如何用awk实现将日期转换为时间戳？

Shell_HAT

1. 
   
2. while read line; do date -d "$line" +%s; done < urfile
   

复制代码

sayigood

回复 11# Shell_HAT


    3ks

Shell_HAT

1. 
   
2. awk -F'[ :-]' '{printf("%d\n",mktime($1" "$2" "$3" "$4" "$5" "0))}' urfile
   

复制代码

sayigood

回复 6# iori809


    我有个新的问题，你给帮忙看看。http://bbs.chinaunix.net/viewthread.php?tid=1709887&extra=

ccjsj1

strftime 将时间戳转日期

1. awk 'BEGIN{print strftime("%Y-%m-%d",systime())}'

复制代码

mktime将日期转为时间戳

1. awk 'BEGIN {printf("%d\n",mktime(2006" "8" "5" "15" "09" "0))}'

复制代码

walykyy

回复 6# iori809

求教，我是用awk获取的时间戳列，应该怎么把时间戳转换成时间格式呢？？我按照您这个方法写了，应该是我水平不行，所以总是失败

wh7211

回复 16# walykyy

如何将文件中的日期转换为时间戳？
例如：输入文件file.in,格式如下：
2006-8-5 15:09
2006-8-5 16:10
2006-8-5 17:13
2006-8-5 18:16
2006-8-5 19:19
2006-8-5 20:22
2006-8-5 21:24
2006-8-5 22:28
2006-8-5 23:33
2006-8-6 0:39
2006-8-6 1:46
2006-8-6 2:55
2006-8-6 4:04
2006-8-6 12:45

1. awk '{split($0,t,"[- :]");a=mktime(""t[1]" "t[2]" "t[3]" "t[4]" "t[5]" 00");b=strftime("%Y-%m-%d %H:%M",a);print $0" -> "a" -> "b}' file.in

复制代码

输出（“时间格式”转成“时间戳”再转成“时间格式”）：
2006-8-5 15:09 -> 1154761740 -> 2006-08-05 15:09
2006-8-5 16:10 -> 1154765400 -> 2006-08-05 16:10
2006-8-5 17:13 -> 1154769180 -> 2006-08-05 17:13
2006-8-5 18:16 -> 1154772960 -> 2006-08-05 18:16
2006-8-5 19:19 -> 1154776740 -> 2006-08-05 19:19
2006-8-5 20:22 -> 1154780520 -> 2006-08-05 20:22
2006-8-5 21:24 -> 1154784240 -> 2006-08-05 21:24
2006-8-5 22:28 -> 1154788080 -> 2006-08-05 22:28
2006-8-5 23:33 -> 1154791980 -> 2006-08-05 23:33
2006-8-6 0:39 -> 1154795940 -> 2006-08-06 00:39
2006-8-6 1:46 -> 1154799960 -> 2006-08-06 01:46
2006-8-6 2:55 -> 1154804100 -> 2006-08-06 02:55
2006-8-6 4:04 -> 1154808240 -> 2006-08-06 04:04
2006-8-6 12:45 -> 1154839500 -> 2006-08-06 12:45