info = {"name":'小明', "id":1234}
id = info.get('age')
上述程序运行将报出KeyError的错误
id = info.get('age')
上述程序运行将报出KeyError的错误
举一反三
- 请看下面一段程序: def info(age=18,name): print("%s的年龄为%d"%(name,age)) info(28,'小明') 运行上述程序,最终输出的结果为
- 请看下面的一段程序:info = {1:'小明', 2:'小黄',3:'小兰'}name = info.get(4,'小红')name2 = info.get(1)print(name)print(name2)运行上述程序,最终输出的结果为 A: 小红,小黄 B: 小红,小明 C: 小黄,小明 D: 小兰,小明
- 请看下面的一段程序:info = {1:'Rose', 2:'Mike',3:'Jack'}name = info.get(4,'Tom')name2 = info.get(1)print(name)print(name2)运行上述程序,最终输出的结果为()。 A: Tom,Mike B: Tom,Rose C: Mike,Rose D: Jack,Rose
- 学生表(id,name,sex,age,depart_id,depart_name),存在函数依赖是id→name,sex,age,depart_id;dept_id→dept_name,其满足
- 以下自定义数据类型的语句中,正确的是 A: A.Type student ID As String * 20 name As String * 10 age As Integer End student B: B.Type student ID As String * 20 name As String * 10 age As Integer End Type C: C.Type student ID As String name As String age As Integer End student D: C.Type ID As String * 20 name As String * 10 age As Integer End Type student