如何使用ajax从servlet接收json对象？

Question

doGet设置项readState新值并将其作为Json对象发送回ajax调用：

PrintWriter out = null;
            try {
                out = resp.getWriter();

                Integer read = Integer.parseInt(req.getParameter("feedReadVal"));

                Gson data = new Gson();
                JsonObject jsonObj = new JsonObject();
                Integer readState = null;
                Item item = new Item();

                readState = 1;
                item.setReadState(readState.byteValue());
                JsonElement element = data.toJsonTree(item);

                jsonObj.add("read", element);
                out.print(jsonObj);
            } catch (Exception e) {
                e.printStackTrace();
                }

通话：       System.out.println("JSON OBJECT: " + jsonObj.toString());显示类似JSON OBJECT: {"read":{"readState":1}}的内容，但当我尝试获取成功函数错误时抛出。我的jQuery：

var data = "feedReadVal=" + itemRead + "&feedItemId=" + feedItemId
        + "&action=" + 'read';
$.ajax({
    type : "Get",
    url : "FeedItemController",
    data : data,
    dataType : "json",

    success : function(data, textStatus, jqXHR) {
        alert(data.read.readState);
    },
    error : function(jqXHR, textStatus, errorThrown) {
        alert("Something really bad happened ");
    }

}).done(function() {

}); 

我一直收到错误提醒。

Answer 1

您需要将其写入流

，而不是打印

将out.print(jsonObj);更改为

out.write(jsonObj.toString());
out.flush();