用bash对逗号描绘文件进行排序

Question

我有以下内容：

    http://www.google.com/site,11/30/2012 6:51:30 PM
    http://www.google.com/site,10/1/2012 6:51:30 PM
    http://www.google.com/site,11/16/2012 6:51:30 PM
    http://www.google.com/site,8/1/2012 6:51:30 PM

，我希望它通过mm/dd/yyyy

对其进行排序

    http://www.google.com/site,8/1/2012 6:51:30 PM
    http://www.google.com/site,10/1/2012 6:51:30 PM
    http://www.google.com/site,11/16/2012 6:51:30 PM
    http://www.google.com/site,11/30/2012 6:51:30 PM

我可以使用sort命令，uniq命令，tr，sed等。我无法访问尴尬。有任何想法吗 ？sort -t "," -k1有点工作。

Answer 1

while read line; do
    s=$(date -d "${line#*,}" +%s)
    echo $s $line
done < input.txt | sort -n | cut -d ' ' -f2-

因此，对于每行创建秒以来，从日期字符串开始（逗号之后），将其用作排序键，然后将其从结果输出中删除。

Answer 2

尝试这样做：

sort -n -t "," -k 2 file.txt

见

man sort

Answer 3

我对sort命令进行了一些工作，发现我的问题，这似乎在起作用：

    cat s.txt | sort -n -t"/" -k 4,4n -k 5,5n
    https://www.virustotal.com/,9/16/2012 8:19:00 AM
    https://www.virustotal.com/,10/6/2012 9:20:59 AM
    https://www.virustotal.com/,11/1/2012 9:20:22 AM
    https://www.virustotal.com/,11/4/2012 9:11:02 AM
    https://www.virustotal.com/,11/6/2012 8:50:27 AM
    https://www.virustotal.com/,11/12/2012 6:51:32 PM

反向：

    cat s.txt | sort -n -t"/" -k 4,4nr -k 5,5nr
    https://www.virustotal.com/,11/12/2012 6:51:32 PM
    https://www.virustotal.com/,11/6/2012 8:50:27 AM
    https://www.virustotal.com/,11/4/2012 9:11:02 AM
    https://www.virustotal.com/,11/1/2012 9:20:22 AM
    https://www.virustotal.com/,10/6/2012 9:20:59 AM
    https://www.virustotal.com/,9/16/2012 8:19:00 AM

但是，我需要在逗号之后执行此操作，因为具有更多"/"的站点无法像http://site.com/go/to/somelink,9/16/2012 8:19:00 AM