变换多个返回值的自定义字符串默认Django管理界面

Question

我从一个模型方法返回三个变量，并希望在管理界面上显示它们：

def time_since(self): 
    time_since = timezone.now() - self.date_opened 
    return time_since.days, time_since.seconds//3600, (time_since.seconds//60)%60 

获得在admin.py

值

class TicketAdmin(admin.ModelAdmin): 
    list_display = (
    'issue', 
    'time_since', 
    ) 

并且显示为（0,1,21）。我如何将它转换为0天，1:21？

Answer 1

假设您已经返回一个元组（0,1,21），则：

x = (0,1,21) 

# Assumes that the values in your tuple will always be an integer. 
s = '{:d} days, {:d}:{:d}'.format(*x) 

print s # 0 days, 1:21 

给出here

编辑字符串格式的文档：在回答您的评论我认为你会调整你的代码返回字符串像这样：

def time_since(self): 
    time_since = timezone.now() - self.date_opened 
    out = [time_since.days, time_since.seconds//3600, (time_since.seconds//60)%60] 

    # As per Aamir Adnan's suggestion check whether day or days is applicable 
    if x[0] == 1: 
     return '{:d} day, {:d}:{:d}'.format(*x) 
    else: 
     return '{:d} days, {:d}:{:d}'.format(*x) 

Answer 2

你可以使用django内置模板fil ter timesince另请参见：

from django.template.defaultfilters import timesince_filter 

def time_since(self): 
    return timesince_filter(self.date_opened) # e.g. yield 4 days, 6 hours