通过 Flask REST API 创建 JSON 文件

Question

我想在我创建的 ID 文件夹中创建一个包含请求的项目名称、客户名称、客户地址、项目描述参数的 JSON 文件。

我是 Python 的新手，我不知道该怎么做。我尝试了很多搜索但找不到任何东西。一切都是运行直到 json.dump 即最后 8 行

import os
import numpy as np
from flask import Flask,jsonify,request
import json

PROJECTS_DIR_NAME       = "Projects"
ID_DIR_NAME = 'ID_variable.npy'
SCRIPT_PATH = os.path.dirname(__file__)
PR_DIR          = os.path.join(SCRIPT_PATH,PROJECTS_DIR_NAME)
ID_DIR = os.path.join(PR_DIR,ID_DIR_NAME)

app = Flask(__name__)

@app.route('/initiate',methods =['POST'])
def initiate():
    path = PR_DIR
    f= ID_DIR
    try:
        ID_variable = np.load(ID_DIR)
        ID_variable = int(ID_variable)
        ID_variable =ID_variable+ 1
        np.save(f,ID_variable)
    except FileNotFoundError:
        np.save(f, 1)
        ID_variable=1
        
        
    os.chdir(PR_DIR)
    NewProjectID= 'ID'+ str(ID_variable)
    os.makedirs(NewProjectID)
    path2= path+'\'+NewProjectID
    os.chdir(path2)
    projectname = request.args.get('projectname')
    customername = request.args.get('customername')
    customeraddress = request.args.get('customeraddress')
    projectdescription = request.args.get('projectdescription')
    json.dump({"Project Name": projectname} {"Customer Name": customername} {"Customer Address": customeraddress} {"Project Description" : projectdescription},path2)
    return jsonify({'ID': ID_variable})



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

Answer 1

代码编写不正确 json 并且您需要 json.dump

的文件指针

您拥有的 json 是 json 的列表。

with open("your_file_name.json", "w") as f:
    json.dump([{"Project Name": projectname}, {"Customer Name": customername},{"Customer Address": customeraddress},{"Project Description" : projectdescription}],f)

通过 Flask REST API 创建 JSON 文件

Create JSON file via Flask REST API

python

json

flask

flask-restful