在Django中流式传输CSV文件

Question

我正在尝试将csv文件作为附件下载流式传输.CSV文件大小为4MB或更多,我需要一种方法让用户主动下载文件,而无需等待创建所有数据并首先提交到内存.

我首先使用自己的基于Django FileWrapper类的文件包装器.那失败了.然后我在这里看到了一个使用生成器来传输响应的方法: 如何使用Django流式传输HttpResponse

当我在生成器中引发错误时,我可以看到我正在使用该get_row_data()函数创建正确的数据,但是当我尝试返回响应时它返回空.我也禁用了Django GZipMiddleware.有谁知道我做错了什么？

编辑:我遇到的问题是ConditionalGetMiddleware.我不得不更换它,代码在下面的答案中.

这是观点:

from django.views.decorators.http import condition

@condition(etag_func=None)
def csv_view(request, app_label, model_name):
    """ Based on the filters in the query, return a csv file for the given model """

    #Get the model
    model = models.get_model(app_label, model_name)

    #if there are filters in the query
    if request.method == 'GET':
        #if the query is not empty
        if request.META['QUERY_STRING'] != None:
            keyword_arg_dict = {}
            for key, value in request.GET.items():
                #get the query filters
                keyword_arg_dict[str(key)] = str(value)
            #generate a list of row objects, based on the filters
            objects_list = model.objects.filter(**keyword_arg_dict)
        else:
            #get all the model's objects
            objects_list = model.objects.all()
    else:
        #get all the model's objects
        objects_list = model.objects.all()
    #create the reponse object with a csv mimetype
    response = HttpResponse(
        stream_response_generator(model, objects_list),
        mimetype='text/plain',
        )
    response['Content-Disposition'] = "attachment; filename=foo.csv"
    return response

这是我用来传输响应的生成器:

def stream_response_generator(model, objects_list):
    """Streaming function to return data iteratively """
    for row_item in objects_list:
        yield get_row_data(model, row_item)
        time.sleep(1)

以下是我创建csv行数据的方法:

def get_row_data(model, row):
    """Get a row of csv data from an object"""
    #Create a temporary csv handle
    csv_handle = cStringIO.StringIO()
    #create the csv output object
    csv_output = csv.writer(csv_handle)
    value_list = [] 
    for field in model._meta.fields:
        #if the field is a related field (ForeignKey, ManyToMany, OneToOne)
        if isinstance(field, RelatedField):
            #get the related model from the field object
            related_model = field.rel.to
            for key in row.__dict__.keys():
                #find the field in the row that matches the related field
                if key.startswith(field.name):
                    #Get the unicode version of the row in the related model, based on the id
                    try:
                        entry = related_model.objects.get(
                            id__exact=int(row.__dict__[key]),
                            )
                    except:
                        pass
                    else:
                        value = entry.__unicode__().encode("utf-8")
                        break
        #if it isn't a related field
        else:
            #get the value of the field
            if isinstance(row.__dict__[field.name], basestring):
                value = row.__dict__[field.name].encode("utf-8")
            else:
                value = row.__dict__[field.name]
        value_list.append(value)
    #add the row of csv values to the csv file
    csv_output.writerow(value_list)
    #Return the string value of the csv output
    return csv_handle.getvalue()

Answer 1

这是一些简单的代码,可以流式传输CSV; 你可以从这个到你需要做的事情:

import cStringIO as StringIO
import csv

def csv(request):
    def data():
        for i in xrange(10):
            csvfile = StringIO.StringIO()
            csvwriter = csv.writer(csvfile)
            csvwriter.writerow([i,"a","b","c"])
            yield csvfile.getvalue()

    response = HttpResponse(data(), mimetype="text/csv")
    response["Content-Disposition"] = "attachment; filename=test.csv"
    return response

这只是将每一行写入内存中的文件,读取行并生成它.

此版本更有效地生成批量数据,但在使用之前一定要了解上述内容:

import cStringIO as StringIO
import csv

def csv(request):
    csvfile = StringIO.StringIO()
    csvwriter = csv.writer(csvfile)

    def read_and_flush():
        csvfile.seek(0)
        data = csvfile.read()
        csvfile.seek(0)
        csvfile.truncate()
        return data

    def data():
        for i in xrange(10):
            csvwriter.writerow([i,"a","b","c"])
        data = read_and_flush()
        yield data

    response = HttpResponse(data(), mimetype="text/csv")
    response["Content-Disposition"] = "attachment; filename=test.csv"
    return response

Answer 2

从Django 1.5开始,中间件问题已经解决,并且引入了StreamingHttpResponse.以下应该做:

import cStringIO as StringIO
import csv

def csv_view(request):
    ...
    # Assume `rows` is an iterator or lists
    def stream():
        buffer_ = StringIO.StringIO()
        writer = csv.writer(buffer_)
        for row in rows:
            writer.writerow(row)
            buffer_.seek(0)
            data = buffer_.read()
            buffer_.seek(0)
            buffer_.truncate()
            yield data
    response = StreamingHttpResponse(
        stream(), content_type='text/csv'
    )
    disposition = "attachment; filename=file.csv"
    response['Content-Disposition'] = disposition
    return response

有一些关于如何从Django输出csv的文档,但它没有利用它,StreamingHttpResponse所以我继续打开一张票以便跟踪它.