83189354 发表于 2006-11-12 11:30:00

在版主提示下写了以下代码(程序运行时间)

<P>(setq star_time1(getvar"cdate"))&nbsp;&nbsp;&nbsp;&nbsp;;开始时间<BR>(setq end_time2(getvar"cdate"))&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;;结束时间<BR>(setq use_time(- end_time2 star_time1))<BR>(setq use_time(rtos use_time 2 6))<BR>(setq use_time_cd(STRLEN use_time))<BR>(setq bc 1)<BR>(setq rq "" sj "" mgzj "")&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;;rq日期&nbsp;&nbsp;sj时间&nbsp;&nbsp;mgzj每个字节<BR>(while (/= bc (1+ use_time_cd))<BR>&nbsp;(setq mgzj(SUBSTR use_time bc 1))<BR>&nbsp; &nbsp;(if (/= mgzj ".")<BR>&nbsp;&nbsp;&nbsp; (progn<BR>&nbsp;&nbsp; (setq rq(STRCAT rq mgzj))<BR>&nbsp;&nbsp; (setq bc(1+ bc))<BR>&nbsp;&nbsp;&nbsp;&nbsp; )<BR>&nbsp;&nbsp; (progn<BR>&nbsp;&nbsp; (setq sj(substr use_time (1+ bc)))<BR>&nbsp;&nbsp;&nbsp; (setq bc(1+ use_time_cd))<BR>&nbsp;&nbsp; )<BR>&nbsp; )<BR>&nbsp; )</P>
<P>(Print rq)<BR>(print sj)<BR>(print "--------------")<BR>(setq rqcd(strlen rq))<BR>(setq xs(substr sj 1 2))&nbsp;;小时xs<BR>(setq fz(substr sj 3 2))&nbsp;;分钟fz<BR>(setq mz(substr sj 5 2))&nbsp;;秒钟mz<BR>(setq ys(strcat "程序用时:" rq "天" xs "小时" fz "分钟" mz "秒钟"))&nbsp;;用时<BR>(print ys)<BR></P>

rongyifei 发表于 2006-11-12 12:23:00

<P>我觉得不够准确,假若开始时间是20061112.121159,结束时间为20061112.121201,</P>
<P>如果说只是简单相减的话得到的时间是:42秒,实际花费时间是:2秒!</P>

FCMhXrJ 发表于 2018-10-13 16:18:30

你那个代码怎样使用呀？都没有命令呢？
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