How to Sort The Output of "Grep -L" Chronologically by Newest Modification Date Last

How to sort the output of grep -l chronologically by newest modification date last?

Try:

ls -rt *.txt | xargs grep -l <pattern>

We first use ls to list *.txt files and sort them by modification time (newest last), then for each entry run them through grep so we only print out files that contain the pattern.

How to sort the output of recursive “grep -lr” chronologically by newest modification date last?

Assuming your file names don't contain newlines:

find dir -type f -printf '%T@\t%p\n' | sort | cut -f2- | xargs grep -l whatever

More robustly using GNU versions of the tools to deal with dir/file names containing exoctic characters:

find dir -type f -printf '%T@\t%p\0' | sort -z | cut -z -f2- | xargs -0 grep -l whatever

How to grep files in date order

You may use this pipeline to achieve this with gnu utilities:

find . -maxdepth 1 -name '*.py' -printf '%T@:%p\0' |
sort -z -t : -rnk1 |
cut -z -d : -f2- |
xargs -0 grep 'pattern'

This will handle filenames with special characters such as space, newline, glob etc.

1. find finds all *.py files in current directory and prints modification time (epoch value) + : + filename + NUL byte
2. sort command performs reverse numeric sort on first column that is timestamp
3. cut command removes 1st column (timestamp) from output
4. xargs -0 grep command searches pattern in each file

UNIX - count unique values based on file modification date

There may be better alternatives, but try this one:

ls -lT | tr -s ' ' | cut -d ' ' -f 9 | sort | uniq | wc -l

ls -lT : Displays detailed listing with modified date and year in full.

tr -s ' ' : Removes excess space

cut -d ' ' -f 9: Collect the year column

sort: Sort the years

uniq: Collect the unique years

wc -l: Count the number of lines.

Sort logs by date field in bash

For GNU sort: sort -k2M -k3n -k4

• -k2M sorts by second column by month (this way "March" comes before "April")
• -k3n sorts by third column in numeric mode (so that " 9" comes before "10")
• -k4 sorts by the fourth column.

See more details in the manual.

how do sort items in a text file within its contained time interval

Since perl is available in any unix flavor, and you can use it from command line, I would use it in this way:

$ perl -n -e 'print "$_" if (/(.+)\|(.+)\|/ && $1 eq "14-Oct-2013" && $2 >= "04:18" && $2 <= "05:16")' | sort
14-Oct-2013|04:16|
14-Oct-2013|04:18|
14-Oct-2013|04:20|
14-Oct-2013|05:16|

This could be simplest:

$ perl -n -e 'print $_ if ($_ ge "14-Oct-2013|04:16" && $_ lt "14-Oct-2013|05:17")' | sort
14-Oct-2013|04:16|
14-Oct-2013|04:18|
14-Oct-2013|04:20|
14-Oct-2013|05:16|

To read sh vars you just hast to use correctly quotes and $, so as the shell can expand variables, here you have two cases using single or double quotes to embrace perl arguments:

$ FROM="14-Oct-2013|04:16"
$ TO="14-Oct-2013|05:17"
$ perl -n -e 'print $_ if ($_ ge "'$FROM'" && $_ lt "'$TO'")' | sort
or
$ perl -n -e "print \$_ if (\$_ ge '$FROM' && \$_ lt '$TO')" | sort

Linux grep and sort log files

Try this:

grep --color=always "myID" file*.log | sort -t : -k2,2 -k3,3n -k4,4n

Output:

file3.log:2015-09-26 15:39:48,788 - ERROR - bar : {'id' : myID}
file1.log:2015-09-26 15:39:50,788 - DEBUG - blabla : {'id' : myID}
file2.log:2015-09-26 15:39:51,788 - ERROR - foo : {'id' : myID}

---

Related Topics