这是我的asynctask部分，我将pin放在从xml解析的某个位置。这对于小型位置数据集来说很好，但是当位置数据集很大时需要很长时间。
private class MapLoader extends AsyncTask<Void, Void, Void> {
ProgressDialog dialog;
ArrayList<POI>[] mappoiList;
@Override
protected void onPreExecute() {
mapOverlays = mapview.getOverlays();
dialog = new ProgressDialog(MainActivity.this);
dialog.show();
}
@Override
protected Void doInBackground(Void... params) {
// TODO Auto-generated method stub
try {
URL url = new URL(MainActivity.baseUrl
+ "latitude=40.9192799&longitude=-74.0657508&distance=20");
SAXParserFactory spf = SAXParserFactory.newInstance();
SAXParser sp = spf.newSAXParser();
XMLReader xr = sp.getXMLReader();
ParserHandler parserHandler = new ParserHandler();
xr.setContentHandler(parserHandler);
xr.parse(new InputSource(url.openStream()));
Log.d("Thread", url.toString());
basicItems = parserHandler.getBasicItems();
featuredItems = parserHandler.getFeaturedItems();
restaurants = parserHandler.getRestaurants();
mapvisibleList = new ArrayList<POI>();
mappoiList = new ArrayList[2];
mappoiList[0] = new ArrayList<POI>();
mappoiList[1] = new ArrayList<POI>();
for (int i = 0; i < featuredItems.size(); i++) {
POI poi = new POI(featuredItems.get(i), 1);
mappoiList[0].add(poi);
}
for (int i = 0; i < restaurants.size(); i++) {
POI poi = new POI(restaurants.get(i), 2);
mappoiList[0].add(poi);
}
for (int i = 0; i < basicItems.size(); i++) {
POI poi = new POI(basicItems.get(i), 0);
mappoiList[1].add(poi);
}
for (int i = 0; i < mappoiList[0].size(); i++) {
if (mappoiList[0] != null) {
mapvisibleList.add(mappoiList[0].get(i));
}
}
for (int i = 0; i < mappoiList[1].size(); i++) {
if (mappoiList[1] != null) {
mapvisibleList.add(mappoiList[1].get(i));
}
}
for (FeaturedItem item : featuredItems) {
Log.d("FEATURED",
item.getName() + "Distance: " + item.getDistance());
}
for (Restaurant item : restaurants) {
Log.d("RESTAURANTS",
item.getName() + "Distance: " + item.getDistance());
}
for (BasicItem item : basicItems) {
Log.d("BASIC",
item.getName() + "Distance: " + item.getDistance());
}
} catch (MalformedURLException e) {
e.printStackTrace();
showErrorDialog("Error", "Malformed URL Error Occurred");
} catch (ParserConfigurationException e) {
e.printStackTrace();
showErrorDialog("Error",
"Parser Configuration Problem Occurred");
} catch (SAXException e) {
e.printStackTrace();
showErrorDialog("Error", "SAX Parser Error Occurred");
} catch (IOException e) {
e.printStackTrace();
showErrorDialog("Error", "IOException Occurred");
} catch (Exception e) {
e.printStackTrace();
}
GlobalData.gBasicItems = basicItems;
GlobalData.gFeaturedItems = featuredItems;
GlobalData.gRestaurants = restaurants;
if (currentLocation != null) {
int curLat = (int) (currentLocation.getLatitude() * 1e6);
int curLon = (int) (currentLocation.getLongitude() * 1e6);
GeoPoint gp = new GeoPoint(curLat, curLon);
mapControl.animateTo(gp);
mapControl.setZoom(14);
OverlayItem item = new OverlayItem(gp, "My Location",
"My Location");
itemizedOverlay1 = new MyItemizedOverlay(drawableFeature,
MainActivity.this, mappoiList[0]);
itemizedOverlay2 = new MyItemizedOverlay(drawableBasic,
MainActivity.this, mappoiList[1]);
itemizedOverlay3 = new MyItemizedOverlay(drawableCurrent,
MainActivity.this, mappoiList[0]);
Log.i("asyncbasic", "" + basicItems.size());
Log.i("asyncfeatured", "" + featuredItems.size());
Log.i("asyncres", "" + restaurants.size());
