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如何使用Java更新JSON文件中的嵌套JSON对象?

io arraylist json arrays java

btrballin · 技术社区 · 6 年前

我有一个包含用户对象数组的JSON文件。每个用户对象包含一个包含对象的目标数组,在每个目标对象中是一个包含1到多个里程碑对象的里程碑数组。

我能够利用JSON Simple轻松地解析它们,并以Java对象的形式存储信息,但我发现在JSON文件中重新写入某些块或将其附加到其中一个数组中很有挑战性。我在网上找到的例子过于简单,没有考虑到这么多的嵌套。

如果我想重写JSON文件中的一个目标对象或里程碑数组中的一个特定里程碑,我将如何根据相应Java对象模型中指定的属性来更新文件?是否可以在其中一个数组中修改/附加JSON文本文件中的特定块?

[
{       
    "username": "example",
    "password": "pwd",  
    "Goals":[
        {   
            "goalId": "G0001",
            "goalName": "COEN 275 Midterm",
            "startDate": "02/14/2017",
            "endDate": "02/28/2017",
            "numMilestones": 5,
            "color": "#4286f4",
            "hoursPerDay": 4,
            "milestones":[
                {   
                    "milestoneName": "Chapter 1",
                    "Difficulty": "E",
                    "dueDate": "02/18/2018",
                    "completed": true
                },
                {   "milestoneName": "Chapter 2",
                    "Difficulty": "E",
                    "dueDate": "02/22/2018",
                    "completed": true
                },
                {   "milestoneName": "Chapter 3",
                    "Difficulty": "M",
                    "dueDate": "02/23/2018",
                    "completed": true
                },
                {   "milestoneName": "Chapter 4",
                    "Difficulty": "M",
                    "dueDate": "02/25/2018",
                    "completed": false
                },
                {   "milestoneName": "Chapter 5",   
                    "Difficulty": "H",
                    "dueDate": "02/28/2018",
                    "completed": false
                }
            ]
        }
    ]
},
{       
    "username": "username",
    "password": "pwd",  
    "Goals":[
        {   
            "goalId": "G0001",
            "goalName": "COEN 279 Midterm",
            "startDate": "02/14/2017",
            "endDate": "02/28/2017",
            "numMilestones": 5,
            "color": "#4286f4",
            "hoursPerDay": 4,
            "milestones":[
                {   
                    "milestoneName": "Chapter 5",
                    "Difficulty": "E",
                    "dueDate": "02/18/2018",
                    "completed": true
                },
                {   "milestoneName": "Chapter 6",
                    "Difficulty": "E",
                    "dueDate": "02/22/2018",
                    "completed": true
                },
                {   "milestoneName": "Chapter 7",
                    "Difficulty": "M",
                    "dueDate": "02/23/2018",
                    "completed": true
                },
                {   "milestoneName": "Chapter 11",
                    "Difficulty": "M",
                    "dueDate": "02/25/2018",
                    "completed": false
                },
                {   "milestoneName": "Chapter 12",  
                    "Difficulty": "H",
                    "dueDate": "02/28/2018",
                    "completed": false
                }
            ]
        }
    ]
}
]

1 回复 | 直到 6 年前

Jose Da Silva Gomes Tarsbir Singh 6 年前

JSON simple中的数组是JSONArray类型的类(这只是一个java util ArrayList),因此一旦获得数组和对象(这是一个java util HashMap),就可以修改它。

JSONArray users = (JSONArray) JSONValue.parse(json);
JSONObject firstUser = (JSONObject) users.get(0);

JSONArray firstUserGoals = (JSONArray) firstUser.get("Goals");
JSONObject firstUserFirstGoal = (JSONObject) firstUserGoals.get(0);

JSONArray firstUserFirstGoalMilestones = (JSONArray) firstUserFirstGoal.get("milestones");
JSONObject firstUserFirstGoalFirstMilestone = (JSONObject) firstUserFirstGoalMilestones.get(0);

因此,您可以编辑里程碑

firstUserFirstGoalFirstMilestone.put("milestoneName", "Chapter 0");

您可以创建新的里程碑

JSONObject milestone1 = new JSONObject();
milestone1.put("milestoneName", "Chapter 1");
milestone1.put("Difficulty", "H");
milestone1.put("dueDate", "02/18/2018");
milestone1.put("completed", true);

然后将其添加到位置1的里程碑列表中。如果没有指定索引,则最后添加。

firstUserFirstGoalMilestones.add(1, milestone1);

最后,您可以将JSON作为字符串获取。

String finalJson = users.toJSONString();

或写入文件(应处理异常)

Writer writer = new FileWriter(path);
users.writeJSONString(writer);
writer.close();