDir". How this is possible?
Using a simple dot without + does the job:
grep 'config_dict\["backup.moduleDir"].' *.py
This will find config_dict["backup.moduleDir"] followed by at least 1 character, in all python scripts.
Okay I have a file that contains numbers like this:
L21479
What I am trying to do is use grep (or a similar tool) to find all the strings in a file that have the format:
L#####
The # will be the number. SO an L followed by 5 numbers.
Is this even possible in grep? Should I load the file and perform regex?
You can do this with grep, for example with the following command:
grep -E -o 'L[0-9]{5}' name_of_file
For example, given a file with the text:
kasdhflkashl143112343214L232134614
3L1431413543454L2342L3523269ufoidu
gl9983ugsdu8768IUHI/(JHKJASHD/(888
The command above will output:
L23213
L14314
L35232
If it is just in a single file, you can do something along the lines of:
grep -e 'L[0-9]{5}' filename
If you need to search all files in a directory for these strings:
find . -type f | xargs grep -e 'L[0-9]{5}'
Suppose I write a grep query to find out the occurrence of a method call on an object like this:
// might not be accurate, but irrelevant
grep -nr "[[:alnum:]]\.[[:alnum:]](.*)" .
This would give many results. How to find out how many such results are obtained?
What about using | wc -l to count the number of result lines?
What about
man grep | grep "count"
It outputs
-c, --count
Suppress normal output; instead print a count of matching lines for each input file. [...]
Previous answers are OK, I just want to put it into command line instructions in order to have copy-paste versions (from explicit to simplest) for the future:
grep --count "PATTERN" FILE
Is exactly the same as:
grep -c "PATTERN" FILE
And it is equivalent to:
grep "PATTERN" FILE | wc -l
As a bonus, below i give you a version where a file with a list of patterns is used.
grep -count --file=PATTERNFILE FILE
or simply
grep -cf PATTERNFILE FILE
I need to find ALL files that have multiple keywords anywhere in the file (not necessarily on the same line), given a starting directory like ~/. Does "grep -ro" do this?
(I'm using Unix, Mac OSX 10.4)
You can use the -l option to get a list of filenames with matches, so it's just a matter of finding all of the files that have the first keyword and then filtering that list down to the files that also have the second keyword:
grep -rl first_keyword basedir | xargs grep -l second_keyword
To search just *.txt
find ~/. -name "*.txt" | xargs grep -l first_keyword | xargs grep -l second_keyword
Thanks Adam!