我有一個未來的電話，我想獲得 2 個串列字串
但我得到的回傳值是 Future<List< dynamic >>
var datasource = const FutureInformation();
getSliderDetailsEvents() async {
SliderDetails futureEvents = await datasource.getSliderDetails();
List<String> images = futureEvents.image; //List<String>
List<String> titles = futureEvents.title; // List<String>
List information = [images,titles]; //return an array
return information;
}
如何訪問此異步呼叫中的兩個串列？我知道因為它是異步的，所以我必須回傳一個未來，我是否必須創建一個 FutureBuilder 來創建串列。
謝謝
uj5u.com熱心網友回復：
我在這里回答以更好地格式化回應，您可以從您的代碼中執行以下操作：
Future<List<String>> getSliderDetailsEvents() async {
SliderDetails futureEvents = await datasource.getSliderDetails();
List<String> images = futureEvents.image; //List<String>
List<String> titles = futureEvents.title; // List<String>
List<String> information = [...images,...titles]; //
return information;
}
uj5u.com熱心網友回復：
通過使用Spread Operator ...
Future<List<String>> getSliderDetailsEvents() async{
SliderDetails futureEvents = await datasource.getSliderDetails();
final images = futureEvents.image;
final titles = futureEvents.title;
return [...images, ...titles];
}
或者你可以這樣做
Future<List<String>> getSliderDetailsEvents() async{
SliderDetails futureEvents = await datasource.getSliderDetails();
return [...futureEvents.image, ...futureEvents.title];
}
uj5u.com熱心網友回復：
您可能需要指定串列存盤的物件型別（我假設您正在代碼中創建串列串列）：
List<List<String>> information = <List<String>>[images,titles]; //return an array
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