Java/Android 从 xml 获取数组
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/4691706/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Java/Android get array from xml
提问by Ashley
I have a list of longitude and longitude points in an xml file that is used throughout my application. I find my self repeating this code to get points often and think there must be a better way?
我在整个应用程序中使用的 xml 文件中有一个经度和经度点列表。我发现自己经常重复此代码以获取积分并认为必须有更好的方法？
String[] mTempArray = getResources().getStringArray(R.array.stations);
int len = mTempArray.length;
mStationArray = new ArrayList<Station>();
for(int i = 0; i < len; i++){
Station s = new Station();
String[] fields = mTempArray[i].split("[\t ]");
s.setValuesFromArray(fields);
Log.i("ADD STATION", ""+s);
mStationArray.add(s);
}
XML is in the format of:
XML 的格式为：
<?xml version="1.0" encoding="utf-8"?>
<resources>
<array name="stations">
<item>
<name>Station name</name>
<longitude>1111111</longitude>
<latitude>11111</latitude>
<code>1</code>
</item>
And another (possible) problem is that to get just one station I have to get all of them and pull the one I want from the array. Is this going to be considerably slower? Can I make this array consistent throughout the app? (But keeping the separate Intent methodology)
另一个（可能的）问题是，要获得一个站，我必须获得所有站并从阵列中拉出我想要的站。这会慢很多吗？我可以使这个数组在整个应用程序中保持一致吗？（但保持单独的意图方法）
采纳答案by Dan Breslau
I had the same thought as MilkJug, to use a utility method to create the stations, but I want to offer a slightly different approach: Move as much of the construction logic as possible into the Stationclass constructor. To keep the example simple, I'm moving the utility method into the Stationclass as well.
我和MilkJug有同样的想法，使用实用方法来创建站，但我想提供一种稍微不同的方法：将尽可能多的构造逻辑移到Station类构造函数中。为了使示例保持简单，我还将实用程序方法移到Station类中。
This provides an overall cleaner design, as outside of the Station class itself, your code should never have to deal with a Station object whose construction/initialization steps haven't been fully completed.
这提供了一个整体更简洁的设计，因为在 Station 类本身之外，您的代码永远不必处理其构造/初始化步骤尚未完全完成的 Station 对象。
(kgiannakakis'ssuggestion to use a database may be a better way to go if you have a lot of Station objects.)
（如果您有很多 Station 对象，kgiannakakis建议使用数据库可能是更好的方法。）
public class Station {
private static List<Station> sStationArray = null;
/**
* Construct a Station from a specially-encoded String. The String
* must have all the necessary values for the Station, separated by tabs.
*/
public Station(String fieldString) {
String[] fields = fieldString.split("[\t ]");
// For safety, setValuesFromArray() should be declared 'final'.
// Better yet, you could just move its body into this constructor.
setValuesFromArray(fields);
// I'm assuming 'mName' is the name field for the Station
Log.i("Station", this.mName);
}
public static Station getStationArray(Context ctx) {
if (sStationArray == null) {
// (Please don't use the prefix 'm' for non-member variables!)
final String[] tempArray =
ctx.getResources().getStringArray(R.array.stations);
final int len = tempArray.length;
// Passing the length into the ArrayList constructor (if it's
// known, or can be guessed at) can be a very simple yet
// effective optimization. In this case the performance boost
// will almost certainly **not** be meaningful, but it's
// helpful to be aware of it.
sStationArray = new ArrayList<Station>(len);
for (int i = 0; i < len; i++) {
Station s = new Station(tempArray[i]);
sStationArray.add(s);
}
}
return sStationArray;
}
}
回答by MilkJug
Why not create a utility method that takes a context as a parameter and returns the station resources? For example:
为什么不创建一个以上下文为参数并返回站点资源的实用方法呢？例如：
public class StatUtil {
private static List<Station> mStationArray = null;
public static Station getStation(Context ctx) {
if (mStationArray == null) {
String[] mTempArray = getResources().getStringArray(R.array.stations);
int len = mTempArray.length;
mStationArray = new ArrayList<Station>();
for(int i = 0; i < len; i++){
Station s = new Station();
String[] fields = mTempArray[i].split("[\t ]");
s.setValuesFromArray(fields);
Log.i("ADD STATION", ""+s);
mStationArray.add(s);
}
}
return mStationArray;
}
}
and call it from your code with:
并从您的代码中调用它：
stationArray = StatUtil.getStation(this);
Repeatedly fetching the stations will be slower than caching them, but not significantly slower unless you are fetching them in a loop. Doing as above will prevent multiple copies from being fetched.
重复获取站点将比缓存它们慢，但除非您在循环中获取它们，否则不会明显更慢。执行上述操作将防止获取多个副本。
回答by kgiannakakis
I could propose two solutions:
我可以提出两种解决方案：
- You could create a Singleton class that initializes once, reads the data from the XML and stores the stations in a List or a Map. Use a Map if you want to quickly find a station based on its name. The Singleton class will provide methods for retrieving all stations or just one of them.
- Create a database table and store the information there. You may need more code, but the advantage will be that you will be able to run more advanced queries.
- 您可以创建一个初始化一次的 Singleton 类，从 XML 读取数据并将站点存储在列表或地图中。如果您想根据名称快速找到车站，请使用地图。Singleton 类将提供检索所有电台或仅其中一个电台的方法。
- 创建一个数据库表并将信息存储在那里。您可能需要更多代码，但好处是您将能够运行更高级的查询。
