bash 在 unix 中对日期字段进行排序
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Sorting date field in unix
提问by user1011046
I have text file which contains hundreds of thousands of records. One of the fields is a date field. Is there is any way to sort the file based on the date field?
我有包含数十万条记录的文本文件。其中一个字段是日期字段。有没有办法根据日期字段对文件进行排序？
09-APR-12 04.08.43.632279000 AM
19-MAR-12 03.53.38.189606000 PM
19-MAR-12 03.56.27.933365000 PM
19-MAR-12 04.00.13.387316000 PM
19-MAR-12 04.04.45.168361000 PM
19-MAR-12 03.54.32.595348000 PM
27-MAR-12 10.28.14.797580000 AM
28-MAR-12 12.28.02.652969000 AM
27-MAR-12 07.28.02.828746000 PM
The Output should come as
输出应该是
19-MAR-12 03.53.38.189606000 PM
19-MAR-12 03.54.32.595348000 PM
19-MAR-12 03.56.27.933365000 PM
19-MAR-12 04.00.13.387316000 PM
19-MAR-12 04.04.45.168361000 PM
27-MAR-12 10.28.14.797580000 AM
27-MAR-12 07.28.02.828746000 PM
28-MAR-12 12.28.02.652969000 AM
09-APR-12 04.08.43.632279000 AM
I have tried the sort command to order the date (taking the date field as a string), but it is not giving the correct output.
我已经尝试使用 sort 命令对日期进行排序（将日期字段作为字符串），但它没有给出正确的输出。
回答by William Pursell
Chronicle's solution is close, but misses the AM/PM distinction, sorting 27-MAR-12 07.28.02.828746000 PMbefore 27-MAR-12 10.28.14.797580000 AM. This can be modified:
Chronicle 的解决方案很接近，但错过了 AM/PM 的区别，排序27-MAR-12 07.28.02.828746000 PM在27-MAR-12 10.28.14.797580000 AM. 这可以修改：
sort -t- -k 3.1,3.2 -k 2M -k 1n -k 3.23,3.24
But that is still very fragile. It would be much better to convert the dates to an epoch time and compare numerically.
但这仍然非常脆弱。将日期转换为纪元时间并进行数字比较会好得多。
回答by Debaditya
Try this :
尝试这个 ：
Input.txt
输入.txt
09-APR-12 04.08.43.632279000 AM
19-MAR-12 03.53.38.189606000 PM
19-MAR-12 03.56.27.933365000 PM
19-MAR-12 04.00.13.387316000 PM
19-MAR-12 04.04.45.168361000 PM
19-MAR-12 03.54.32.595348000 PM
27-MAR-12 10.28.14.797580000 AM
28-MAR-12 12.28.02.652969000 AM
27-MAR-12 07.28.02.828746000 PM
Code
代码
sort -t "-" -k 3 -k 2M -nk 1 Input.txt
Output
输出
19-MAR-12 03.53.38.189606000 PM
19-MAR-12 03.54.32.595348000 PM
19-MAR-12 03.56.27.933365000 PM
19-MAR-12 04.00.13.387316000 PM
19-MAR-12 04.04.45.168361000 PM
27-MAR-12 07.28.02.828746000 PM
27-MAR-12 10.28.14.797580000 AM
28-MAR-12 12.28.02.652969000 AM
09-APR-12 04.08.43.632279000 AM
回答by Andrey
This script sorts by Epoch time with nanosecond resolution:
此脚本以纳秒分辨率按纪元时间排序：
awk '{
t = gensub(/\.([0-9]{2})\./, ":\1:", 1, while read a; do
grep "^${a}" input.txt;
done < <(sed 's/\./:/;s/\./:/' input.txt | xargs -n3 -I{} date -d"{}" +%s | sort | xargs -n1 -I{} date -d @'{}' +'%d-%^h-%y %I.%M.%S')
);
command = "date +%s%N -d \x022" t "\x022";
command | getline t;
close(command);
print t, ##代码##;
}' unsorted.txt | sort -n -k 1 | cut -d ' ' -f 2- > sorted.txt
回答by Andrey
You could use date, which is generally probably a decent idea, especially if you don't need to worry about the microseconds, otherwise you could probably clip the microseconds off and sort that as a secondary sorting field.
您可以使用日期，这通常可能是一个不错的主意，特别是如果您不需要担心微秒，否则您可能会剪掉微秒并将其作为辅助排序字段进行排序。
##代码##