GREP: is there a way to use grep inserting a text between filename and the pattern? - shell
I have grep --color -EH "^([^,]*\,){3}5" try.csv
and the output it does is this:
try.csv:410,30151010,K,5001,,,,,,,,,,,,,,,,,,,,,,,,,,,,,512
try.csv:652,20151010,K,5001,,,,,,,,,,,,,,,,,,,,,,,,,,,,,41
try.csv:109,30151010,R,5005,,,,,,,,,,,,,,,,,,,,,,,,,,,,,455
I tried grep --color -EH "^([^,]*,){3}5" try.csv | perl -ne 'print ",$_"'
but the output looks like this :
,try.csv:410,30151010,K,5001,,,,,,,,,,,,,,,,,,,,,,,,,,,,,512
,try.csv:652,20151010,K,5001,,,,,,,,,,,,,,,,,,,,,,,,,,,,,41
,try.csv:109,30151010,R,5005,,,,,,,,,,,,,,,,,,,,,,,,,,,,,455
Expected output:
try.csv:,410,30151010,K,5001,,,,,,,,,,,,,,,,,,,,,,,,,,,,,512
try.csv:,652,20151010,K,5001,,,,,,,,,,,,,,,,,,,,,,,,,,,,,41
try.csv:,109,30151010,R,5005,,,,,,,,,,,,,,,,,,,,,,,,,,,,,455
I am very new to Perl and shell. I'm searching in the CSV files.
You may insert a comma by using sed,
$ grep --color -EH "^([^,]*\,){3}5" try.csv | sed 's/:/&,/'
s/:/&,/: the the special character & in the replacement refers to that portion of the string which matched. And you may add a comma behind & to meet your requirement.
Related
How do I grep only the word I gave as a variable?
I want grep a word in my file, but only the part I gave to grep. Example : in my file, i've "hell\nhell:o", i want grep hell but not hello. How can I do that?
Give this a try: grep -E "\bhell(\s|$)" file add o if you just want the matched word: kent$ echo "hell\nhell:o"|grep -oE "\bhell(\s|$)" hell or: grep -oP '\bhell(?=\s|$)'
Using the -o option of grep : grep -o hell your_file From grep manual page : -o, --only-matching Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
Your question (and comments) aren't overly clear but you could try either: "negative lookahead" using the "end of line anchor" Negative Lookahead The following regex will match any hell that isn't followed by a : (adapt as suitable): hell(?!:) grep doesn't support this, so you'll need to use perl or something else: echo -e "hell\nhell:o" \ | perl -ne 'print if /hell(?!:)/' End of Line Anchor The following will work with grep, matching only where the hell touches the end of the line ($): hell$ echo -e "hell\nhell:o" \ | grep 'hell$'
Try to use -oP echo "hell hello" | grep -oP '\bhell\b'
Grep multiple strings from text file
Okay so I have a textfile containing multiple strings, example of this - Hello123 Halo123 Gracias Thank you ... I want grep to use these strings to find lines with matching strings/keywords from other files within a directory example of text files being grepped - 123-example-Halo123 321-example-Gracias-com-no 321-example-match so in this instance the output should be 123-example-Halo123 321-example-Gracias-com-no
With GNU grep: grep -f file1 file2 -f FILE: Obtain patterns from FILE, one per line. Output: 123-example-Halo123 321-example-Gracias-com-no
You should probably look at the manpage for grep to get a better understanding of what options are supported by the grep utility. However, there a number of ways to achieve what you're trying to accomplish. Here's one approach: grep -e "Hello123" -e "Halo123" -e "Gracias" -e "Thank you" list_of_files_to_search However, since your search strings are already in a separate file, you would probably want to use this approach: grep -f patternFile list_of_files_to_search
