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Trim21 Sep 2, 2020 via Android
用 datetime （ 0 点）循环加 timedelta （ hours=1 ），直到大于当前时间，然后把符合条件的的序列化成你想要的格式
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Eyon OP @Trim21 谢谢
```python from datetime import datetime,timedelta now = datetime.now() thismonth_start = now.strftime('%Y%m01') start = datetime(2020,9,1) while True: if start < now: print(start.strftime('%Y%m%d%H')) start = start+timedelta(hours=1) ``` |
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tairan2006 Sep 2, 2020 via Android
zfill 函数了解一下
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TEwrc Sep 3, 2020
@tairan2006 看了一下没太懂 请教一下怎么用到题主说的这个问题上呢？
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tairan2006 Sep 3, 2020
@TEwrc
now = datetime.now() prefix = now.strftime('%Y%m%d') return [prefix + str(x).zfill(2) for x in range(now.hour)] |
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TEwrc Sep 4, 2020
@tairan2006 秒啊！感谢
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biglazycat Sep 11, 2020
from datetime import datetime
time_list = [] now = datetime.now() start_day_hour = int(now.strftime('%Y%m%d00')) stop_day_hour = int(now.strftime('%Y%m%d%H')) for i in range(start_day_hour, stop_day_hour + 1): time_list.append(i) print(time_list) |
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