无法在 python 中转换 13 位 UNIX 时间戳

Question

对这个还是陌生的。我试图将 13 位时间戳转换为您可以阅读但没有运气的内容。我正在做的是从楼宇自动化系统中采样温度数据。这是我正在使用的代码。我错过了一些非常简单的东西。我可以朝正确的方向戳一下吗？

when_updated=driver.find_element_by_name("_lastUpdated")
timestamp=when_updated.get_attribute("value")
print("Timestamp:", timestamp)

nos=10  # Number of samples
freq=5  # Sampling frequency
print("\nTemperature & timestamp, sampled every", freq, "seconds:")
while True:
     ts=driver.find_element_by_name("_lastUpdated").get_attribute("value")
     print("\nTemperature: ", element.text, "read at:", ts)
     time.sleep(freq)


driver.switch_to_default_content()

输出

Temperature:

Timestamp: 0

Temperature & timestamp, sampled every 5 seconds:

Temperature:   read at: 0

Temperature:  21.0 ºC read at: 1520912895599

Temperature:  21.0 ºC read at: 1520912901432

Temperature:  21.0 ºC read at: 1520912907070

Answer 1

#!python2

# Epoch time needs to be converted to a human readable format.
# Also, epoch time uses 10 digits, yours has 13, the last 3 are milliseconds

import datetime, time

epoch_time = 1520912901432

# truncate last 3 digits because they're milliseconds
epoch_time = str(epoch_time)[0: 10]

# print timestamp without milliseconds
print datetime.datetime.fromtimestamp(float(epoch_time)).strftime('%m/%d/%Y -- %H:%M:%S')
'''
03/12/2018 -- 21:48:21
'''

# %   %   %   %   %   %   %   %   %   %   %   %   %   %   %   %   %   %   %   %   %   %

# If you need milliseconds
epoch_time = 1520912901432

# get first 10 digits
leading = str(epoch_time)[0: 10]

# get last 3 digits
trailing = str(epoch_time)[-3:]

# print timestamp with milliseconds
print datetime.datetime.fromtimestamp(float(leading)).strftime('%m/%d/%Y -- %H:%M:%S.') + ('%s' % int(trailing))
'''
03/12/2018 -- 21:48:21.432
'''

Python 2time模块：https : //docs.python.org/2/library/time.html