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I use the following 'grep' command to get the count of the string alert in each of my files at the given path:
grep 'alert' -F /usr/local/snort/rules/* -c
How do I sort the resulting output in desired order- say ascending order, descending order, ordered by name, etc. An answer specific to these cases is sufficient.
You may freely suggest a command other than grep as well.
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To sort anything you need to use the sort command. You can use grep to identify the lines to sort, then pipe the output to sort and use the -h switch for a numeric sort with -k identifying what column to sort on.
The grep command prints entire lines when it finds a match in a file. To print only those lines that completely match the search string, add the -x option. The output shows only the lines with the exact match.
For BSD or GNU grep you can use -B num to set how many lines before the match and -A num for the number of lines after the match. If you want the same number of lines before and after you can use -C num . This will show 3 lines before and 3 lines after.
To use grep to search for words in a file, type grep, the word or words you want to search for, the files you want to look in, and press <Enter>. If you want to look for more than one word, you need to put ``double quotes'' around the words.
Pipe it into sort. Assuming your filenames have no colons, use the "-t" option to specify the colon as field saparator. Use -n for numerical sorting.
Example:
grep 'alert' -F /usr/local/snort/rules/* -c | sort -t: -n -k2
should split lines into fields separated by ":", use the second field for sorting, and treat this as numbers (so 21 is actually later than 3).
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