相关疑难解决方法(0)

我有一个拉链文件(new.txt)的Python脚本.

tofile =  "/root/files/result/"+file
targetzipfile = new.zip   # This is how I want my zip to look like
zf = zipfile.ZipFile(targetzipfile, mode='w')
try:
    #adding to archive
    zf.write(tofile)
finally:
    zf.close()

当我这样做时,我得到了zip文件.但是当我尝试解压缩文件时,我得到的文件文件位于与文件路径对应的一系列目录中,即我看到目录root中有一个文件夹result,里面有更多目录,即我有

/root/files/result/new.zip

当我解压缩时,new.zip我有一个看起来像的目录结构

/root/files/result/root/files/result/new.txt.

有没有一种方法可以拉链,这样当我解压缩时我才会得到new.txt.

换句话说,我有/root/files/result/new.zip,当我解压缩时new.zip,它应该看起来像

/root/files/results/new.txt

我正在寻找一种方法将zipfile发送到客户端,该方法是从请求响应生成的.在此示例中,我将一个JSON字符串发送到url,该url返回已转换的JSON字符串的zip文件.

@app.route('/sendZip', methods=['POST'])
def sendZip():
    content = '{"type": "Point", "coordinates": [-105.01621, 39.57422]}'
    data = {'json' : content}
    r = requests.post('http://ogre.adc4gis.com/convertJson', data = data)
    if r.status_code == 200:
        zipDoc = zipfile.ZipFile(io.BytesIO(r.content))
        return Response(zipDoc,
                mimetype='application/zip',
                headers={'Content-Disposition':'attachment;filename=zones.zip'})

但是我的zip文件是空的,而烧瓶返回的错误是

Debugging middleware caught exception in streamed response at a point where response 
headers were already sent