将时代(秒)格式的日期替换为日志文件中的正常日期

Question

我在一台ubuntu机器上试图从日志文件中读取数据，并发现日志包含日期和时间的时代格式(秒)。如何从命令行将这些数字转换成任何可读的格式？

1411622206, HOST ALERT, host-001,DOWN,SOFT,1,CHECK_NRPE, Socket timeout after 10 seconds.
1411622586, HOST ALERT, host-001,DOWN,SOFT,1,CHECK_NRPE, Socket timeout after 10 seconds.
1411623976, HOST ALERT, host-021,DOWN,SOFT,1,CHECK_NRPE, Socket timeout after 10 seconds.
1411624986, HOST ALERT, host-055,DOWN,SOFT,1,CHECK_NRPE, Socket timeout after 10 seconds.
1411625076, HOST ALERT, host-023,DOWN,SOFT,1,CHECK_NRPE, Socket timeout after 10 seconds.
1411625356, HOST ALERT, host-032,DOWN,SOFT,1,CHECK_NRPE, Socket timeout after 10 seconds.
1411625736, HOST ALERT, host-044,DOWN,SOFT,1,CHECK_NRPE, Socket timeout after 10 seconds.

Answer 1

来自date手册(info date)：

若要将这种难以处理的秒数转换为更易读的形式，请使用如下命令：

# local time zone used
date -d '1970-01-01 UTC 946684800 seconds' +"%Y-%m-%d %T %z"
1999-12-31 19:00:00 -0500

或者，如果您不介意依赖于Coreutils5.3.0中存在的@特性，则可以将其缩短为：

date -d @946684800 +"%F %T %z"
1999-12-31 19:00:00 -0500

通常情况下，最好输出UTC-相对日期和时间：

date -u -d '1970-01-01 946684800 seconds' +"%Y-%m-%d %T %z"
2000-01-01 00:00:00 +0000