您可以使用Jackson将Map转换为JSON，如下所示:

Map<String,Object> map = new HashMap<>();
//You can convert any Object.
String[] value1 = new String[] { "value11", "value12", "value13" };
String[] value2 = new String[] { "value21", "value22", "value23" };
map.put("key1", value1);
map.put("key2", value2);
map.put("key3","string1");
map.put("key4","string2");

String json = new ObjectMapper().writeValueAsString(map);
System.out.println(json);

Jackson的Maven依赖:

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-core</artifactId>
    <version>2.5.3</version>
    <scope>compile</scope>
</dependency>

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
    <version>2.5.3</version>
    <scope>compile</scope>
</dependency>

If you are using `JSONObject` library, you can convert map to `JSON` as follows: JSONObject Library: import org.json.JSONObject; Map<String, Object> map = new HashMap<>(); // Convert a map having list of values. String[] value1 = new String[] { "value11", "value12", "value13" }; String[] value2 = new String[] { "value21", "value22", "value23" }; map.put("key1", value1); map.put("key2", value2); JSONObject json = new JSONObject(map); System.out.println(json); Maven Dependencies for `JSONObject` : <dependency> <groupId>org.json</groupId> <artifactId>json</artifactId> <version>20140107</version> </dependency> Hope this will help. Happy coding.

首先将所有对象转换为有效的string

HashMap<String, String> params = new HashMap<>();
params.put("arg1", "<b>some text</b>");
params.put("arg2", someObject.toString());

然后将整个映射插入到org.json.JSONObject中

JSONObject postData = new JSONObject(params);

现在您可以通过调用对象的toString来获取JSON

postData.toString()
//{"arg1":"<b>some text<\/b>" "arg2":"object output"}

创建一个新的JSONObject

JSONObject o = new JSONObject(postData.toString());

或者作为通过HTTP发送的字节数组

postData.toString().getBytes("UTF-8");

我使用阿里巴巴fastjson，简单明了:

<dependency>
    <groupId>com.alibaba</groupId>
    <artifactId>fastjson</artifactId>
    <version>VERSION_CODE</version>
</dependency>

和导入:

import com.alibaba.fastjson.JSON;

然后:

String text = JSON.toJSONString(obj); // serialize
VO vo = JSON.parseObject("{...}", VO.class); //unserialize

一切都好。

如果使用复杂对象，应该应用enableComplexMapKeySerialization()，如https://stackoverflow.com/a/24635655/2914140和https://stackoverflow.com/a/26374888/2914140中所述。

Gson gson = new GsonBuilder().enableComplexMapKeySerialization().create();
Map<Point, String> original = new LinkedHashMap<Point, String>();
original.put(new Point(5, 6), "a");
original.put(new Point(8, 8), "b");
System.out.println(gson.toJson(original));

输出将是:

{
 "(5,6)": "a",
 "(8,8)": "b"
}

我找到了另一种处理方法。

Map obj=new HashMap();    
obj.put("name","sonoo");    
obj.put("age",new Integer(27));    
obj.put("salary",new Double(600000));   
String jsonText = JSONValue.toJSONString(obj);  
System.out.print(jsonText);

希望这能有所帮助。

谢谢。

这个解决方案适用于复杂的json:

public Object toJSON(Object object) throws JSONException {
    if (object instanceof HashMap) {
        JSONObject json = new JSONObject();
        HashMap map = (HashMap) object;
        for (Object key : map.keySet()) {
            json.put(key.toString(), toJSON(map.get(key)));
        }
        return json;
    } else if (object instanceof Iterable) {
        JSONArray json = new JSONArray();
        for (Object value : ((Iterable) object)) {
            json.put(toJSON(value));
        }
        return json;
    }
    else {
        return object;
    }
}