Django-tables2反向查找
Django-tables2反向查找
提问于 2014-01-02 01:08:12
我有一个表格，其中显示了渲染良好的"leads“列表。还有一个被称为"Leadupdate“的相关模型，它与表中使用的"lead”模型相关。从Leadupdate到具有外键的lead存在多对一关系。我想要在其中一个更新列中显示单个"leads“的所有相关更新。网上有几个通过外键跟踪正向关系的例子，但还没有找到一个反向关系。以下是所述关系Accessor forward look up.的一个示例
编辑:查找将在Django-tables2模块实例表上完成。我不是在模型上要求反向查找，而是在Django-tables2的上下文中这样做。
Models.py：
class lead(models.Model):
slug = models.SlugField(unique=True,blank=True, null=True)
first_name = models.CharField(max_length=100)
last_name = models.CharField(max_length=100, blank=True, null=True)
business_name = models.CharField(max_length=100,blank=True, null=True)
email = models.EmailField(max_length=75, blank=True, null=True)
phone_number = models.CharField(max_length=20, blank=True, null=True)
address = models.CharField(max_length=150, blank=True, null=True)
city = models.CharField(max_length=50, blank=True, null=True)
state = models.CharField(max_length=10, blank=True, null=True)
zipcode = models.CharField(max_length=5, blank=True, null=True)
submission_date = models.DateTimeField(auto_now_add=True, blank=True)
assigned_to = models.ManyToManyField(Listing,blank=True, null=True, related_name="leads")
requested_software = models.CharField(max_length=50, blank=True, null=True)
type_of_business = models.CharField(max_length=30, choices=TYPE_OF_BUSINESS, default='Bar', blank=True, null=True)
time_frame = models.CharField(max_length=10, choices=TIME_FRAME, default='1')
comments = models.TextField(blank=True, null=True)
def __unicode__(self):
return self.business_name
@models.permalink
def get_absolute_url(self):
return('listing_detail', (),{'slug' :self.slug,})
def save(self, *args, **kwargs):
if not self.slug:
self.slug = slugify(self.business_name)
super(lead, self).save(*args, **kwargs)
class Leadupdate(models.Model):
CONFIDENCE_LEVEL = (
('HOT', 'HOT'),
('COLD', 'COLD'),
)
LEAD_VALUE = (
('1K3K', '1K-3K'),
('5K10K', '5K-10K'),
('10K20K', '10K-20K'),
('20K50K', '20K-50K'),
('50KUP', '5OK-UP'),
)
ESTIMATED_CLOSING = (
('1w4w', '1-4 Weeks'),
('1m3m', '1-3 Months'),
('3m6m', '3-6 Months'),
('6m+', '6+ Months'),
)
updatedate = models.DateTimeField(auto_now_add=True)
update = models.TextField(blank=True, null=True)
updatefrom = models.ForeignKey(Listing, related_name="update_from", blank=True, null=True)
lead = models.ForeignKey(lead, related_name="related_update",blank=True, null=True)
lead_confidence_level = models.CharField(max_length=10, choices=CONFIDENCE_LEVEL, default='COLD', blank=True, null=True)
estimated_lead_value = models.CharField(max_length=10, choices=LEAD_VALUE, default='1K3K', blank=True, null=True)
estimated_closing_frame = models.CharField(max_length=10, choices=ESTIMATED_CLOSING, default='1-4 Weeks', blank=True, null=True)
def __unicode__(self):
return u" %s - %s " % (self.update, self.updatedate)表：
class LeadTable(tables.Table):
business_name = tables.LinkColumn('lead-detail', args=[A('slug')])
updates = tables.Column(accessor='lead.related_update')
class Meta:
model = lead
fields = ("business_name","first_name", "last_name","number_of_pos","submission_date","updates")
attrs = {"class":"paleblue"}回答 2
Stack Overflow用户
发布于 2016-01-24 14:12:06
这是一个迟来的答案，但这里是我在Django 1.8.6和django-tables2 1.1.0 (基于Django-Tables2 Issue 156和This answer)中工作的方法。要通过外键关系访问一组一对多的对象，只需在访问器中使用related_name，然后创建一个render方法来生成写入列单元格的内容。在该方法中，您可以获取所有外来模型对象，并在for循环中访问它们的字段。
class LeadTable(tables.Table):
business_name = tables.LinkColumn('lead-detail', args=[A('slug')])
updates = tables.Column(accessor='related_update')
def render_updates(self, value, table):
updates = ""
uFirst = True
updatesList = list(value.all())
for u in updatesList:
if not uFirst:
updates += ", "
else:
uFirst = False
updates += u.update
return updates
class Meta:
model = lead
fields = ("business_name","first_name", "last_name","number_of_pos","submission_date","updates")
attrs = {"class":"paleblue"}Stack Overflow用户
发布于 2014-01-02 03:15:29
根据django docs的说法
在视图中，您可以这样访问它们(假设lead_instance是lead类的一个实例)：
all_leadtables_for_lead = lead_instance.leadtable_set附注:为了遵守python ，对类(class Lead(models.Model):)使用大写的名称。
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原文链接：
https://stackoverflow.com/questions/20871026
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