Python how to render data with d3.js?

Question

JSON

Good afternoon, you need to transfer data from the backend and accept it to d3.js.

import pymysql, json
from flask import Flask, render_template

app = Flask(__name__)


@app.route('/get_data')
def index():
  #Подключаемся к MYSQL
    conn = pymysql.connect(host='localhost', port=3306, 
        user='root', passwd='nemate666', db='data_map')
    cursor = conn.cursor(pymysql.cursors.DictCursor)
  
  #Вытаскиваем данные и сериализуем в json
    query_data = cursor.execute("SELECT project_name, countryname, lendprojectcost FROM   data_load")
    for row in cursor:
        data = json.dumps(row)
        data_json = [data]


  #Закрываем соединение
    cursor.close()
    conn.close()

    return render_template('index.html', data_json=data_json)


if __name__=="__main__":
    app.run(host='0.0.0.0', port=5000, debug='true')

And accept d3.js

var width = 900, height = 600;

var projection = d3.geo.mercator();

      
var svg = d3.select("body").append("svg")
        .attr("width", width)
        .attr("height", height);
var path = d3.geo.path().projection(projection);
var g = svg.append("g");

    d3.json("/get_data", function(error, data_json){
        g.selectAll("content")
      .data(topojson.object(data_json, data_json.objects.countries)
          .geometries)
  .enter()
    .append("content")
    .attr("d", path)
    });

Accordingly, the error data_json is undefined. How to accept d3.js data?

Answer 1

do you return mimetype JSON?

Answer 2

Now the function formats have changed, and json() returns a Promise with data that must be caught using the then() function:
d3.json("/get_data")
.then(function(data_json){
. . .
});