如何在模板中請求 GET 參數以獲取文件列表？（Django Zip 文件下載問題）

Question

我想通過壓縮下載 function 創建的所有單個或多個文件。

問題出在模板上。 請正確建議為我收到錯誤的文件列表傳遞 GET 參數：

FileNotFoundError at /test/
[Errno 2] No such file or directory: '['

這是錯誤放置引用文件列表的查詢字符串的錯誤。

我的看法如下：

def submit(request):
    def file_conversion(input_file,output_file_pattern,chunk_size):
            output_filenames = []
            with open(input_file,"r+") as fin:
    # ignore headers of input files
                for i in range(1):
                    fin.__next__()
            
                reader = csv.reader(fin, delimiter=',')
                
                
                for i, chunk in enumerate(chunked(reader, chunk_size)):
                    output_filename = output_file_pattern.format(i)
                    with open(output_filename, 'w', newline='') as fout:
                        output_filenames.append(output_filename)
                        writer = csv.writer(fout, reader, delimiter='^')
                        writer.writerow(fed_headers)
                        writer.writerows(chunk)
                            # print("Successfully converted into", output_file)
                return output_filenames

    paths = file_conversion(input_file,output_file+'{01}.csv',10000)
    # paths return a list of filenames that are created like output_file1.csv,output_file2.csv can be of any limit
    # when i tried paths[0], it returns output_file1.csv
    context = {'paths' :paths}

def test_download(request):
    paths = request.GET.get('paths')
    context ={'paths': paths}
    response = HttpResponse(content_type='application/zip')
    zip_file = zipfile.ZipFile(response, 'w')
    for filename in paths:
        zip_file.write(filename)
    zip_file.close()
    response['Content-Disposition'] = 'attachment; filename='+'converted files'
    return response

模板

<p> 
<a href ="{% url 'test_download' %}?paths={{ paths|urlencode }} " download>Converted Files</a> </p>
<br>

幫我在這里找到問題。

問題：

the?path 正在尋找文件，但它找到了列表。

然后我嘗試了：

def test_download(request):
    paths = request.GET.getlist('paths')
    context ={'paths': paths}
    response = HttpResponse(content_type='application/zip')
    zip_file = zipfile.ZipFile(response, 'w')
    for filename in paths:
         zip_file.write(filename)
    zip_file.close()
    response['Content-Disposition'] = 'attachment; filename='+'converted files'
    return response

模板：

<p> <a href ="{% url 'test_download' %}?paths=path1 " download>Converted Files</a> </p>
<br>

它查找文件為

FileNotFoundError at /test/
[Errno 2] No such file or directory: '/home/rikesh/Projects/FedMall/main/temp/47QSHA19D003A_UPDATE_20210113_{:01}.csv'

文件路徑應該是：

/home/rikesh/Projects/FedMall/main/temp/47QSHA19D003A_UPDATE_20210113_0.csv

Answer 1

request.GET.get('paths')返回路徑列表的字符串表示形式，而不是列表 object。 返回的值將是這樣"['/path/file1', 'path/file2']" 。 當您遍歷路徑時，它實際上會遍歷字符串中的每個字符。 這就是為什么它首先嘗試查找名稱為[的目錄。

要將文件路徑列表傳遞給GET請求，您需要將 url 更改為此

<your_url>?paths=path1&paths=path2&paths=path3...

在您的 Python 代碼中，使用此獲取文件路徑

request.GET.getlist('paths')