在java中对属性进行排序

Question

有没有办法在java中对Properties对象进行排序？

我有字符串对属性进行分组，并检查数据是否以地图格式可用。

Answer 1

请使用此示例：

来自链接：http://www.java2s.com/Tutorial/Java/0140__Collections/SortPropertieswhensaving.htm

import java.io.FileOutputStream;
import java.util.Collections;
import java.util.Enumeration;
import java.util.Properties;
import java.util.Vector;

public class Main{
  public static void main(String[] args) throws Exception {
    SortedProperties sp = new SortedProperties();
    sp.put("B", "value B");
    sp.put("C", "value C");
    sp.put("A", "value A");
    sp.put("D", "value D");
    FileOutputStream fos = new FileOutputStream("sp.props");
    sp.store(fos, "sorted props");
  }

}
class SortedProperties extends Properties {
  public Enumeration keys() {
     Enumeration keysEnum = super.keys();
     Vector<String> keyList = new Vector<String>();
     while(keysEnum.hasMoreElements()){
       keyList.add((String)keysEnum.nextElement());
     }
     Collections.sort(keyList);
     return keyList.elements();
  }

}

Answer 2

您可以找到一个没有对属性进行子类化并使用Java 8/9/10的示例

通过这种方式，来自Properties的keySet，keys和entrySet方法返回排序键。然后方法存储也保存已排序的属性文件。

这段代码是here

Properties properties = new Properties() {

private static final long serialVersionUID = 1L;

@Override
public Set<Object> keySet() {
    return Collections.unmodifiableSet(new TreeSet<Object>(super.keySet()));
}

@Override
public Set<Map.Entry<Object, Object>> entrySet() {

    Set<Map.Entry<Object, Object>> set1 = super.entrySet();
    Set<Map.Entry<Object, Object>> set2 = new LinkedHashSet<Map.Entry<Object, Object>>(set1.size());

    Iterator<Map.Entry<Object, Object>> iterator = set1.stream().sorted(new Comparator<Map.Entry<Object, Object>>() {

        @Override
        public int compare(java.util.Map.Entry<Object, Object> o1, java.util.Map.Entry<Object, Object> o2) {
            return o1.getKey().toString().compareTo(o2.getKey().toString());
        }
    }).iterator();

    while (iterator.hasNext())
        set2.add(iterator.next());

    return set2;
}

@Override
public synchronized Enumeration<Object> keys() {
    return Collections.enumeration(new TreeSet<Object>(super.keySet()));
    }
};

Answer 3

考虑到TreeMap是一个有序的Map，那么你可以这样做：

//properties as they are:
System.out.println(System.getProperties());

//now sorted:
TreeMap<Object, Object> sorted = new TreeMap<>(System.getProperties());
System.out.println(sorted);