Question

在我的项目中，我使用JSON将readed数据从数据库传递到客户端。在此代码中，我将JSONObject放入JSONArray并传递给客户端。

这是JsonProcessing.java servlet的代码：

JSONObject obj = new JSONObject();
JSONArray objarr = new JSONArray();
//read  from DB
sql = "SELECT id, name FROM test";
ResultSet rs = stmt.executeQuery(sql);
while (rs.next()) {
    // Retrieve by column name
    int id = rs.getInt("id");
    String first = rs.getString("name");
    obj.put("name", first);
    obj.put("id", id);
    objarr.add(obj);
    System.out.print("\nobj: "+obj);
}
response.setContentType("application/json");
response.setCharacterEncoding("UTF-8");

PrintWriter out = response.getWriter();
out.print(objarr);
out.flush();

这是我的javascript代码：

$.ajax({

    url : "/test/JsonProcessing",
    type : "get",
    data : {
        id : 30
    },
    success : function(info) {

        //alert(info.name);
        console.log(info);
        for (var i = 0; i < info.length; i++) {
            alert(i + ":  " + info[i].name + "  " + info[i].id);
        }

        // data: return data from server
    },
    error : function() {
        alert('Faild');
        // if fails
    }

});

我不知道为什么我会在警报中重复相同的结果？

Answer 1

更新您的java代码，如下所示：

JSONArray objarr = new JSONArray();
 //read  from DB
sql = "SELECT id, name FROM test";
ResultSet rs = stmt.executeQuery(sql);
while (rs.next()) {

        JSONObject obj = new JSONObject(); // updated

        // Retrieve by column name
        int id = rs.getInt("id");
        String first = rs.getString("name");

        // Display values
        System.out.print("\nID: " + id);
        System.out.print(", Name: " + first);

        obj.put("name", first);
        obj.put("id", id);
        objarr.add(obj);
        System.out.print("\nobj: "+obj);
    }
    response.setContentType("application/json");
response.setCharacterEncoding("UTF-8");

PrintWriter out = response.getWriter();
out.print(objarr);
out.flush();

JSONArray重复相同的结果

1 个答案: