我正在尝试在我的web应用程序上上传一个基本的文本/csv文件,该应用程序运行flask来处理http请求。我试图遵循flasks文档中在本地主机here上运行的baby示例。但当我在我的页面上尝试这段代码时,它看起来像是上传的,但随后只是挂起,事实上我的flask服务器冻结了,我不得不关闭终端再试一次……Ctrl+C甚至不起作用。
我执行run.py
:
#!/usr/bin/env python
from app import app
if __name__ == '__main__':
app.run(host='0.0.0.0', port=5000, debug=False, use_reloader=False)
并且app
是同一目录中的一个目录,其中run.py
具有以下__init__.py
:
import os
from flask import Flask
from werkzeug import secure_filename
#Flask object initialization
#app flask object has to be created before importing views below
#because it calls "import app from app"
UPLOAD_FOLDER = '/csv/upload'
ALLOWED_EXTENSIONS = set(['txt', 'csv'])
app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER
这是我的views.py
文件,它包含了我的所有路由:
from flask import render_template, request, redirect, url_for
from app import app
import os
#File extension checking
def allowed_filename(filename):
return '.' in filename and filename.rsplit('.',1)[1] in ALLOWED_EXTENSIONS
@app.route('/', methods=['GET', 'POST'])
@app.route('/index.html', methods=['GET', 'POST'])
def index():
if request.method == 'POST':
submitted_file = request.files['file']
if submitted_file and allowed_filename(submitted_file):
filename = secure_filename(submitted_file.filename)
submitted_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 action="" method=post enctype=multipart/form-data>
<p><input type=file name=file>
<input type=submit value=Upload>
</form>
'''
问题是你把错误的东西传递给了
allowed_filename()
。你应该传递submitted_file.filename
而不是submitted_file
本身有一个库*用来处理用Flask上传的文件:
https://github.com/joegasewicz/flask-file-upload
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