Python Flask:原生文件上传
来自CloudWiki
简单来说，只有三个步骤：
1、创建一个上传表单：
<form method="POST" enctype="multipart/form-data">
<input type="file" name="file">
<input type="submit" value="Upload">
</form>
2、当按下提交键后，通过request对象上的files获取文件。和以前用request获取表单值一样，使用input字段的name值获取：
file = request.files['file']
3、使用save()方法保存文件，指定保存的地址及文件名：
file.save(path + filename)
当然，除了这些，还有很多东西要考虑。
代码
# -*- coding: utf-8 -*-
import os
from flask import Flask, request, url_for, send_from_directory
from werkzeug.utils import secure_filename
#允许的扩展名
ALLOWED_EXTENSIONS = set(['png', 'jpg', 'jpeg', 'gif'])
app = Flask(__name__)
app.config['UPLOAD_FOLDER'] = os.getcwd()
app.config['MAX_CONTENT_LENGTH'] = 16 * 1024 * 1024
#模板
html = '''
<!DOCTYPE html>
<title>Upload File</title>
<h1>图片上传</h1>
<form method=post enctype=multipart/form-data>
<input type=file name=file>
<input type=submit value=上传>
</form>
'''
#检验文件后缀名是否合法
def allowed_file(filename):
return '.' in filename and \
filename.rsplit('.', 1)[1] in ALLOWED_EXTENSIONS
#配置一个函数来获取上传文件的url：
@app.route('/uploads/<filename>')
def uploaded_file(filename):
return send_from_directory(app.config['UPLOAD_FOLDER'],
filename)
@app.route('/', methods=['GET', 'POST'])
def upload_file():
if request.method == 'POST':
file = request.files['file']
if file and allowed_file(file.filename):
filename = secure_filename(file.filename)#导入Werkzeug提供的secure_filename()函数来检查文件名
file.save(os.path.join(app.config['UPLOAD_FOLDER'], filename))
file_url = url_for('uploaded_file', filename=filename)
return html + '<br><img src=' + file_url + '>'
return html
if __name__ == '__main__':
app.run()
