【发布时间】:2023-01-26 22:39:11
【问题描述】:
您可以使用索引直接访问它
print(mainList[0].orders[0].id);
将打印第一个 pointOfService 的第一个订单的 Id
注意:这里的 mainList 是包含所有 pointOfService 的列表的名称,我假设你在每个订单中都有 id
【讨论】:
-
这是一个很大的帮助,谢谢!
您的问题不清楚您想要实现什么,但是要在 list 中访问 list ,您可以参考这个,
class PointOfService {
final String name;
final List<Order> orders;
PointOfService({this.name, this.orders});
}
class Order {
final String name;
Order({this.name});
}
void main() {
List<PointOfService> pointofServices = [
PointOfService(
name: "PointOfService 1",
orders: [
Order(name: "Order 1"),
Order(name: "Order 2"),
]),
PointOfService(
name: "PointOfService 2",
orders: [
Order(name: "Order 3"),
Order(name: "Order 4"),
])
];
for (var pointOfService in pointofServices) {
print("PointOfService name: ${pointOfService.name}");
for (var order in pointOfService.orders) {
print("Order name: ${order.name}");
}
}
}
这将输出
PointOfService name: PointOfService 1
Order name: Order 1
Order name: Order 2
PointOfService name: PointOfService 2
Order name: Order 3
Order name: Order 4
【讨论】: