bgm*_*ter 16 python django django-templates timedelta
我无法让我的django模板始终显示timedelta对象.我尝试在模板中使用时间过滤器,但是当我这样做时没有显示任何内容.如果我使用Assert False,timedelta对象在错误页面上显示如下:
time datetime.timedelta(0, 38, 132827)
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这会将时差显示为:
0:00:38.132827
我想只显示每个timedelta对象的小时,分钟和秒.有没有人对我如何做到这一点有任何建议?
我按照彼得的建议写了一个自定义模板过滤器。
这是我采取的步骤。
首先,我按照本指南创建自定义模板过滤器。
请务必阅读本节有关代码布局的内容。
这是我的过滤器代码
from django import template
register = template.Library()
@register.filter()
def smooth_timedelta(timedeltaobj):
"""Convert a datetime.timedelta object into Days, Hours, Minutes, Seconds."""
secs = timedeltaobj.total_seconds()
timetot = ""
if secs > 86400: # 60sec * 60min * 24hrs
days = secs // 86400
timetot += "{} days".format(int(days))
secs = secs - days*86400
if secs > 3600:
hrs = secs // 3600
timetot += " {} hours".format(int(hrs))
secs = secs - hrs*3600
if secs > 60:
mins = secs // 60
timetot += " {} minutes".format(int(mins))
secs = secs - mins*60
if secs > 0:
timetot += " {} seconds".format(int(secs))
return timetot
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然后在我的模板中
{% load smooth_timedelta %}
{% timedeltaobject|smooth_timedelta %}
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输出示例
据我所知,您必须为此编写自己的模板标签。下面是我根据 Django 核心 timesince/timeuntil 代码编写的代码,它应该输出您想要的内容:
@register.simple_tag
def duration( duration ):
"""
Usage: {% duration timedelta %}
Returns seconds duration as weeks, days, hours, minutes, seconds
Based on core timesince/timeuntil
"""
def seconds_in_units(seconds):
"""
Returns a tuple containing the most appropriate unit for the
number of seconds supplied and the value in that units form.
>>> seconds_in_units(7700)
(2, 'hour')
"""
unit_totals = OrderedDict()
unit_limits = [
("week", 7 * 24 * 3600),
("day", 24 * 3600),
("hour", 3600),
("minute", 60),
("second", 1)
]
for unit_name, limit in unit_limits:
if seconds >= limit:
amount = int(float(seconds) / limit)
if amount != 1:
unit_name += 's' # dodgy pluralisation
unit_totals[unit_name] = amount
seconds = seconds - ( amount * limit )
return unit_totals;
if duration:
if isinstance( duration, datetime.timedelta ):
if duration.total_seconds > 0:
unit_totals = seconds_in_units( duration.total_seconds() )
return ', '.join([str(v)+" "+str(k) for (k,v) in unit_totals.iteritems()])
return 'None'
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