I want to grep line.
Example I have #chargen in file and i want to command to grep output this word with #
cat /etc/inetd.conf | grep '*#' chargen
no result, any idea
Related
I am trying to search for files with specific text but excluding a certain text and showing only the files.
Here is my code:
grep -v "TEXT1" *.* | grep -ils "ABC2"
However, it returns:
(standard input)
Please suggest. Thanks a lot.
The output should only show the filenames.
Here's one way to do it, assuming you want to match these terms anywhere in the file.
grep -LZ 'TEXT1' *.* | xargs -0 grep -li 'ABC2'
-L will match files not containing the given search term
use -LiZ if you want to match TEXT1 irrespective of case
The -Z option is needed to separate filenames with NUL character and xargs -0 will then separate out filenames based on NUL character
If you want to check these two conditions on same line instead of anywhere in the file:
grep -lP '^(?!.*TEXT1).*(?i:ABC2)' *.*
-P enables PCRE, which I assume you have since linux is tagged
(?!regexp) is a negative lookahead construct, so ^(?!.*TEXT1) will ensure the line doesn't have TEXT1
(?i:ABC2) will match ABC2 case insensitively
Use grep -liP '^(?!.*TEXT1).*ABC2' if you want to match both terms irrespective of case
(standard input)
This error is due to use of grep -l in a pipeline as your second grep command is reading input from stdin not from a file and -l option is printing (standard input) instead of the filename.
You can use this alternate solution in a single awk command:
awk '/ABC2/ && !/TEXT1/ {print FILENAME; nextfile}' *.* 2>/dev/null
How can I grep all lines starting by a specific letter (for instance "n")in a file ?
for the file /var/log/messages, I tried
# cat /var/log/messages | grep n*
it didn't work
All lines starting with n:
cat /var/log/messages | grep "^n"
use regular expressions for pattern
^ Indicates the beginning of an input string
$ Indicates the end of an input string
cat /var/log/messages | grep ^n
Have a look at https://www.linux.com/topic/desktop/introduction-regular-expressions-new-linux-users/
Please, accept my apologies, if this question was asked before. I am new and do not know how to do it. I have a file containing the data like this:
name=1|surname=2|phone=3|email=4
phone=5|surname=6|name=7|email=8
surname=9|phone=10|email=11|name=12
phone=13|email=14|name=15|surname=6
I would like to have a file like this:
name=1
name=7
name=12
name=15
Thanks in advance!
Say names.txt is your file, then use something like :
cat names.txt | tr "|" "\n" | grep "^name="
tr transforms | to newlines
grep filters for the lines with name
And here is a one command solution with GNU awk:
awk -v RS="[|\n]" '/^name=/' names.txt
the -v RS="[|\n]' set the record separatro to|` or newline
the /^name=/ filters for records starting with name= (and implicitly prints them)
I would go for the solution of #Lars, but I wanted to test this with "lookbehind".
With grep you can get the matches only with grep -o, but the following line will also find surname:
grep -o "name=[0-9]*" names.txt
You can fix this a little by looking for the character before name (start of line with ^ or |).
grep -o "(^|\|)name=[0-9]*" names.txt
What a fix! Now you get the right names, but sometimes with an extra |.
With \K (and grep option -P) you can tell grep to use something for the matching but skip it during output.
grep -oP "(^|\|)\Kname=[0-9]*" names.txt
How can I apply the following command to only a part of a text file? For example from the beginning to the line 5000.
grep "^ A : 11 B : 10" filename | wc -l
I cannot use head and then apply the above command since the text file is huge.
You could try using the sed command, which I believe does better for large files, from this question and pipe to grep.
sed -n 1,5000p file | grep ...
You can try combination of -n (prefixing each line of output with line number) and -m (limiting number of matching lines). Something like this:
grep -n -m 5000 pattern file.txt | grep -B 5000 "^5000:" | wc -l
First grep search for pattern, add line numbers and limit output to first 5000 matching lines (worst case scenario: all lines from range match pattern). Second grep match line number 5000, and print all lines before this line.
I don't know if it is more efficient solution
In a very large file I need to find the position (line number) of a string, then extract the 2 lines above and below that string.
To do this right now - I launch vi, find the string, note it's line number, exit vi, then use sed to extract the lines surrounding that string.
Is there a way to streamline this process... ideally without having to run vi at all.
Maybe using grep like this:
grep -n -2 your_searched_for_string your_large_text_file
Will give you almost what you expect
-n : tells grep to print the line number
-2 : print 2 additional lines (and the wanted string, of course)
You can do
grep -C 2 yourSearch yourFile
To send it in a file, do
grep -C 2 yourSearch yourFile > result.txt
Use grep -n string file to find the line number without opening the file.
you can use cat -n to display the line numbers and then use awk to get the line number after a grep in order to extract line number:
cat -n FILE | grep WORD | awk '{print $1;}'
although grep already does what you mention if you give -C 2 (above/below 2 lines):
grep -C 2 WORD FILE
You can do it with grep -A and -B options, like this:
grep -B 2 -A 2 "searchstring" | sed 3d
grep will find the line and show two lines of context before and after, later remove the third one with sed.
If you want to automate this, simple you can do a Shell Script. You may try the following:
#!/bin/bash
VAL="your_search_keyword"
NUM1=`grep -n "$VAL" file.txt | cut -f1 -d ':'`
echo $NUM1 #show the line number of the matched keyword
MYNUMUP=$["NUM1"-1] #get above keyword
MYNUMDOWN=$["NUM1"+1] #get below keyword
sed -n "$MYNUMUP"p file.txt #display above keyword
sed -n "$MYNUMDOWN"p file.txt #display below keyword
The plus point of the script is you can change the keyword in VAL variable as you like and execute to get the needed output.