javascript - 如何修复 Django 中的 "object() takes no parameters"错误-6ren

javascript - 如何修复 Django 中的 "object() takes no parameters"错误

转载作者：行者123 更新时间：2023-11-27 23:48:02

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4

我正在尝试访问名为 Active_Project_View 的数据库。这里面是我试图在 ag-grid 中显示的信息。我已经使用 objects.all() 来获取信息，但它不起作用。

class getActiveGrid():
    def get_queryset(self, request, format=None):
        queryset = Active_Project_View.objects.all()
        return queryset

    def get(self, request, format=None):
         filters = request.data
         serializer = ActiveGridViewSerializer(self.get_queryset(filters), many=True)
         return HttpResponse(serializer.data)


<!DOCTYPE html>

<html lang="en" xmlns="http://www.w3.org/1999/xhtml">
<head>
<link rel="stylesheet" href="static\app\content\agGrid.css" >
<script src="https://unpkg.com/ag-grid-community/dist/ag-grid- 
 community.min.noStyle.js"></script>
<link rel="stylesheet" href="https://unpkg.com/ag-grid- 
  community/dist/styles/ag-grid.css">
<link rel="stylesheet" href="https://unpkg.com/ag-grid- 
    community/dist/styles/ag-theme-balham.css">

</head>
<body>

<h1>Project Grids</h1>

<div class="box c">

活跃元素

  <script type="text/javascript" charset="utf-8">

    const Active_Project = [
        { headerName: "Last Name", field: "app.Customer_Last_Name", sort: 'asc' },
        { headerName: "First Name", field: "Customer_First_Name", },
        { headerName: "Phone Number", field: "Customer_Phone_Number", },
        { headerName: 'Employee', field: 'Employee_First_Name' },
        { headerName: 'Zip Code', field: 'Customer_Zip' },
       ];

   const ActiveProjectGrid = {
        columnDefs: Active_Project,
      };


    var eGridDiv1 = document.querySelector('#ActiveProjectGrid');     
    new agGrid.Grid(eGridDiv1, ActiveProjectGrid);

      fetch('getActiveGrid').then(function (response) {
          return response.data();
      }).then(function (data) {
          ActiveProjectGrid.api.setRowData(data);
          })
  </script>
</body>
</html>


url(r'^getActiveGrid$',app.views.getActiveGrid, name='postActive'),


class ActiveGridViewSerializer():
    class Meta:
    model = Active_Project_View
    fields= '__all__'

类 Active_Project_View(models.Model):

id = models.UUIDField(primary_key=True, default=uuid.uuid4, editable=False)
Customer_First_Name= models.CharField(max_length=255, blank = False)
Customer_Last_Name = models.CharField(max_length=255, blank = False)
Customer_Phone_Number = models.CharField(max_length=10, blank=True) 
Customer_Zip = models.CharField(max_length = 5, blank = False)
Employee_First_Name=models.CharField(max_length=255)
Quote_Status = models.CharField(max_length = 16, default = None)

class Meta:
    managed = False

我希望得到要显示的客户姓名、电话号码、邮政编码和员工是网格。我目前收到“正在加载...”框，但我不确定自己做错了什么。

最佳答案

您的实现存在多个问题。您可以像这样一一修复它们:

from rest_framework.views import APIView
from rest_framework.response import Response
from rest_framework import serializers

#serializer
class ActiveGridViewSerializer(<b>serializers.ModelSerializer</b>):
    class Meta:
        model = Active_Project_View
        fields= '__all__'

# view
class getActiveGrid(<b>APIView</b>):
    def get_queryset(self):
        queryset = Active_Project_View.objects.all()
        return queryset

    def get(self, request, format=None):
         serializer = ActiveGridViewSerializer(<b>self.get_queryset()</b>, many=True)
         return <b>Response(serializer.data)</b>

# url
url(r'^getActiveGrid$',app.views.getActiveGrid<b>.as_view()</b>, name='postActive'),

关于javascript - 如何修复 Django 中的 "object() takes no parameters"错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/56678712/

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javascript - 如何修复 Django 中的 "object() takes no parameters"错误