1
public class MapDeserialize 
{
2
public static void main(String[] args) {
3
Map<String, String> map = new HashMap<String, String>();
4 map.put("key1", "value1");
5 map.put("key2", null);
6 map.put("key3", "");
7
8 System.out.println(map);
9
10 Map<String, String> emptyMap = new HashMap<String, String>();
11 System.out.println(emptyMap);
12
13 MapDeserialize deserialize = new MapDeserialize();
14 String str1 = "{key3=, key2=null, key1=value1}";
15 String str2 = "{}";
16 Map<String, String> map1 = deserialize.str2Map(str1);
17 System.out.println("map1: " + map1);
18 Map<String, String> map2 = deserialize.str2Map(str2);
19 System.out.println("map2: " + map2);
20
}
21
22 // We are assuming that the str is generated by map.toString(), so the str will be something like:
23 // '{key3=, key2=null, key1=value1}' or '{}'
24 public Map<String, String> str2Map(String str) {
25 Map<String, String> map = new HashMap<String, String>();
26 // The parameters map is empty
27 if("{}".equals(str) || str == null || str.length() == 0) {
28 return map;
29 }
30
31 // Remove the '{' prefix and '}' suffix
32 str = str.substring(1, str.length() - 1);
33 String[] entries = str.split(",");
34 for(String entry : entries) {
35 String[] pair = entry.split("=");
36 String key = pair[0].trim();
37 if(pair.length == 1) {
38 map.put(key, "");
39 } else {
40 String value = pair[1].trim();
41 if("null".equals(value)) {
42 map.put(key, null);
43 } else {
44 map.put(key, value);
45 }
46 }
47 }
48
49 return map;
50 }
51
} 這段代碼貌似沒什么價值，只是保留著，以后再遇到相應的情況，可以再做改進。
posted on 2011-09-20 15:54
DLevin 閱讀(522)
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