函数源代码:
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说明:若$varTime在今天的范围内,返回ture,否则返回false.
function bTodayToTomorrow($varTime)
{
;;;;$nowYear=date("Y");
;;;;$nowMonth=date("n");
;;;;$nowDay=date("j");
;;;;$aryTodayAndTomorrow["Today"]=date("U",mktime(0,0,0,$nowMonth,$nowDay,$nowYear));
;;;;$aryTodayAndTomorrow["Tomorrow"]=date("U",mktime(0,0,0,$nowMonth,($nowDay+1),$nowYear));
;;;;if($varTime>=$aryTodayAndTomorrow["Today"]&&$varTime<$aryTodayAndTomorrow["Tomorrow"])
;;;;{
return true;
;;;;}
;;;;else
;;;;{
return false;
;;;;}
}
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类似的你可以进行任意时间段的判断。;
标签: 函数