Firebase 如何减少颤振中for循环的时间？
我有一个名为getNearByPlaces（）的方法，然后我有一个for循环，循环遍历每个place_id，并向google API发送请求，以获取place_id的名称， 所以这个操作大约需要15秒，我怎样才能让它更快呢Firebase 如何减少颤振中for循环的时间？,firebase,flutter,dart,asynchronous,google-cloud-firestore,Firebase,Flutter,Dart,Asynchronous,Google Cloud Firestore,我有一个名为getNearByPlaces（）的方法，然后我有一个for循环，循环遍历每个place_id，并向google API发送请求，以获取place_id的名称， 所以这个操作大约需要15秒，我怎样才能让它更快呢 Future<void> getNearByPlaces(double latitude, double longitude) async { List results = []; List placesId = []; List near
Future<void> getNearByPlaces(double latitude, double longitude) async {
List results = [];
List placesId = [];
List nearbyPlaces = [];
String nearbyUrl =
"https://maps.googleapis.com/maps/api/place/nearbysearch/json?location=${latitude},${longitude}&radius=500&key=${mapKey}";
var nearbyResponse =
await RequestAssistant.getRequest(Uri.parse(nearbyUrl));
if (nearbyResponse == "Failed.") {
return;
}
results = nearbyResponse["results"];
for (int i = 0; i < results.length; i++) {
placesId.add(results[i]['place_id']);
String placeDetailsUrl =
"https://maps.googleapis.com/maps/api/place/details/json?place_id=${results[i]['place_id']}&key=$mapKey";
var response =
await RequestAssistant.getRequest(Uri.parse(placeDetailsUrl));
if (response == "Failed.") {
return;
}
if (response["status"] == "OK") {
await nearbyPlaces.add(response["result"]["name"]);
}
}
print(nearbyPlaces);
await FirebaseFirestore.instance
.collection("nearbyPlaces")
.doc(uid)
.set({'nearbyPlaces': nearbyPlaces});
}
Future getNearByPlaces（双纬度、双经度）异步{
列出结果=[]；
列表placesId=[]；
列出邻近位置=[]；
字符串nearbyUrl=
"https://maps.googleapis.com/maps/api/place/nearbysearch/json?location=${纬度}，${经度}&半径=500&key=${mapKey}”；
var nearbyResponse=
wait RequestAssistant.getRequest（Uri.parse（nearbyUrl））；
if（nearbyResponse==“失败”）{
返回；
}
结果=接近响应[“结果”]；
for（int i=0；i
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