如何通过特定列中的 asc 或 desc 顺序重新排序文件答案

【问题标题】：How can reorder a file by asc or desc order from specific column如何通过特定列中的 asc 或 desc 顺序重新排序文件
【发布时间】：2013-06-24 17:35:07
【问题描述】：

如何从特定列日期和时间或仅日期列按 asc 或 desc 顺序重新排序文件。

{"hostname":"sg2","role":"epay","eventid":"ALERT_ORACLE","date":"02/08/2011","time":"01:30:00"},
{"hostname":"sg1","role":"pay","eventid":"ALERT_DF","date":"11/24/2010","time":"19:33:01"},
{"hostname":"sg3","role":"ipay","eventid":"ALERT_SMF","date":"12/11/2012","time":"02:30:00"},

通常使用“sort -gk（列号）”在这种情况下无法捕获日期和时间列或仅日期，可能需要一些分隔符字符串或相关。

预期的观点是：

{"hostname":"sg1","role":"pay","eventid":"ALERT_DF","date":"11/24/2010","time":"19:33:01"},
{"hostname":"sg2","role":"epay","eventid":"ALERT_ORACLE","date":"02/08/2011","time":"01:30:00"},
{"hostname":"sg3","role":"ipay","eventid":"ALERT_SMF","date":"12/11/2012","time":"02:30:00"},

【问题讨论】：

标签： perl sed awk grep tr

【解决方案1】：

使用sort 和: 作为分隔符并按列排序5M 进行日期比较：

$ sort -t: -k5M file
{"hostname":"sg1","role":"pay","eventid":"ALERT_DF","date":"11/24/2010","time":"19:33:01"},
{"hostname":"sg2","role":"epay","eventid":"ALERT_ORACLE","date":"02/08/2011","time":"01:30:00"},
{"hostname":"sg3","role":"ipay","eventid":"ALERT_SMF","date":"12/11/2012","time":"02:30:00"},

来源：Sort logs by date field in bash。

【讨论】：

【解决方案2】：

如果您选择了 ISO 日期格式 (YYYY-mm-dd)，您的生活会更轻松，但是

sort_by_date() {
    awk -F \" '{
        split($16, date, "/")
        split($20, time, ":")
        timestamp = mktime(date[3]" "date[1]" "date[2]" "time[1]" "time[2]" "time[3])
        print timestamp "\t" $0
    }' "$1" |
    sort -k1,1n |
    cut -f2-
}
sort_by_date filename

降序使用sort -k1,1nr

【讨论】：