在python Flask应用程序中使用json对象上传文件的POST方法 [英] POST method to upload file with json object in python flask app

问题描述

我遇到了一个问题，我正在尝试构建单个 API，该 API 将上传文件和 json 对象.我需要这个 API 来创建 webhook.

I am stuck in a problem where I am trying to build single API which will upload file along with json object. I need this API to create webhook.

使用多部分，我可以上传文件，在选项文件中我可以发送 json 对象.

Using multi part, I am able to upload file and in option filed I am able to send json object.

在 Flask 应用程序中，当我尝试检索 json 对象时，将其转换为 blob 类型.我尝试将其转换为 base64，然后再次转换为字符串，但整个过程不起作用.

In flask app when I am trying to retrieve json object its converting as blob type. I tried to convert it base64 then again converting into string but whole process is not working.

如果有人有好的解决方案，请告诉我，我可以将文件和 json 对象组合在一起，并通过 Flask python 应用程序获取它.

Let me know if anyone has good solution I can combine file and json object together and fetch it by flask python app.

zz 是我的代码中的变量，我试图在其中存储我的 json 对象.name 是我将 json 对象与文件一起传递的字段.

zz is the variable in my code where I am trying to store my json object. name is the field where I am passing my json object with file.

提前致谢.

我当前的代码

app.config['UPLOAD_FOLDER'] = UPLOAD_FOLDER

@app.route('/upload/',methods = ['POST'])
def upload():
    customer_name='abc'
    if request.method == 'POST':

        zz=base64.b64encode(request.files['name'].read())
        try:
            file = request.files['file']
            if file:
                file.filename=customer_name+'_'+str(datetime.now())+'.'+file.filename.split('.')[-1]
                filename = secure_filename(file.filename)
                path=os.path.join(app.config['UPLOAD_FOLDER'], filename)
                file.save(path)

            return jsonify({
                'status':success,
                'junk_data':[],
                'message':error
                })
        except Exception as err:
            logger.error(str(datetime.now())+' '+str(err))
            return jsonify({
                'status':False,
                'junk_data':[],
                'message':str(err)
                })
if __name__ == '__main__':
    app.run(host='localhost',debug=True, use_reloader=True,port=5000)

推荐答案

经过大量研发，我得到了答案.

I have got the answer after lot R&D.

请求格式

//user any client-side

content-type:multipart/form-data
 file: file need to upload 
 data: {"name":"abc","new_val":1}

从请求对象中获取的python代码

data=json.loads(request.form.get('data'))
file = request.files['file']

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