在python中将SQL表返回为JSON

  • Z时代
  • 2024-01-10
  • 分类：问答

我在web.py中玩一个小型Web应用程序，并且正在设置一个URL以返回JSON对象。使用python将SQL表转换为JSON的最佳方法是什么？

回答：

就个人而言，我更喜欢使用SQLObject进行此类操作。我改编了一些必须快速完成的测试代码：

import simplejson
from sqlobject import *
# Replace this with the URI for your actual database
connection = connectionForURI('sqlite:/:memory:')
sqlhub.processConnection = connection
# This defines the columns for your database table. See SQLObject docs for how it
# does its conversions for class attributes <-> database columns (underscores to camel
# case, generally)
class Song(SQLObject):
    name = StringCol()
    artist = StringCol()
    album = StringCol()
# Create fake data for demo - this is not needed for the real thing
def MakeFakeDB():
    Song.createTable()
    s1 = Song(name="B Song",
              artist="Artist1",
              album="Album1")
    s2 = Song(name="A Song",
              artist="Artist2",
              album="Album2")
def Main():
    # This is an iterable, not a list
    all_songs = Song.select().orderBy(Song.q.name)
    songs_as_dict = []
    for song in all_songs:
        song_as_dict = {
            'name' : song.name,
            'artist' : song.artist,
            'album' : song.album}
        songs_as_dict.append(song_as_dict)
    print simplejson.dumps(songs_as_dict)
if __name__ == "__main__":
    MakeFakeDB()
    Main()

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