如何在Java中将List<com.amazonaws.util.json.JSONObject>转为JSON字符串

Got it, let's walk through how to turn your List<JSONObject> into a JSON string using Jackson's ObjectMapper. I'll cover two common approaches depending on whether you want to add an extra Jackson module or not.

Approach 1: Use Jackson's JSON-Org Module (Simplest) #

Jackson has a dedicated module to handle org.json.JSONObject types directly. First, make sure you have the jackson-datatype-json-org dependency in your project. Then, register the module with your ObjectMapper and you're good to go:

// Register the module to handle org.json types
objectMapper.registerModule(new JsonOrgModule());

// Serialize the list directly to JSON string
String jsonString = objectMapper.writeValueAsString(jsonObjlist);

This is the cleanest method if you don't mind adding the extra dependency.

Approach 2: Convert to List

If you want to avoid adding another module, you can manually convert each JSONObject to a Map<String, Object> before serializing. Here's how:

List<Map<String, Object>> mapList = new ArrayList<>();
for (JSONObject jsonObj : jsonObjlist) {
    Map<String, Object> map = new HashMap<>();
    Iterator<String> keys = jsonObj.keys();
    while (keys.hasNext()) {
        String key = keys.next();
        map.put(key, jsonObj.get(key));
    }
    mapList.add(map);
}

// Now serialize the map list to JSON
String jsonString = objectMapper.writeValueAsString(mapList);

This works because Jackson knows how to serialize Map instances out of the box.

Handling Empty Values #

I noticed your list has an empty string for mx_Age. By default, Jackson will include this in the output. If you want to exclude empty (or null) values, you can configure your ObjectMapper like this:

objectMapper.setSerializationInclusion(JsonInclude.Include.NON_EMPTY);

This will omit any entries where the value is an empty string, null, or an empty collection.

Full Working Example #

Here's a complete code snippet that uses Approach 2 (no extra modules) with your exact input list:

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.annotation.JsonInclude;
import org.json.JSONObject;
import java.util.*;

public class JsonConversionExample {
    public static void main(String[] args) throws Exception {
        // Your input list
        List<JSONObject> jsonObjlist = new ArrayList<>();
        jsonObjlist.add(new JSONObject("{\"Attribute\":\"EmailAddress\",\"Value\":\"abc@yahoo.com\"}"));
        jsonObjlist.add(new JSONObject("{\"Attribute\":\"Source\",\"Value\":\"Missing_Fixed\"}"));
        jsonObjlist.add(new JSONObject("{\"Attribute\":\"mx_Lead_Status\",\"Value\":\"Registered User\"}"));
        jsonObjlist.add(new JSONObject("{\"Attribute\":\"mx_Age\",\"Value\":\"\"}"));
        jsonObjlist.add(new JSONObject("{\"Attribute\":\"mx_LoginID\",\"Value\":\"abc@yahoo.com\"}"));
        jsonObjlist.add(new JSONObject("{\"Attribute\":\"mx_Registration_Source\",\"Value\":\"EMAIL\"}"));

        // Initialize ObjectMapper
        ObjectMapper objectMapper = new ObjectMapper();
        // Optional: Uncomment to exclude empty values
        // objectMapper.setSerializationInclusion(JsonInclude.Include.NON_EMPTY);

        // Convert List<JSONObject> to List<Map>
        List<Map<String, Object>> mapList = new ArrayList<>();
        for (JSONObject jsonObj : jsonObjlist) {
            Map<String, Object> map = new HashMap<>();
            Iterator<String> keys = jsonObj.keys();
            while (keys.hasNext()) {
                String key = keys.next();
                map.put(key, jsonObj.get(key));
            }
            mapList.add(map);
        }

        // Generate JSON string
        String jsonString = objectMapper.writeValueAsString(mapList);
        System.out.println(jsonString);
    }
}

Running this will output the JSON string matching your input list. If you enable the NON_EMPTY inclusion, the mx_Age entry will be excluded from the result.

内容的提问来源于stack exchange，提问作者Vinay Sawant