如何在Django的views.py中应用循环

Question

研究/模型

RECRUITING_CHOICES = [
    ('Recruiting', 'Recruiting'),
    ('Not yet recruiting', 'Not yet recruiting'),
    ('Completed', 'Completed'),
    ('Holding', 'Holding'),
]
RECRUITING_CHOICES_INV = {val: val for val, _ in RECRUITING_CHOICES}

class Research(models.Model):
    is_deleted = models.BooleanField(default=False)
    is_recruiting = models.CharField(choices=RECRUITING_CHOICES, null=True, max_length=50)
    research_name = models.CharField(max_length=2000, null=True)
    teacher= models.CharField(max_length=2000, null=True)

我正在使用Django，并且我试图创建一个列表，用于使用类模型'Research‘在views.py中生成顶点图(图)。

我试图将is_recruiting字段的4个值(​​)中的每个值添加到筛选中，但是添加招聘(而不是招聘)只需使用多行代码，使代码看起来效率低下。有使用for语句的解决方法吗？

研究/观点

counts = Research.objects.values('teacher') \
    .annotate(A=Count('id', filter=Q(type='A')),
              B=Count('id', filter=Q(type='B')),
              C=Count('id', filter=Q(type='C')),
              D=Count('id', filter=Q(type='D')),
              E=Count('id', filter=Q(type='E')),
              F=Count('id', filter=Q(type='F')),
              r_A=Count('id', filter=Q(type='A', is_recruiting='Recruiting')),
              r_B=Count('id', filter=Q(type='B', is_recruiting='Recruiting')),
              r_C=Count('id', filter=Q(type='C', is_recruiting='Recruiting')),
              r_D=Count('id', filter=Q(type='D', is_recruiting='Recruiting')),
              r_E=Count('id', filter=Q(type='E', is_recruiting='Recruiting')),
              r_F=Count('id', filter=Q(type='F', is_recruiting='Recruiting')),
              N_A=Count('id', filter=Q(type='A', is_recruiting='Not yet recruiting')),
              N_B=Count('id', filter=Q(type='B', is_recruiting='Not yet recruiting')),
              N_C=Count('id', filter=Q(type='C', is_recruiting='Not yet recruiting')),
              N_D=Count('id', filter=Q(type='D', is_recruiting='Not yet recruiting')),
              N_E=Count('id', filter=Q(type='E', is_recruiting='Not yet recruiting')),
              N_F=Count('id', filter=Q(type='F', is_recruiting='Not yet recruiting'))) \
    .values('teacher', 'A', 'B', 'C', 'D', 'E', 'F',
            'r_A', 'r_B', 'r_C', 'r_D', 'r_E', 'r_F',
            'N_A', 'N_B', 'N_C', 'N_D', 'N_E', 'N_F')

views.py - ADD

teacher = [];
A= [];
B= [];
C= [];
D= [];
E= [];
F= [];
r_A = [];
r_B = [];
r_C = [];
r_D = [];
r_E = [];
r_F = [];
for count in counts:
    teacher.append(str(count['teacher']))
    A.append(str(count['A']))
    B.append(str(count['B']))
    C.append(str(count['C']))
    D.append(str(count['D']))
    E.append(str(count['E']))
    Ftc.append(str(count['F']))
    r_A.append(str(count['r_A']))
    r_B.append(str(count['r_B']))
    r_C.append(str(count['r_C']))
    r_D.append(str(count['r_D']))
    r_E.append(str(count['r_E']))
    r_F.append(str(count['r_F']))

is_recruiting = request.GET.get('is_recruiting')

graph.html

<script>
    var options3 = {
    series: [{% if is_recruiting == 'Recruiting' %}
    {
      name: 'A',
      data: {{ r_A | safe }}
    }, {
      name: 'B',
      data: {{ r_B | safe }}
    }, {
      name: 'C',
      data: {{ r_C | safe }}
    }, {
      name: 'D',
      data: {{ r_D | safe }}
    }, {
      name: 'E',
      data: {{ r_E | safe }}
    }, {
      name: 'F',
      data: {{ r_F | safe }}
    },{% elif is_recruiting == 'ALL' %}
    {
      name: 'A',
      data: {{ A | safe }}
    }, {
      name: 'B',
      data: {{ B | safe }}
    }, {
      name: 'C',
      data: {{ C | safe }}
    }, {
      name: 'D',
      data: {{ D | safe }}
    }, {
      name: 'E',
      data: {{ E | safe }}
    }, {
      name: 'F',
      data: {{ F | safe }}
    },{% endif %}
    ],
</script>

我是否必须在JavaScript部分中一次又一次地列出if语句？

Answer 1

您可以通过创建参数字典并在向**传递时解压该字典，从而将动态关键字参数构建为函数。

循环遍历您的类型，并为每个类型的注释创建键和值。

还可以将创建的动态注释解压缩为字段名，以便使用values()传递给*。

annotations = {}
types = ('A', 'B', 'C', 'D', 'E', 'F')
for type in types:
    annotations[type] = Count('id', filter=Q(type=type))
    annotations[f'r_{type}'] = Count('id', filter=Q(type=type, is_recruiting='Recruiting'))
    annotations[f'N_{type}'] = Count('id', filter=Q(type=type, is_recruiting='Not yet recruiting'))
counts = Research.objects.values('teacher').annotate(**annotations).values('teacher', *annotations.keys())

Answer 2

Kwargs可以作为dict传递，所以您所需要做的就是用以下键创建一个dict : value对。

你可以这样做：

recruiting_values = (
    ("", dict()),
    ("r_", dict(is_recruting="Recruiting")),
    ("n_", dict(is_recruting="Not"),
)
types = ("A", "B", "C")

...

.annotate(**{
    f"{prefix}{type}": Count("id", filter=Q(type=type, **recruiting)) for prefix, recruiting in recruiting_value for type in types
})