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无法打开使用 python 和 Django 创建的 zip 文件

[英]Can't open zip file created with python and Django

I create a set of pdf files and want to add them to zip archive.我创建了一组 pdf 文件,并希望将它们添加到 zip 存档中。 Everything seems fine, but when I download my zip file It can't be open.一切似乎都很好,但是当我下载我的 zip 文件时,它无法打开。

So I create pdf with create_pdf function所以我用create_pdf function 创建 pdf

def create_pdf(child):
    buffer = io.BytesIO()
    canvas = Canvas(buffer, pagesize=A4)
    p = staticfiles_storage.path('TNR.ttf')
    pdfmetrics.registerFont(TTFont('TNR', p))
    canvas.setFont('TNR', 14)
    t = canvas.beginText(-1 * cm, 29.7 * cm - 1 * cm)
    t.textLines(create_text(child), trim=0)

    canvas.drawText(t)
    canvas.save()
    pdf = buffer.getvalue()
    return pdf

Then I create zip file and pack it to response然后我创建 zip 文件并将其打包以响应

def create_zip(pdfs):
    mem_zip = io.BytesIO()
    i = 0
    with zipfile.ZipFile(mem_zip, mode='w', compression=zipfile.ZIP_DEFLATED)\
         as zf:
        for f in pdfs:
            i += 1
            zf.writestr(f'{str(i)}.pdf', f)
    return mem_zip.getvalue()


def get_files(request, children):
    pdfs = []
    for child in children:
        pdfs.append(create_pdf(child))
    zip = create_zip(pdfs)
    response = FileResponse(zip,
                            content_type='application/zip',
                            filename='zayavleniya.zip')
    response['Content-Disposition'] = 'attachment; filename=files.zip'
    return response

Please help to find where I am wrong.请帮助找出我错在哪里。

In the documentation , you can see the write_str method expects data as second argument.文档中,您可以看到write_str方法需要data作为第二个参数。 Here, you are providing a filename.在这里,您提供了一个文件名。 So the content of the pdf files are just "i.pdf", which is of course not the content that you expect for a pdf file.所以 pdf 文件的内容只是“i.pdf”,这当然不是您期望的 pdf 文件的内容。

Try with something like this:尝试这样的事情:

def create_zip(pdfs):
    mem_zip = io.BytesIO()
    i = 0
    with zipfile.ZipFile(mem_zip, mode='w', compression=zipfile.ZIP_DEFLATED)\
         as zf:
        for filename in pdfs:
            i += 1
            with open(filename, 'rb') as f:
                zf.writestr(f'{i}.png', f.read())
    return mem_zip.getvalue()

NB: try to avoid using zip as a variable name, since it is already a builtin python function注意:尽量避免使用zip作为变量名,因为它已经是内置 python function

Update更新

If you isolate the archive creation to get a minimal working example, you get this, which create an zipfile as you wanted:如果您隔离存档创建以获得最小的工作示例,您会得到这个,它会根据需要创建一个 zipfile:

def create_zip(pdfs):
    i = 0
    with zipfile.ZipFile(HERE / "my_archive.zip", mode='w', compression=zipfile.ZIP_DEFLATED)\
         as zf:
        for filename in pdfs:
            i += 1
            with open(filename, 'rb') as f:
                zf.writestr(f'{str(i)}.png', f.read())

create_zip(["icon.png"])

After I post this question I managed to find the answer myself.在我发布这个问题之后,我设法自己找到了答案。 I changed create_zip我改变了create_zip

def create_zip(pdfs):
    mem_zip = io.BytesIO()
    i = 0
    with zipfile.ZipFile(mem_zip, mode='w', compression=zipfile.ZIP_DEFLATED)\
         as zf:
        for f in pdfs:
            i += 1
            zf.writestr(f'{str(i)}.pdf', f)
    mem_zip.seek(0)
    return mem_zip

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