Question

我正在python flask中创建api。我正在将MySQL用于数据库，并使用SELECT查询成功获取了数据。我已经使用flask的jsonify以json格式返回了数据。我想问的是，在使用整数作为索引的jsonify之后，我正在以以下格式获取json

int main()
{
cv::Mat image1, flat_image;
cv::Mat initial_image = cv::imread("D:\mémoire\DB\to use in training\images_sorted\\image1.jpg", 0);
cv::Mat trainData = initial_image.reshape(1, 1);
 ////// trainData
    for (int i = 2; i < 30; i++)
    {
        std::string filename = "D:\mémoire\DB\to use in training\images_sorted\\image";
        filename = filename + std::to_string(i);
        filename = filename + ".jpg";
        image1 = cv::imread(filename, 0);
        flat_image = image1.reshape(1, 1);
        trainData.push_back(flat_image);
    }
}

这很好，我可以使用它，但是我正在寻找具有数据库表属性作为其索引的json，就像我们可以在nodejs中获取那样，因此在字体结尾工作将更加容易。我想使用以下格式的json，其中index是我的数据库表列名称。

[
 [
  1, 
  "FURNITURE", 
  "INDOOR FURNITURE"
 ], 
 [
  2, 
  "AUTOMOBILES", 
  "CARS, BIKES"
 ], 
 [
  3, 
  "LAPTOP & ACCESSORIES", 
  "LAPTOP, MOUSE, KEY BOARD, PENDRIVE"
 ]
]

Answer 1

该问题未指定您的SQL适配器，但是如果您将SQLAlchemy与Flask一起使用，则以下是如何查询数据库并以所需格式输出结果的示例：

from flask import Flask
from models import db
import json

app = Flask(__name__)

app.config['SQLALCHEMY_DATABASE_URI'] = 'your_database_connection_string'
db.init_app(app)

result = db.session.execute("SELECT * FROM categories WHERE id = :id", {"id":999})

# If no rows were returned (e.g., an UPDATE or DELETE), return an empty list
if result.returns_rows == False:
    response = []

# Convert the response to a plain list of dicts
else:
    response = [dict(row.items()) for row in result]

# Output the query result as JSON
print(json.dumps(response))

Answer 2

我找到了使用sql-alchemy更好的解决方案。如果仍然需要按照问题陈述的解决方案，那么这里是一个解决方案

import json 

def conv_func(list_data):
    dic ={ "CATEGORY_ID":list_data[0],
          "CATEGORY_NAME":list_data[1],
          "DESCRIPTION":list_data[2]
          }
    return dic


data = '''[[
  1, 
  "FURNITURE", 
  "INDOOR FURNITURE"
 ], 
 [
  2, 
  "AUTOMOBILES", 
  "CARS, BIKES"
 ], 
 [
  3, 
  "LAPTOP & ACCESSORIES", 
  "LAPTOP, MOUSE, KEY BOARD, PENDRIVE"
 ]
]'''

data = json.loads(data)



new_data=[]
for i in data:
        new_data.append(conv_func(i))

print(new_data)

Answer 3

import pymysql
cur = mysql.connect().cursor(pymysql.cursors.DictCursor)
cur.execute(sql)
row = cur.fetchall()
print row

[{u'symbol': u'AAPL'}, {u'symbol': u'SQ'}]

将结果转换为python flask中的json

3 个答案: