Question

我必须通过curl命令将文件上传到flask。我尝试使用来自Folder/Files Upload with Flask的curl命令进行发送，但无法识别请求。

curl --form "fileupload=@filename.txt" http://hostname/resource

以及Python代码：

from flask import Flask,abort,render_template,request,redirect,url_for
from werkzeug import secure_filename
import os
app = Flask(__name__)
UPLOAD_FOLDER = './uploads'
app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER 


@app.route('/upload/',methods = ['POST'])
def upload_file():
    if request.method =='POST':
        file = request.files['file']
        print request.files
        if file:
            filename = secure_filename(file.filename)
            file.save(os.path.join(app.config['UPLOAD_FOLDER'],filename))



if __name__ == '__main__':
    app.run(debug = True)

数据应在通过CURL命令发布时保存

Answer 1

好吧，您可能需要返回一些内容，如果没有它，Flask会为您添加尝试错误提供

return jsonify({'message':'File uploaded!'})

然后尝试：)请记住从flask导入jsonify！

在Flask中卷曲上传的文件

1 个答案: