Calculating Time Difference

前端 未结 5 2021
情话喂你
情话喂你 2020-12-02 16:47

at the start and end of my program, I have

from time import strftime
print int(strftime(\"%Y-%m-%d %H:%M:%S\")



Y1=int(strftime(\"%Y\"))
m1=int(strftime(\         


        
5条回答
  •  鱼传尺愫
    2020-12-02 17:17

    Here is a piece of code to do so:

    def(StringChallenge(str1)):

    #str1 = str1[1:-1]
    h1 = 0
    h2 = 0
    m1 = 0
    m2 = 0
    
    def time_dif(h1,m1,h2,m2):
        if(h1 == h2):
            return m2-m1
        else:
            return ((h2-h1-1)*60 + (60-m1) + m2)
    count_min = 0
    
    if str1[1] == ':':
        h1=int(str1[:1])
        m1=int(str1[2:4])
    else:
        h1=int(str1[:2])
        m1=int(str1[3:5])
    
    if str1[-7] == '-':
        h2=int(str1[-6])
        m2=int(str1[-4:-2])
    else:
        h2=int(str1[-7:-5])
        m2=int(str1[-4:-2])
    
    if h1 == 12:
        h1 = 0
    if h2 == 12:
        h2 = 0
    
    if "am" in str1[:8]:
        flag1 = 0
    else:
        flag1= 1
    
    if "am" in str1[7:]:
        flag2 = 0
    else:
        flag2 = 1
    
    if flag1 == flag2:
        if h2 > h1 or (h2 == h1 and m2 >= m1):
            count_min += time_dif(h1,m1,h2,m2)
        else:
            count_min += 1440 - time_dif(h2,m2,h1,m1)
    else:
        count_min += (12-h1-1)*60
        count_min += (60 - m1)
        count_min += (h2*60)+m2
    
    
    return count_min
    

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