问个问题,忽略日期,就用时间段对比,如何判断当前时间段是否在 08:00:00 跟 08:30:00 的范围内? - V2EX
basstk
V2EX    PHP

问个问题,忽略日期,就用时间段对比,如何判断当前时间段是否在 08:00:00 跟 08:30:00 的范围内?

  •  
  •   basstk Dec 1, 2018 5521 views
    This topic created in 2722 days ago, the information mentioned may be changed or developed.

    数据库 2 个字段存储了 2 个时间段: t1=08:00:00 跟 t2=08:30:00

    忽略日期,就用时间段对比,如:如何判断当前时间段 09:30:00 是否在 08:00:00 跟 08:30:00 的范围内?

    15 replies    2018-12-02 08:11:16 +08:00
    Midnight
        1
    Midnight  
       Dec 1, 2018
    大于小于啊
    ysc3839
        2
    ysc3839  
       Dec 1, 2018 via Android
    比较当天的秒数。
    jisxu
        3
    jisxu  
       Dec 1, 2018 via Android   1
    java 的话,去看看 LocalTime 这个类,parse,然后直接 compareTo 完事
    chinvo
        4
    chinvo  
       Dec 1, 2018 via iPhone
    推荐一个库,叫 Carbon,PHP 的日期时间增强
    Kbyte
        5
    Kbyte  
       Dec 1, 2018
    一减不就完了
    947211232
        6
    947211232  
       Dec 1, 2018
    strtotime
    zxcvsh
        7
    zxcvsh  
       Dec 1, 2018 via iPhone
    如果是取出来对比好像方法挺多的
    gowa2017
        8
    gowa2017  
       Dec 1, 2018 via iPhone
    sql hour(date) =8 and minute(date) <30
    jugelizi
        9
    jugelizi  
       Dec 1, 2018
    不就是 8000 和 8300 对比吗
    lhx2008
        10
    lhx2008  
       Dec 1, 2018 via Android
    楼上正解,转成四位的就好了
    dante3imin
        11
    dante3imin  
       Dec 1, 2018
    time() > mktime(8, 0, 0) && time() < mktime(8, 30, 0)
    xpresslink
        12
    xpresslink  
       Dec 1, 2018
    把 9:30 转成 用字符串 "8000"/"8300" 或整数 8000/8300 直接比较就可以了。
    realpg
        13
    realpg  
    PRO
       Dec 1, 2018
    PHP 的 STRTOTIME 比较强悍
    honeycomb
        14
    honeycomb  
       Dec 1, 2018 via Android
    不想写 SQL 的话,导出成 localtime 类以后可以直接比较。
    qinrui
        15
    qinrui  
       Dec 2, 2018 via iPhone
    case when substr(time_str,12,8)>“ 08:00:00 ” and
    substr(time_str,12,8)<“ 08:30:00 ” then 1 else 0 end
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