在Python SQLite中执行INSERT后,表中没有结果

在Python SQLite中执行INSERT后,表中没有结果,python,sqlite,Python,Sqlite,我使用Python和SQLite包创建表并将数据插入表中。但是,在我启动执行后,表中没有任何内容。有人能帮我弄清楚吗?谢谢 def conSqlite(): conn = sqlite3.connect('C:\\Users\jet.cai\Documents\Logsitic.db') json_path = r'C:\Users\jet.cai\PycharmProjects\VJSF\txtToJson.json' try: create_table = ('''

我使用Python和SQLite包创建表并将数据插入表中。但是,在我启动执行后,表中没有任何内容。有人能帮我弄清楚吗?谢谢

def conSqlite():
  conn = sqlite3.connect('C:\\Users\jet.cai\Documents\Logsitic.db')
  json_path = r'C:\Users\jet.cai\PycharmProjects\VJSF\txtToJson.json'
  try:
      create_table = ('''
          CREATE TABLE IF NOT EXISTS CODE2
          (Delivery            TEXT,
          Customer_Name        NCHAR(50),
          Shipment_Priority    TEXT
          )''')
      conn.execute(create_table)
  except:
      print("Table Failed")
      return False

  with open(json_path, 'r') as jsonf:
      lines = json.load(jsonf)
      for line in lines:
          sql = "insert into CODE2(Delivery,Customer_Name,Shipment_Priority) values('%s','%s','%s')"%(line['Delivery'],line['Customer Name'],line['Shipment Priority'])

          conn.execute(sql)


  # No results can be selected out
  df = pd.read_sql("select Delivery from CODE2", conn)

  print(df)

commit
是处理数据库时要记住的一个非常重要的概念。伙计们,我发现了,我没有提交insert操作,如果发生了,很抱歉打扰您Hi Alan,没错,谢谢您的评论!我们同时看到了它。一切都很好。Stack应该允许我们用拇指仰视:)