python - send_file() 调用时返回文本文档而不是图像

Question

我想从服务器端向客户端发送一个图像文件。我正在使用 flask 框架。

但问题是每当我调用 send_file() 所在的路由时，响应返回的是一个文件。当我单击此文件时，gedit 打开它时该文件中没有任何内容。这意味着它必须是文本文件写入。

我引用了 flask 文档 send_file() .

这是我在代码中所做的:

@app.route('/try')
def trial():
    todown = 'https://igcdn-photos-e-a.akamaihd.net//hphotos-ak-xaf1//t51.2885-15//e35//12093691_1082288621781484_1524190206_n.jpg'  
    resp = requests.get(todown)
    return send_file(resp,mimetype="image/jpeg",attachment_filename="img.jpg",as_attachment=True)

每当我加载 localhost:5000/try 时，都会下载一个文件，但不会下载我想要下载的图像文件。

我在终端中遇到的错误是 AttributeError: 'Response' object has no attribute 'read' error。

一定是什么问题。上面的代码片段中是否缺少任何内容？

Answer 1

1. resp 是一个 requests.models.Response 对象，不是字符串也不是字节:

   >>> import requests
   >>> todown = 'https://igcdn-photos-e-a.akamaihd.net//hphotos-ak-xaf1//t51.2885-15//e35//12093691_1082288621781484_1524190206_n.jpg'
   >>> resp = requests.get(todown)
   >>> resp
   <Response [200]>
   >>> type(resp)
   <class 'requests.models.Response'>
   

2. Flask.send_file() 发送文件。

---

所以首先你需要使用 resp.content 来获取对象的内容，它会返回 bytes 对象(顺便说一句，resp.text返回字符串对象。如果您要下载图片、视频或其他非文本内容，请始终使用 .content。

>>> import requests
>>> todown = 'https://igcdn-photos-e-a.akamaihd.net//hphotos-ak-xaf1//t51.2885-15//e35//12093691_1082288621781484_1524190206_n.jpg'
>>> resp = requests.get(todown)
>>> type(resp.content)
<class 'bytes'>

请查看the document了解更多详情。

---

然后，因为Flask.send_file() 发送一个文件，所以在发送之前需要将图像写入文件。

但由于您不需要在您的服务器上使用此图像，我建议使用 io.BytesIO在这种情况下，您不需要在发送后删除该图像。并注意使用 io.StringIO如果您要发送文本文件。

例如:

import requests
from io import BytesIO
from flask import Flask, send_file

app = Flask(__name__)

@app.route('/')
def tria():
    todown = 'https://igcdn-photos-e-a.akamaihd.net//hphotos-ak-xaf1//t51.2885-15//e35//12093691_1082288621781484_1524190206_n.jpg'
    resp = requests.get(todown)
    return send_file(BytesIO(resp.content), mimetype="image/jpeg", attachment_filename="img2.jpg", as_attachment=True)

app.run(port=80, debug=True)

---

不过，如果你想把图片写入文件然后发送，当然也可以。我们可以使用 tempfile.NamedTemporaryFile()创建一个 tempfile 而不是仅仅创建一个文件以避免重写您的重要文件。

来自文档:

This function operates exactly as TemporaryFile() does, except that the file is guaranteed to have a visible name in the file system (on Unix, the directory entry is not unlinked).

That name can be retrieved from the name attribute of the file object. Whether the name can be used to open the file a second time, while the named temporary file is still open, varies across platforms (it can be so used on Unix; it cannot on Windows NT or later). If delete is true (the default), the file is deleted as soon as it is closed.

The returned object is always a file-like object whose file attribute is the underlying true file object. This file-like object can be used in a with statement, just like a normal file.

例如:

import tempfile
import requests
from flask import Flask, send_file

app = Flask(__name__)


@app.route('/')
def tria():
    todown = 'https://igcdn-photos-e-a.akamaihd.net//hphotos-ak-xaf1//t51.2885-15//e35//12093691_1082288621781484_1524190206_n.jpg'

    resp = requests.get(todown)

    with tempfile.NamedTemporaryFile() as f:  
        # create a file-like object use `NamedTemporaryFile()` and `with` 
        # as the basic usage which said in the document     

        f.write(resp.content)
        # write the content of the image into it

        return send_file(f.name, mimetype="image/jpeg",
                         attachment_filename="img2.jpg", as_attachment=True)                             
        # `f.name` is the temp file's filename

app.run(port=80, debug=True)