Django上传xlsx文件直接转化为DataFrame或直接保存的方法

from django.urls import path

from . import views

# 为 URL 名称添加命名空间
app_name = 'report'

urlpatterns = [
    # 上传报告
    path('/upload_report/', views.upload_report, name='upload_report'),
]

def handle_uploaded_file(file, path):
    """
    保存上传的文件

    :param file: 上传的文件
    :param path: 存储路径(包含文件名)
    :return:
    """

    with open(path, 'wb+') as destination:
        # 将文件分块
        for chunk in file.chunks():
            # 写文件
            destination.write(chunk)

            
def upload_report(request):
    """
    上传报表

    :param request:
    :return:
    """

    # 获取文件
    file = request.FILES.get('file')
    
    # 报告存储路径
    path = 'reports/' + file.name

 # 1.直接转化为DataFrame
    file_df = pandas.read_excel(file)
    
    # 数据处理
    
    # DataFrame保存为xlsx
 file_df.to_excel(path, index=False, header=None, float_format='%.2f')


    # 2。直接保存上传的文件
    handle_uploaded_file(request.FILES['file'], path)

    return JsonResponse({'reports': 1})