Properties類是HashTable的子類,HashTable又實現了Map.

你像往常一樣加載它：

Properties properties = new Properties();

try {

properties.load(new FileInputStream("filename.properties"));

} catch (IOException e) {

}

編輯：好的,所以你想將它轉換為Map< String,Datasource> 😉

//First convert properties to Map

Map m = Maps.fromProperties(properties);

//Sort them so that password < url < user for each datasource and dataSource1.* < dataSource2.*. In your case default string ordering is ok so we can take a normal treemap

Map sorted = Maps.newTreeMap();

sorted.putAll(m);

//Create Multimap> mapping datasourcename->[password,url, user ]

Function, String> propToList = new Function() {

@Override

public String apply(Map.Entry entry) {

return entry.getKey().split("\\.")[0];

}

};

Multimap nameToParamMap = Multimaps.index(m.entrySet(), propToList);

//Convert it to map

Map> mm = nameToParamMap.asMap();

//Transform it to Map

Map mSD = Maps.transformEntries(mm, new EntryTransformer, DataSource>() {

public DataSource transformEntry(String key, Collection value) {

// Create your datasource. You know by now that Collection is actually a list so you can assume elements are in order: [password, url, user]

return new Datasource(.....)

}

};

//Copy transformed map so it's no longer a view

Map finalMap = Maps.newHashMap(mSD);

可能有一種更簡單的方法,但這應該工作:)

還是你用json或xml更好.您還可以從不同的文件加載不同數據源的屬性.

edit2：少用番石榴,更多java：

//Sort them so that password < url < user for each datasource and dataSource1.* < dataSource2.*. In your case default string ordering is ok so we can take a normal SortedSet

SortedSet sorted = new SortedSet();

sorted.putAll(m.keySet);

//Divide keys into lists of 3

Iterable> keyLists = Iterables.partition(sorted.keySet(), 3);

Map m = new HashMap();

for (keyList : keyLists) {

//Contains datasourcex.password, datasroucex.url, datasourcex.user

String[] params = keyList.toArray(new String[keyList.size()]);

String password = properties.get(params[0]);

String url = properties.get(params[1]);

String user = properties.get(params[2]);

m.put(params[0].split("\\.")[0], new DataSource(....)

}

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