IndexOutOfBoundsException异常的名单<Address>

Question

在我的Android应用程序，我有这样的代码：

LatLng[] branches; 
String[] branchesArray = HomeActivity.branches.toArray(new String[HomeActivity.branches.size()]); 

for (int i = 0; i < HomeActivity.branches.size(); i++) { 
    branches[i] = getLocationFromAddress(branchesArray[i]); 
} 

getLocationFromAddress方法：

public LatLng getLocationFromAddress(String strAddress) { 

    Geocoder coder = new Geocoder(this); 
    List<Address> address; 
    LatLng p1 = null; 

    try { 
     address = coder.getFromLocationName(strAddress, 1); 

     if (address == null) { 
      return null; 
     } 

     Address location = address.get(0); 
     location.getLatitude(); 
     location.getLongitude(); 

     p1 = new LatLng((double) (location.getLatitude()), (double) (location.getLongitude())); 
    } catch (IOException e) { 
     Log.e("Error", e.getMessage()); 
    } 

    return p1; 
} 

此代码应该创建的LatLng数组，从字符串地址数组提取。问题在于，无论何时我运行此代码，我都会在日志中获得java.lang.IndexOutOfBoundsException: Invalid index 0, size is 0。它是指行137作为问题行，这是这条线：

Address location = address.get(0); 

我该如何解决这个问题？

Answer 1

getLocationFromName文件说：

返回地址对象的列表。如果找不到 匹配项或没有可用的后端服务，则返回空列表或空列表。

你的情况，它返回一个空列表，所以你应该添加一个额外的检查：

public LatLng getLocationFromAddress(String strAddress) { 

    Geocoder coder = new Geocoder(this); 
    List<Address> address; 
    LatLng p1 = null; 

    try { 
     address = coder.getFromLocationName(strAddress, 1); 

     if (address == null || address.isEmpty()) { 
      return null; 
     } 

     Address location = address.get(0); 

     p1 = new LatLng(location.getLatitude(), location.getLongitude()); 
    } catch (IOException e) { 
     Log.e("Error", e.getMessage()); 
    } 

    return p1; 
} 

Answer 2

的probleme是您忘了初始化“分支”变量与正确的尺寸，这就是为什么你越来越“大小0索引0”

String[] branches = HomeActivity.branches.toArray(new String[HomeActivity.branches.size()]);