如何在python Flask框架中发送zip文件？ [英] How to send zip files in the python Flask framework?

问题描述

我有一个烧录服务器，它从数据库中抓取几个不同文件的二进制数据，并把它们放到一个python的zipfile对象中。我想发送生成的zip文件与我的代码使用flask的send_file方法。

我最初能够通过使用BytesIO（bin ）作为send_file的第一个参数，但出于某种原因，我不能用我生成的zip文件做同样的事情。它给出了错误：

ZipFile没有缓冲区接口。

如何发送这个zip文件对象与Flask用户？

这是我的代码：

 <$ c $ b $ def def downloadFiles（）：$ b $如果request.method =='POST'：
 mongo = MongoDAO（'localhost'，27017）
 identifier = request.form ['CapsuleName'] 
 password = request.form ['CapsulePassword'] 
 result = mongo.getCapsuleByIdentifier标识符，密码）
 zf = zipfile.ZipFile（'capsule.zip'，'w'）
 files = result ['files'] 
用于文件中的个人文件：
数据= zipfile.ZipInfo（individualFile ['fileName']）
 data.date_time = time.localtime（time.time（））[：6] 
 data.compress_type = zipfile.ZIP_DEFLATED 
 zf .writestr（data，individualFile ['fileData']）
 return send_file（BytesIO（zf），attachment_filename ='capsule.zip'，as_attachment = True）
 return render_template（'download.html'）
  需要传递字节数据  ，但是 ZipFile（）对象不是bytes-data;你实际上在你的硬盘上创建了一个文件 。
 $ b $你可以创建一个 ZipFile（）在内存中使用 BytesIO（） 作为基础：
 
 
  memory_file = BytesIO（）
 with zipfile.ZipFile（memory_file，'w'）as zf：
 files = result ['files'] 
 files for individualFile： 
 data = zipfile.ZipInfo（individualFile ['fileName']）
 data.date_time = time.localtime（time.time（））[：6] 
 data.compress_type = zipfile.ZIP_DEFLATED 
 zf.writestr（data，individualFile ['fileData']）
 memory_file.seek（0）
 return send_file（memory_file，attachment_filename ='capsule.zip'，as_attachment = True）
  
使用语句的确保 ZipFile（）对象在完成添加条目时正确关闭，导致它将所需的尾部写入内存中的文件对象。需要 memory_file.seek（0）来将文件对象的读写位置倒带回到开始位置。
 
I have a flask server that grabs binary data for several different files from a database and puts them into a python 'zipfile' object.  I want to send the generated zip file with my code using flask's "send_file" method.

I was originally able to send non-zip files successfully by using the BytesIO(bin) as the first argument to send_file, but for some reason I can't do the same thing with my generated zip file.  It gives the error:

'ZipFile' does not have the buffer interface.

How do I send this zip file object to the user with Flask?

This is my code:
@app.route("/getcaps",methods=['GET','POST'])
def downloadFiles():
    if request.method == 'POST':
        mongo = MongoDAO('localhost',27017)
        identifier = request.form['CapsuleName']
        password = request.form['CapsulePassword']
        result = mongo.getCapsuleByIdentifier(identifier,password)
        zf = zipfile.ZipFile('capsule.zip','w')
        files = result['files']
        for individualFile in files:
            data = zipfile.ZipInfo(individualFile['fileName'])
            data.date_time = time.localtime(time.time())[:6]
            data.compress_type = zipfile.ZIP_DEFLATED
            zf.writestr(data,individualFile['fileData'])
        return send_file(BytesIO(zf), attachment_filename='capsule.zip', as_attachment=True)
    return render_template('download.html')

解决方案
BytesIO() needs to be passed bytes data, but a ZipFile() object is not bytes-data; you actually created a file on your harddisk.

You can create a ZipFile() in memory by using BytesIO() as the base:
memory_file = BytesIO()
with zipfile.ZipFile(memory_file, 'w') as zf:
    files = result['files']
    for individualFile in files:
        data = zipfile.ZipInfo(individualFile['fileName'])
        data.date_time = time.localtime(time.time())[:6]
        data.compress_type = zipfile.ZIP_DEFLATED
        zf.writestr(data, individualFile['fileData'])
memory_file.seek(0)
return send_file(memory_file, attachment_filename='capsule.zip', as_attachment=True)
The with statement ensures that the ZipFile() object is properly closed when you are done adding entries, causing it to write the required trailer to the in-memory file object. The memory_file.seek(0) call is needed to 'rewind' the read-write position of the file object back to the start.

                        
这篇关于如何在python Flask框架中发送zip文件？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！