我想以最佳方式向同一服务器发送多个请求。所以我有
Future<List<Item>> getAllItems() async {
var client = new http.Client();
List<String> itemsIds = ['1', '2', '3']; //different ids
List<Item> itemList = [];
for (var item in itemsIds) {
//make call to server eg: 'sampleapi/1/next' etc
await client.get('sampleapi/' + item + '/next').then((response) {
//Do some processing and add to itemList
});
}
client.close();
return itemList;
}
现在,API呼叫是一个又一次的。但是API呼叫彼此独立。什么是实施的最佳方法以避免异步等待地狱?
您可以使用Future.wait(...)
等待一组Future
s完成:
Future<List<Item>> getAllItems() async {
var client = new http.Client();
List<String> itemsIds = ['1', '2', '3']; //different ids
return Future.wait<Item>(['1', '2', '3'].map((item) =>
client.get('sampleapi/' + item + '/next').then((response) {
//Do some processing and add to itemList
return foo; // some Item that is the result of this request
});
);
}
另请参见https://api.dartlang.org/stable/1.24.3/dart-ashync/future/wait.html
günter击败了我几分钟,但是我已经输入了它,所以这里有一个轻微的选择,也可以使用,并避免使用'then''then'完全。
Future<List<Item>> getAllItems() async {
var client = new Client();
List<String> itemsIds = ['1', '2', '3']; //different ids
List<Response> list = await Future.wait(itemsIds.map((itemId) => client.get('sampleapi/$itemId/next')));
return list.map((response){
// do processing here and return items
return new Item();
}).toList();
}