angularjs – 使用DjangoREST Angular上传模型ImageField

2024年1月30日 187次阅读

我是REST的新手,尝试使用一个模型和两个字段CharField,
ImageField创建简单的应用程序.

模型：

class Photo(models.Model):

    img = models.ImageField(upload_to='photos/', max_length=254)
    text = models.CharField(max_length=254, blank=True)

串行：

class PhotoSerializer(serializers.ModelSerializer):

    class Meta:
        model = Photo
        fields = ('img','text')

观点：

class PhotoViewSet(viewsets.ModelViewSet):

    queryset = Photo.objects.all()
    serializer_class = PhotoSerializer

photo_router = DefaultRouter()
photo_router.register(r'photo', PhotoViewSet)

网址：

urlpatterns = [
    url(r'^admin/', admin.site.urls),
    url(r'^api/', include(photo_router.urls)),]

POST和GET在/ api / photo /中运行良好
我之前问过类似的问题,我当然肯定,问题出在前端,使用AngularJS,我也是第一次使用AngularJS.但我想现在,我的服务器应用可能有问题.
角度代码：

var app = angular.module('myApp', ['ngRoute', 'angularFileUpload']);

app.config(function ($routeProvider) {

    $routeProvider
        .when('/', {
            templateUrl: 'http://127.0.0.1:8000/static/js/angular/templates/home.html'
    })
});
app.config(['$httpProvider', function($httpProvider) {
    $httpProvider.defaults.xsrfCookieName = 'csrftoken';
    $httpProvider.defaults.xsrfHeaderName = 'X-CSRFToken';
}]);

GET控制器运行良好,我收到了所有图像：

app.controller('getController', ['$scope', '$http', function($scope, $http){
    $http.get('/api/photo/').success(function(data){
        $scope.photos = data;
    }).error(function(data) {
        $scope.photos = data;
    });
}]);

对于文件上传,我使用了this guide
以下是指南中的代码：

app.directive('fileModel', ['$parse', function ($parse) {
return {
    restrict: 'A',
    link: function(scope, element, attrs) {
        var model = $parse(attrs.fileModel);
        var modelSetter = model.assign;

        element.bind('change', function(){
            scope.$apply(function(){
                modelSetter(scope, element[0].files[0]);
            });
        });
    }
};
}]);
app.service('fileUpload', ['$http', function ($http) {
    this.uploadFileToUrl = function(file, uploadUrl){
        var fd = new FormData();
        fd.append('file', file);
        $http.post(uploadUrl, fd, {
            transformRequest: angular.identity,
            headers: {'Content-Type': undefined}
        })
        .success(function(){
        })
        .error(function(){
        });
    }
}]);
app.controller('myCtrl', ['$scope', 'fileUpload', function($scope,     fileUpload){

    $scope.uploadFile = function(){
        var file = $scope.myFile;
        console.log('file is ' );
        console.dir(file);
        var uploadUrl = "/api/photo/";
        fileUpload.uploadFileToUrl(file, uploadUrl);
    };

}]);

HTML代码：

<div ng-controller = "myCtrl">
    <input type="file" file-model="myFile"/>
    <button ng-click="uploadFile()">upload me</button>
</div>

当我通过此表单上传文件时,我有：

POST 07001 400 (Bad Request)
in responce:
img: [“No file was submitted.”]

请至少帮助理解我的问题所在,在我的REST或Angular中.我怎么调试这个？我花了一整天时间来寻找解决方案.我想我显然不明白REST api需要看起来如何.
UPD
我正在尝试使用ng-file-upload,仍然有400个响应状态.
我决定是序列化程序的问题,我在视图中覆盖POST方法,但我认为我做错了：

class PhotoViewSet(viewsets.ModelViewSet):  
    queryset = Photo.objects.all()
    serializer_class = PhotoSerializer

    def post(self, request, format=None):
        serializer = PhotoSerializer(data=request.DATA)
        if serializer.is_valid():
            serializer.save()
            return Response(serializer.data,  status=status.HTTP_201_CREATED)
        return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)

以下是来自Web工具的完整请求数据：

General
Request URL:07001
Request Method:POST
Status Code:400 Bad Request
Remote Address:127.0.0.1:8000
Response Headers
Allow:GET, POST, HEAD, OPTIONS
Content-Type:application/json
Date:Wed, 09 Mar 2016 14:46:44 GMT
Server:WSGIServer/0.2 CPython/3.4.3
Vary:Accept, Cookie
X-Frame-Options:SAMEORIGIN
Request Headers
Accept:application/json, text/plain, /
Accept-Encoding:gzip, deflate
Accept-Language:ru-RU,ru;q=0.8,en-US;q=0.6,en;q=0.4
Connection:keep-alive
Content-Length:16635
Content-Type:multipart/form-data; boundary=—- WebKitFormBoundarytci4JfMISlMXwzFw
Cookie:sessionid=f5nyqdclupgi44zyqbmo3gkenum9k0cj; csrftoken=E0DuVDs5KrV0rSMuD5wC86RPO0QP4AgT
Host:127.0.0.1:8000
Origin:07004
Referer:07005
User-Agent:Mozilla/5.0 (X11; Linux x86_64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/47.0.2526.106 Safari/537.36
X-CSRFToken:E0DuVDs5KrV0rSMuD5wC86RPO0QP4AgT
Request Payload
——WebKitFormBoundarytci4JfMISlMXwzFw
Content-Disposition: form-data; name=”file”; filename=”2.jpg”
Content-Type: image/jpeg
——WebKitFormBoundarytci4JfMISlMXwzFw–

最佳答案 查看页面
Django Rest Framework File Upload上提供的示例：

您需要在视图中添加相关的parser_classes,并覆盖perform_create函数以从POST数据中获取文件：

class PhotoViewSet(viewsets.ModelViewSet):

    queryset = Photo.objects.all()
    serializer_class = PhotoSerializer
    parser_classes = (MultiPartParser, FormParser,)

    def perform_create(self, serializer):
        serializer.save(img=self.request.data.get('img'))
        # do note that at this time `text` information is not available

因为,我们发布FormData,我们使用以下解析器(link)：

parser_classes = (MultiPartParser, FormParser,)

另外,请注意serializer.is_valid()抛出错误,因为它没有获取post数据中的img字段.您需要设置img而不是文件字段：

fd.append('img', file);
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