if (featuredItems != null) {
int featuredLength = featuredItems.size();
for (int i = 0; i < featuredLength; i++) {
FeaturedItem fItem = featuredItems.get(i);
int lat = (int) (Double
.parseDouble(fItem.getLatitude()) * 1e6);
int lon = (int) (Double.parseDouble(fItem
.getLongitude()) * 1e6);
OverlayItem oItem = new OverlayItem(new GeoPoint(lat,
lon), fItem.getName(), "Feature");
itemizedOverlay1.addOverlay(oItem);
Log.i("Map over lay", "Finished one featured");
}
}
if (basicItems != null) {
int basicLength = basicItems.size();
for (int i = 0; i < basicLength; i++) {
BasicItem bItem = basicItems.get(i);
int lat = (int) (Double
.parseDouble(bItem.getLatitude()) * 1e6);
int lon = (int) (Double.parseDouble(bItem
.getLongitude()) * 1e6);
OverlayItem oItem = new OverlayItem(new GeoPoint(lat,
lon), bItem.getName(), "Basic");
itemizedOverlay2.addOverlay(oItem);
Log.i("Map over lay", "Finished one Basic");
}
}
if (restaurants != null) {
int resLength = restaurants.size();
for (int i = 0; i < resLength; i++) {
Restaurant res = restaurants.get(i);
int lat = (int) (Double.parseDouble(res.getLatitude()) * 1e6);
int lon = (int) (Double.parseDouble(res.getLongitude()) * 1e6);
OverlayItem oItem = new OverlayItem(new GeoPoint(lat,
lon), res.getName(), "Restaurant");
itemizedOverlay1.addOverlay(oItem);
Log.i("Map over lay", "Finished one Restaurant");
}
}
itemizedOverlay3.addOverlay(item);
}
return null;
}
@Override
protected void onPostExecute(Void result) {
// TODO Auto-generated method stub
mapOverlays.clear();
mapOverlays.add(itemizedOverlay1);
mapOverlays.add(itemizedOverlay2);
mapOverlays.add(itemizedOverlay3);
mapview.postInvalidate();
dialog.dismiss();
}
}
我了解了如何使用getLatitudeSpan和getLongitudeSpan来解决这个问题，但我不太清楚在异步任务中何处以及如何使用它。有什么建议吗？
答案 0 :(得分：0)
请提供精确的数据大小，其中所用的时间很长，同时尝试分析代码以查看每个代码块的时间。
但是看一下代码，如果运行的数据增加，循环遍历所有这些变量会对时间产生重大影响，
尝试简化/组合循环，
例如这个循环
for (int i = 0; i < featuredItems.size(); i++) {
POI poi = new POI(featuredItems.get(i), 1);
mappoiList[0].add(poi);
}
和这一个
if (featuredItems != null) {
int featuredLength = featuredItems.size();
for (int i = 0; i < featuredLength; i++) {
FeaturedItem fItem = featuredItems.get(i);
int lat = (int) (Double
.parseDouble(fItem.getLatitude()) * 1e6);
int lon = (int) (Double.parseDouble(fItem
.getLongitude()) * 1e6);
OverlayItem oItem = new OverlayItem(new GeoPoint(lat,
lon), fItem.getName(), "Feature");
itemizedOverlay1.addOverlay(oItem);
Log.i("Map over lay", "Finished one featured");
}
}
无法合并？
也是这个循环
for (int i = 0; i < basicItems.size(); i++) {
POI poi = new POI(basicItems.get(i), 0);
mappoiList[1].add(poi);
}
和这个
if (basicItems != null) {
int basicLength = basicItems.size();
for (int i = 0; i < basicLength; i++) {
BasicItem bItem = basicItems.get(i);
int lat = (int) (Double
.parseDouble(bItem.getLatitude()) * 1e6);
int lon = (int) (Double.parseDouble(bItem
.getLongitude()) * 1e6);
OverlayItem oItem = new OverlayItem(new GeoPoint(lat,
lon), bItem.getName(), "Basic");
itemizedOverlay2.addOverlay(oItem);
Log.i("Map over lay", "Finished one Basic");
}
}
也可以合并，
答案 1 :(得分：0)
我认为问题不在于叠加。你正在使用sax解析器来获取所有项目，这将花费你的时间。您尝试获取所有元素，然后将其添加到地图中。如果您修改代码，因为您需要覆盖单个元素更新并填充它，这可能会加快执行速度。 如果您仍然遇到同样的问题，请尝试使用叠加的单独线程或画布对象添加元素（通过投影地理点）。
答案 2 :(得分：0)
您可能希望了解如何实现Producer-Consumer模型，其中使用者将是地图本身，生产者将是为引脚生成信息的逻辑。
答案 3 :(得分：0)
您正在使用非常耗时的 sax解析器，使用像json这样的对象模型，可以很容易地解析，因此可以在地图上绘制。
//很少有建议本身可以减少您面临的时间问题。