【发布时间】:2020-01-02 02:22:41
【问题描述】:
我正在开发一个审计管理信息系统,我可以在其中记录与审计相关的所有发现。我有具有外键关系的模型。如何查看具有特定作业和 audit_title 和单元的所有结果? 请参阅下面的相关代码。
model.py 内容
class Unit(models.Model):
unit_name = models.CharField(max_length=30, blank=True, null=True)
def __unicode__(self):
return self.unit_name
class Assignment(models.Model):
assignment_name = models.CharField(max_length=30, blank=True, null=True)
def __unicode__(self):
return self.assignment_name
class Task(models.Model):
task_title = models.CharField(max_length=35, blank=True, null=True)
return self.task_title
class Finding(models.Model):
assignment = models.ForeignKey(Assignment, blank=True, null=True)
audit_title = models.ForeignKey(Task, blank=True, null=True)
auditor = models.ManyToManyField(User, blank=True)
unit = models.ForeignKey(Unit, blank=True, null=True)
audit_period = models.DateField(auto_now_add=False, auto_now=False, blank=True, null=True)
contact_person = models.CharField('Contact Person', max_length=500, blank=True, null=True)
finding = models.TextField('Detail Finding', max_length=500, blank=True, null=True)
be = models.CharField(max_length=30, blank=True, null=True)
form.py
class FindingSearchForm(forms.ModelForm):
class Meta:
model = Finding
fields = ['assignment',
'audit_title',
'unit',
'be',
]
我的views.py中有以下内容,但我有这个错误invalid literal for int() with base 10: ''
views.py 内容
def finding_list(request):
title = 'List of Finding'
queryset = Finding.objects.all()
queryset_count = queryset.count()
form = FindingSearchForm(request.POST or None)
context = {
"title": title,
"form": form,
"queryset_count": queryset_count,
}
if request.method == 'POST':
unit = form['unit'].value()
audit_title = form['audit_title'].value()
assignment = form['assignment'].value()
queryset = Finding.objects.all().order_by('-timestamp').filter(be__icontains=form['be'].value(),
unit_id=unit,
assignment_id=assignment,
audit_title_id=audit_title,)
【问题讨论】:
-
你能举一个你想要过滤的值的例子吗?
标签: django foreign-keys django-queryset