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在这里，我使用天气 API。在进行调试时会出现错误，因为，

错误：XMLHttpRequest 错误。C:/b/s/w/ir/cache/builder/src/out/host_debug/dart-sdk/lib/_internal/js_dev_runtime/patch/core_patch.dart 908:28 获取当前包/http/src/browser_client.dart 69:22 C:/b/s/w/ir/cache/builder/src/out/host_debug/dart-sdk/lib/async/zone.dart 1687:54 runUnary C:/b/s/w/ir/ cache/builder/src/out/host_debug/dart-sdk/lib/async/future_impl.dart 160:18 handleValue C:/b/s/w/ir/cache/builder/src/out/host_debug/dart-sdk/ lib/async/future_impl.dart 767:44 handleValueCallback C:/b/s/w/ir/cache/builder/src/out/host_debug/dart-sdk/lib/async/future_impl.dart 796：

我的颤振代码是：

    String searchApiUrl = 'https://www.metaweather.com/api/location/search/?query=';
    String locationApiUrl = 'https://www.metaweather.com/api/location/';


    void fetchSearch(String input) async {
     var searchResult = await http.get(Uri.parse(searchApiUrl + input), headers: {"Accept": 
      "application/json","Access-Control-Allow-Origin": "*"});
  
     var result = json.decode(searchResult.body)[0];

   setState(() {
    location = result["title"];
    woeid = result["woeid"];
});

}

    void fetchLocation() async {
     var locationResult = await http.get(Uri.parse(locationApiUrl + woeid.toString()), 
      headers: {"Accept": "application/json","Access-Control-Allow-Origin": "*"});
     var result = json.decode(locationResult.body);
     var consolidated_weather = result["consolidated_weather"];
     var data = consolidated_weather[0];

   setState(() {
    temperature = data["the_temp"].round();
    weather = data["weather_state_name"].replaceAll(' ','').toLowerCase();
});

}

   void onTextFieldSubmitted(String input){
    fetchSearch(input);
    fetchLocation();

}

4

1 回答 1

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在fetchLocation你取决于来自的woeid值fetchSearch，但你不等待结果onTextFieldSubmitted。尝试这个：

void onTextFieldSubmitted(String input) async {
    await fetchSearch(input);
    await fetchLocation(); // await in this line might not be needed
}
于 2021-12-13T15:33:54.977 回答