如何将正确的 JSON 从服务器端 java 返回到 DataTables？

Question

How do I concatenate two arrays and then return the result as JSON to DataTables?

I am converting a working DataTables to include server-side processing and am stuck on returning the correct array from the java server-side. Currently the array returned is:

List<YthMmbrSectDtls> ymList;

Example data is:

[{"youthMemberID":"MTQ5","surname":"Tendon","firstName":"Achilles"}]

With the DataTables server-side I need to include extra information at the start of the returned JSON such as:

{"draw":9, "recordsTotal:57", "recordsFiltered:57"[...]}

So that I return:

{"draw":9, "recordsTotal:57", "recordsFiltered:57","data":[{"youthMemberID":"MTQ5","surname":"Tendon","firstName":"Achilles"}]}

At least that is my understanding from reading the manuals and watching the videos.

My current code is:

List<YthMmbrSectDtls> ymList;
String[] dtInfo = {"draw", "recordsTotal", "recordsFiltered"};

ymList = MySQLConnection.getYouthMmbrAllDtls(archived);
        
//Test to be replaced by database call, as per above
dtInfo[0] = "9";
dtInfo[1] = "57";
dtInfo[2] = "57";

if (ymList == null || ymList.isEmpty()) {
    response.sendError(HttpServletResponse.SC_BAD_REQUEST, "No members.");
} else {
    System.out.println("ymList: " + ymList);
    String json = new Gson().toJson(ymList);
    response.setContentType("application/json");
    response.setCharacterEncoding("UTF-8");
    response.getWriter().write(json);
}

Answer 1

There are multiple different ways this can be solved:

Modify JsonElement tree

You can first use Gson.toJsonTree(Object) to convert ymList to an in-memory JsonArray and then wrap it inside a JsonObject to which you add your dtInfo properties:

JsonObject dtInfoJsonObj = new JsonObject();
dtInfoJsonObj.addProperty​("draw", dtInfo[0]);
dtInfoJsonObj.addProperty​("recordsTotal", dtInfo[1]);
dtInfoJsonObj.addProperty​("recordsFiltered", dtInfo[2]);

Gson gson = new Gson();
JsonArray ymJsonArray = gson.toJsonTree(ymList).getAsJsonArray();
dtInfoJsonObj.add("data", ymJsonArray);

String json = gson.toJson(dtInfoJsonObj);

Wrap data in separate class

Alternatively you can create a separate class, eg DtInfo , which contains the properties and the data and then have Gson serialize that. This approach is likely more efficient since there is no intermediate step creating a JsonElement tree.

class DtInfo {
    // Gson will use the field names during serialization; you can also customize them
    // using `@SerializedName`
    // Additionally you can make the fields private if you want
    public final int draw;
    public final int recordsTotal;
    public final int recordsFiltered;
    public final List<YthMmbrSectDtls> data;

    public DtInfo(int draw, int recordsTotal, int recordsFiltered, List<YthMmbrSectDtls> data) {
        this.draw = draw;
        this.recordsTotal = recordsTotal;
        this.recordsFiltered = recordsFiltered;
        this.data = data;
    }
}

And then create the JSON like this:

DtInfo dtInfoObj = new DtInfo(dtInfo[0], dtInfo[1], dtInfo[2], ymList);
String json = new Gson().toJson(dtInfoObj);

---

Also note that you can probably increase performance by storing the Gson instance in a static final field. Gson is thread-safe (see class documentation ) and this way it will cache the type adapters it uses internally for converting the objects to JSON.

Additionally Gson provides toJson overloads which accept an Appendable . You could therefore pass the response.getWriter() to these Gson methods which avoids creating the intermediate String representation of the JSON.