如何使用ksh（或另一个shell脚本）获得现在和不同日期（分钟）之间的区别？

编程入门行业动态更新时间:2024-10-18 14:26:35

本文介绍了如何使用ksh（或另一个shell脚本）获得现在和不同日期（分钟）之间的区别？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！问题描述

使用以下值给出两个变量：

Giving two variables with the followind values:

date1 = Mon Feb 27 16:21:34 WET 2012 date2 = Mon Feb 27 16:29:34 WET 2012

如何使差异分钟）使用ksh？

How can I make the difference (in minutes) between them using ksh?

我正在使用Solaris 10。

我做了你所说的，这是错误：

$ function d { echo $((($(date -d"$2" +%s)-$(date -d"$1" +%s))/60)); } $ d "Mon Feb 27 16:21:34 WET 2012" "Mon Feb 27 16:29:34 WET 2012" date: illegal option -- d date: illegal option -- M date: illegal option -- o date: illegal option -- n date: illegal option -- date: illegal option -- F date: illegal option -- e date: illegal option -- b date: illegal option -- date: illegal option -- 2 date: illegal option -- 7 date: illegal option -- date: illegal option -- 1 date: illegal option -- 6 date: illegal option -- : date: illegal option -- 2 date: illegal option -- 9 date: illegal option -- : date: illegal option -- 3 date: illegal option -- 4 date: illegal option -- date: illegal option -- W date: illegal option -- E date: illegal option -- T date: illegal option -- date: illegal option -- 2 date: illegal option -- 0 date: illegal option -- 1 date: illegal option -- 2 usage: date [-u] mmddHHMM[[cc]yy][.SS] date [-u] [+format] date -a [-]sss[.fff] date: illegal option -- d date: illegal option -- M date: illegal option -- o date: illegal option -- n date: illegal option -- date: illegal option -- F date: illegal option -- e date: illegal option -- b date: illegal option -- date: illegal option -- 2 date: illegal option -- 7 date: illegal option -- date: illegal option -- 1 date: illegal option -- 6 date: illegal option -- : date: illegal option -- 2 date: illegal option -- 1 date: illegal option -- : date: illegal option -- 3 date: illegal option -- 4 date: illegal option -- date: illegal option -- W date: illegal option -- E date: illegal option -- T date: illegal option -- date: illegal option -- 2 date: illegal option -- 0 date: illegal option -- 1 date: illegal option -- 2 usage: date [-u] mmddHHMM[[cc]yy][.SS] date [-u] [+format] date -a [-]sss[.fff] 0 $

推荐答案

好的，这是一个C程序，可能会工作。它是（轻轻地）在solaris上测试。

well, here's a C program that might work. it's (lightly) tested on solaris.

它使用POSIX标准 strptime（3）功能来解析给定的字符串，并在epoch秒内打印结果。

it uses the POSIX standard strptime(3) function to parse a given string and prints the result in epoch seconds.

在理论上应该是是

$ ./strptime "Mon Feb 27 16:21:34 WET 2012" "%a %b %e %H:%M:%S %Z %Y"

但我还没有得到识别WET时区，所以你可能需要把它放在一个字面上：

but i haven't been able to get it to recognize the "WET" timezone, so you may need to just put that in as a literal:

$ ./strptime "Mon Feb 27 16:21:34 WET 2012" "%a %b %e %H:%M:%S WET %Y" 1330377694

这里是代码：

#define _GNU_SOURCE #include <stdlib.h> #include <stdio.h> #include <string.h> #include <errno.h> #include <time.h> #define ARGS 3 #define USAGE printf("Usage: %s string format\n",argv[0]); #define EARGS { printf("%s: %s arguments (%d)\n",argv[0],argc<ARGS?"Insufficient":"Too many",argc-1); USAGE; exit(2); } #define err(_s,_p,_f) {fprintf(stderr,"%s: %s: %s\n",(_p),(_f),strerror(errno));exit(_s);} int main(int argc,char**argv){ struct tm tm; time_t t; if(argc!=ARGS)EARGS; strptime(argv[1],argv[2],&tm); if(-1==(t=mktime(&tm)))err(1,argv[0],"mktime"); if(0>printf("%lld\n",(long long)t))err(1,argv[0],"printf"); return 0; }

按照