Flask 'tuple‘对象在保存文件时没有'write’属性
Flask 'tuple‘对象在保存文件时没有'write’属性
提问于 2020-07-16 17:29:08
当我想要保存在post请求中接收到的文件时，我在flask上得到了这个有线错误。我试着调试，但我不理解这个错误，因为我使用flask save，在google和其他堆栈溢出问题上，我发现这必须与python文件API一样缺少打开或错误的标志，但我没有任何标志，或不需要在这里打开任何文件。
我是如何发送文件的：
const uploadFile = async (file) =>{
const formData = new FormData();
formData.append("file", file);
fetch("http://localhost:5000/files/upload", {method: "POST", body: formData});
}我是如何接收文件的：
@app.route('/files/upload', methods = ['POST'])
def recive_upload_file():
file = request.files['file']
filename = file.filename
root_dir = os.path.dirname(os.getcwd())
file.save((os.path.join(root_dir,'backend', 'upload'), filename))
return "done"据我所知，文件发送是正确的，因为如果我尝试在recive_uploaded_file函数中打印文件名，我会得到正确的名称。
错误：
Traceback (most recent call last):
File "c:\Python37\lib\site-packages\flask\app.py", line 2447, in wsgi_app
response = self.full_dispatch_request()
File "c:\Python37\lib\site-packages\flask\app.py", line 1952, in full_dispatch_request
rv = self.handle_user_exception(e)
File "c:\Python37\lib\site-packages\flask_cors\extension.py", line 161, in wrapped_function
return cors_after_request(app.make_response(f(*args, **kwargs)))
File "c:\Python37\lib\site-packages\flask\app.py", line 1821, in handle_user_exception
reraise(exc_type, exc_value, tb)
File "c:\Python37\lib\site-packages\flask\_compat.py", line 39, in reraise
raise value
File "c:\Python37\lib\site-packages\flask\app.py", line 1950, in full_dispatch_request
rv = self.dispatch_request()
File "c:\Python37\lib\site-packages\flask\app.py", line 1936, in dispatch_request
return self.view_functions[rule.endpoint](**req.view_args)
File "index.py", line 239, in upload_file
file.save((os.path.join(root_dir,'backend', 'upload'), filename))
File "c:\Python37\lib\site-packages\werkzeug\datastructures.py", line 3070, in save
copyfileobj(self.stream, dst, buffer_size)
File "c:\Python37\lib\shutil.py", line 82, in copyfileobj
fdst.write(buf)
AttributeError: 'tuple' object has no attribute 'write'回答 2
Stack Overflow用户
回答已采纳
发布于 2020-07-16 17:37:01
您指定的文件路径不正确。您正在尝试以文件路径的形式写入元组：
>>> root_dir = '/root/'
>>> filename = 'test.png'
>>> (os.path.join(root_dir,'backend', 'upload'), filename)
('/root/backend/upload', 'test.png')您应该将文件名移动到os.path.join调用中。
>>> os.path.join(root_dir,'backend', 'upload', filename)
'/root/backend/upload/test.png'Stack Overflow用户
发布于 2020-07-16 17:35:49
我找到问题了。我有错误的路径，文件名应该在os.path.join()中
file.save((os.path.join(root_dir,'backend', 'upload', filename)))页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/62931611
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