文件上传自动生成日期文件夹

Posted 2020-12-29 52-qq

前端：

<div class="container">
    <form action="">
        {% csrf_token %}
        <div class="col-md-4 col-md-offset-4">
        <div class="form-group">
            <input type="file" name="file" id="file_upload">
        </div>
    </div>
    </form>
    <button type="button" class="btn btn-default" id="btn_file">点击上传</button>
</div>

<script>
    $("#btn_file").click(function () {
        var form_data = new FormData();
        var file_info = $(‘#file_upload‘)[0].files[0];
        form_data.append(‘file‘, file_info);
        if (file_info == undefined){
            console.log(‘没有选择任何文件‘);
            return false
        }
        $.ajax({
            url: ‘{% url "uploadFile" %}‘,
            type: ‘POST‘,
            headers:{"X-CSRFToken":$.cookie(‘csrftoken‘)},
            data: form_data,
            contentType:false,
            processData:false,
            success:function (data) {
                console.log(data)
            }
        })
    })
</script>

后端：

def uploadFile(request):
    from finanical import settings
    file_obj = request.FILES.get(‘file‘)
    file_name = file_obj.name
    time_path = time.strftime(‘%Y-%m-%d‘, time.localtime())
    path = os.path.join(settings.MEDIA_ROOT, time_path)
    if not os.path.exists(path):
        os.mkdir(path)
    path = os.path.join(path + ‘/‘ + file_name)
    with open(path, ‘wb‘) as f:
        for i in file_obj.chunks():
            f.write(i)
    response = {
        ‘error‘: 0,
        ‘url‘: ‘media/‘ + time_path + file_name + ‘/‘
    }
    return HttpResponse(json.dumps(response))

 

还需要配置url

from django.conf.urls import url
from django.contrib import admin
from django.views.static import serve
from finanical import settings
from api import views
import xadmin

urlpatterns = [
    url(r‘^admin/‘, admin.site.urls),
    url(r‘^xadmin/‘, xadmin.site.urls),
    url(r‘^index/‘, views.index, name=‘index‘),
    url(r‘^media/(?P<path>.*)‘, serve, {‘document_root‘: settings.MEDIA_ROOT}),
    url(r‘^uploadFile/$‘, views.uploadFile, name=‘uploadFile‘),
]