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# 得到数据前一天得到时间，避免直接减1是错的
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from datetime import datetime, timedelta
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import logging
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import calendar
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# 获取日志记录器
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logger = logging.getLogger(__name__)
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# 获取前一天的日期
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def effective_date(year, month, day):
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try:
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logger.info("开始组装获取前一天的时间")
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# 拿到一个日期
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date_now = year + month + day
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# 转成有效时间
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date = datetime.strptime(date_now, "%Y%m%d")
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# 计算前一天
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day_before = date - timedelta(days=1)
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# 获得年月日并返回
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year = day_before.year
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month = day_before.month
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day = day_before.day
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return year, month, day
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except AttributeError:
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logger.exception(f"获取前一天时间失败：{AttributeError}")
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# 获取后一天的日期
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def is_valid_date(year, month, day):
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try:
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datetime(year=year, month=month, day=day)
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return True
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except ValueError:
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return False
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def get_next_day(year, month, day):
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try:
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if not is_valid_date(year, month, day):
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raise ValueError("输入的日期无效")
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current_date = datetime(year=year, month=month, day=day)
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next_date = current_date + timedelta(days=1)
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return next_date.year, next_date.month, next_date.day
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except ValueError:
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logger.exception(f"获取后一天时间失败：{ValueError}")
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if __name__ == "__main__":
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# 示例使用
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year, month, day = 2025, 6, 5 # 闰年2月28日
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next_year, next_month, next_day = get_next_day(year, month, day)
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print(f"后一天是: {next_year}-{next_month}-{next_day}") # 输出: 2020-2-29
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