I can think of two possible solutions for your question: Use multiple regular expressions - a regular expression for each word you want to find, for example: grep -e Hello123 -e Halo123 file_to_search.txt Use a single regular expression with an "or" operator. Using Perl regular expressions, it will look like the following: grep -P "Hello123|Halo123" file_to_search.txt EDIT: As you mentioned in your comment, you want to use a list of words to find from a file and search in a full directory. You can manipulate the words-to-find file to look like -e flags concatenation: cat words_to_find.txt | sed 's/^/-e "/;s/$/"/' | tr '\n' ' ' This will return something like -e "Hello123" -e "Halo123" -e "Gracias" -e" Thank you", which you can then pass to grep using xargs: cat words_to_find.txt | sed 's/^/-e "/;s/$/"/' | tr '\n' ' ' | dir_to_search/* As you can see, the last command also searches in all of the files in the directory. SECOND EDIT: as PesaThe mentioned, the following command would do this in a much more simple and elegant way: grep -f words_to_find.txt dir_to_search/*
How to exclude hyphen as word separator in bash
I am unable to grep for exact word match containing hyphen as in /home/imper-home,3,0,0,0,jim.imper,NONE,NONE,NONE,http://sanjose /home/imper,15,10,3,30,jim.imper,NONE,NONE,NONE,http://sanjose-age I tried grep -w imper but it returns both /home/imper-home and /home/imper. I want only /home/imper-home to returned by using, grep -wv /home/imper
This will work in general: word=imper grep -w "$word" file | grep -v "$word-"
How to grep with a pattern that includes a " quotation mark?
I want to grep a line that includes a quotation mark, more specifically I want to grep lines that include a " mark. more specifically I want to grep lines like: #include "something.h" then pipe into sed to just return something.h
A single grep will do this job. grep -oP '(?<=")[^"]*(?=")' file Example: $ echo '#include "something.h"' | grep -oP '(?<=")[^"]*(?=")' something.h
sed '#n
/"/ s/.*"\([^"]*\)" *$/\1/p' YourFile
No need of grep (unless performance on huge file is wanted) with a sed. Sed could filter and adapt directly the content
In your case, /"/ is certainly modified by /#include *"/
in case of several string between quote
sed '#n
/"/ {s/"[^"]*$/"/;s/[^"]*"\([^"]*\)" */\1/gp;}' YourFile
You can use awk to get included filename:
awk -F'"' '{print $2}' file.c
something.h
Text Manipulation using sed or AWK
I get the following result in my script when I run it against my services. The result differs depending on the service but the text pattern showing below is similar. The result of my script is assigned to var1. I need to extract data from this variable $var1=HOST1*prod*gem.dot*serviceList : svc1 HOST1*prod*kem.dot*serviceList : svc3, svc4 HOST1*prod*fen.dot*serviceList : svc5, svc6 I need to strip the name of the service list from $var1. So the end result should be printed on separate line as follow: svc1 svc2 svc3 svc4 svc5 svc6 Can you please help with this? Regards
Using sed and grep: sed 's/[^ ]* :\|,\|//g' <<< "$var1" | grep -o '[^ ]*' sed deletes every non-whitespace before a colon and commas. Grep just outputs the resulting services one per line.
Using gnu grep and gnu sed: grep -oP ': *\K\w+(, \w+)?' <<< "$var1" | sed 's/, /\n/' svc1 svc3 svc4 svc5 svc6
grep is the perfect tool for the job.
From man grep:
-o, --only-matching
Print only the matched (non-empty) parts of a matching line, with each such part on a separate output line.
Sounds perfect!
As far as I'm aware this will work on any grep:
echo "$var1" | grep -o 'svc[0-9]\+'
Matches "svc" followed by one or more digits. You can also enable the "highly experimental" Perl regexp mode with -P, which means you can use the \d digit character class and don't have to escape the + any more:
grep -Po 'svc\d+' <<<"$var1"
In bash you can use <<< (a Here String) which supplies "$var1" to grep on the standard input.
By the way, if your data was originally on separate lines, like:
HOST1*prod*gem.dot*serviceList : svc1
HOST1*prod*kem.dot*serviceList : svc3, svc4
HOST1*prod*fen.dot*serviceList : svc5, svc6
This would be a good job for awk:
awk -F': ' '{split($2,a,", "); for (i in a) print a[i]}'