我创建了一个myObject
类型的列表,当我通过list.add
添加对象实例,然后传递myObject
的构造函数时,我可以引用的第一个属性,后续属性为null
// my class and its constructor method
class Channel {
final String channelName;
final String channelDesc;
final String channelEvent;
final String channelThumbLink;
const Channel(this.channelName, this.channelDesc, this.channelEvent,
this.channelThumbLink);
}
// I construct a list of type<Channel>
final List<Channel> channelList = <Channel>[]
..add(Channel(
'Funkycooking',
'Welcome to the Funkycooking channel, we cook, you get funky',
'Sizzling Salsa Saturday!',
'1'))
// Referencing these elements from the list collection in any widget
Text('${channelList[0].channelName}'),
Text('${channelList[0].channelDesc}'),
Text('${channelList[0].channelName}')
按预期返回“ Funkycooking”
Text('${channelList[0].channelDesc}')
返回null,我希望“欢迎来到Funkycooking频道,我们做饭,您会变得时髦”
答案 0 :(得分:0)
抱歉,我错误地认为热重装可以正确反映我的对象编码更改,但不能,但是通过停止并重新启动该应用程序,它的行为符合我的预期。