客户端正在发送任意名称的文件。我正在通过以下实现来处理该请求。
@app.route('/', methods=['GET', 'POST'])
def upload_file():
if request.method == 'POST':
# 'file-name' is the file name here
if 'file-name' not in request.files:
flash('No file part')
return 'no file found'
file = request.files['file-name']
我应该询问另一个定义文件名的查询/路径参数吗？
使用
file = request.files['file']file.filename文档提供了以下完整示例。 （请注意，
'file'def allowed_file(filename):
return '.' in filename and \
filename.rsplit('.', 1)[1].lower() in ALLOWED_EXTENSIONS
@app.route('/', methods=['GET', 'POST'])
def upload_file():
if request.method == 'POST':
# check if the post request has the file part
if 'file' not in request.files:
flash('No file part')
return redirect(request.url)
file = request.files['file']
# if user does not select file, browser also
# submit a empty part without filename
if file.filename == '':
flash('No selected file')
return redirect(request.url)
if file and allowed_file(file.filename):
filename = secure_filename(file.filename)
file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
return redirect(url_for('uploaded_file',
filename=filename))
return '''
<!doctype html>
<title>Upload new File</title>
<h1>Upload new File</h1>
<form method=post enctype=multipart/form-data>
<p><input type=file name=file>
<input type=submit value=Upload>
</form>
'''