【发布时间】:2021-07-18 10:43:13
【问题描述】:
目前我正在尝试以 BLoC 模式从 API 获取数据。但是在通话之后,它会抛出这个消息。 'Future' 不是 'String' 类型的子类型
这里是相关代码。
集团
Stream<NewsState> mapEventToState(NewsEvent event) async* {
if (event is FetchNews) {
yield event.isFeatured == true
? NewsFeaturedLoading()
: NewsCommonLoading();
try {
print("http req->" + event.isFeatured.toString());
final List<News> newsList =
await _fetchNews(event.isFeatured, userRepository.getToken());
yield event.isFeatured == true
? NewsFeaturedSuccess(newsList: newsList)
: NewsCommonSuccess(newsList: newsList);
} catch (error) {
print(error);
yield event.isFeatured == true
? NewsFeaturedFailure(error: error.toString())
: NewsCommonFailure(error: error.toString());
}
}
}
}
HttpCall
Future<List<News>> _fetchNews(isFeatured, accessToken) async {
print("before httprequest->>" + accessToken);
final http.Response response = await http.post(
Uri.parse(Constant.baseUrl + "/api/news"),
headers: {
'Content-type': 'application/json',
'Accept': 'application/json',
"x-access-token": "Bearer " + accessToken,
},
body: {
"isFeatured": isFeatured,
},
);
print("response->>>>" + response.body);
if (response.statusCode == 200) {
print("news-> " + response.body);
var obj = json.decode(response.body);
final data = obj["data"] as List;
return data.map((rawPost) {
return News(
id: rawPost['_id'],
title: rawPost['Title'],
content: rawPost['Description'],
);
}).toList();
} else {
throw Exception(json.decode(response.body));
}
}
查看
SingleChildScrollView(
scrollDirection: Axis.vertical,
child: Column(
children: <Widget>[
SizedBox(height: 25.0),
Align(
alignment: Alignment.topLeft,
child: Padding(
padding: EdgeInsets.only(left: 19.0),
child: Text("Common news",
style: Constant.newsCommonTextStyle),
),
),
if (state is NewsCommonLoading) CircularProgressIndicator(),
if (state is NewsCommonSuccess) CommonNews(),
if (state is NewsCommonFailure)
Text(state.error, style: TextStyle(color: Colors.red)),
],
),
),
这个异常是从哪里来的?我该如何防止这种异常?感谢您的帮助!
【问题讨论】:
-
您能否更新问题以引导我找到确切发生此错误的区域?
-
最终列表
newsList = await _fetchNews(event.isFeatured, userRepository.getToken());在这一行,抛出异常。 -
userRepository.getToken() 是异步函数吗?
-
是的。它正在使用异步
-
@RohanThacker thx 伙计,我想通了。