对日期时间的数据进行排序在UNIX中使用sort

Question

I want to sort my text file data using sort command. My data are in below format.

01-03-17  10:30AM                 1367 data_03012017_10_30_02.csv
01-03-17  11:00AM                   32 data_03012017_11_00_02.csv
01-03-17  11:30AM                 7532 data_03012017_11_30_03.csv
01-03-17  12:00AM                 3442 data_03012017_00_00_02.csv
01-03-17  12:00PM                 9715 data_03012017_12_00_03.csv
01-03-17  12:30AM                 1753 data_03012017_00_30_00.csv
01-03-17  12:30PM                 5857 data_03012017_12_30_03.csv

Above is my file data. Please help me to sort this data. I am getting stuck in AM & PM data. Please help on this. Using sort -n -t"," -k1.7,1.10 -k1.1,1.2 -k1.4,1.5 command data will sort by date but not with AM & PM. Let me know how to do this.

Answer 1

Assuming your data is in unsorted.dat you could use the following code

cat unsorted.dat | while read line ; do
   d=$( echo "$line" | cut -d" "  -f1-2 )
   s=$( echo "$d" | \
    sed 's/^\([0-9]*\)-\([0-9]*\)-\([0-9]*\)\(\s.*\)$/\3-\1-\2\4/' | \
        xargs -0 date +%s -d )
   echo "$s $line"
done | sort -n -k1,1 | cut -d" " -f2-

What it does is

1. Read the file line by line
2. Extract the date field from the first two columns into a variable $d
3. Change the order of the fields from MM-DD-YY to YY-MM-DD as recognized by the GNU date utility
4. Convert the date to seconds since 01-01-1970
5. Add the number of seconds to the line
6. Sort by the number of seconds
7. Cut the first field containing the number of seconds out.

Result:

01-03-17  12:00AM                 3442 data_03012017_00_00_02.csv
01-03-17  12:30AM                 1753 data_03012017_00_30_00.csv
01-03-17  10:30AM                 1367 data_03012017_10_30_02.csv
01-03-17  11:00AM                   32 data_03012017_11_00_02.csv
01-03-17  11:30AM                 7532 data_03012017_11_30_03.csv
01-03-17  12:00PM                 9715 data_03012017_12_00_03.csv
01-03-17  12:30PM                 5857 data_03012017_12_30_03.csv

Answer 2

I would convert your dates to something more standard.

Assuming you're using mm-dd-yy (an old fashioned USA convention), you might do the following using the BSD date command in POSIX or bash shell:

$ while read d t s f; do printf "%s %20d %s\n" "$(date -jf '%m-%d-%y %I:%M%p' "$d $t" '+%F %T')" "$s" "$f"; done < input.txt | sort
2017-01-03 00:00:43                 3442 data_03012017_00_00_02.csv
2017-01-03 00:30:43                 1753 data_03012017_00_30_00.csv
2017-01-03 10:30:43                 1367 data_03012017_10_30_02.csv
2017-01-03 11:00:43                   32 data_03012017_11_00_02.csv
2017-01-03 11:30:43                 7532 data_03012017_11_30_03.csv
2017-01-03 12:00:43                 9715 data_03012017_12_00_03.csv
2017-01-03 12:30:43                 5857 data_03012017_12_30_03.csv

Or, split apart for easier reading:

while read d t s f; do
  printf "%s %20d %s\n" \
    "$(date -jf '%m-%d-%y %I:%M%p' "$d $t" '+%F %T')" \
    "$s" \
    "$f"
done < input.txt | sort

This uses the date command to interpret and reassemble your dates, converting them from your current format into something that sorts naturally. Note that by doing this, you avoid the need for ANY options for the sort command.

Tested in OS X and FreeBSD.