以下JSON有一些父子关系(示例数据(
- ">
- id":1050 是子级,因为它有 "parentId":1051 ">
- id":1051 是父级,因为它有 "parentId":空
像这样我有巨大的数据
{
"statusCode": 200,
"statusMessage": "Success",
"dataCount": 0,
"data": null,
"dataList": [
{
"id": 1050,
"parentId": 1051,
"questionInfo": {
"id": 1050,
"description": "--",
"question": "--",
"answersInfo": [
{
"id": 2041,
"description": "--",
"isCorrect": "--",
"name": "--",
"subQuestions": []
},
{
"id": 2040,
"description": "--",
"isCorrect": "--",
"name": "--",
"subQuestions": []
}
]
}
},
{
"id": 1051,
"parentId": null,
"questionInfo": {
"id": 1051,
"description": "--",
"question": "--",
"answersInfo": [
{
"id": 2024,
"description": "--",
"isCorrect": "--",
"name": "--",
"subQuestions": []
},
{
"id": 2023,
"description": "--",
"isCorrect": "--",
"name": "--",
"subQuestions": [
{
"id": 1050,
"parentId": 1051,
"questionInfo": {
"id": 1050,
"description": "--",
"question": "--",
"answersInfo": [
{
"id": 2041,
"description": "--",
"isCorrect": "--",
"name": "--",
"subQuestions": []
},
{
"id": 2040,
"description": "--",
"isCorrect": "--",
"name": "--",
"subQuestions": []
}
]
}
}
]
}
]
}
},
{
"-------": "-------"
}
],
"dataMap": null
}
我的要求是需要从parentId具有某些值("parentId":1051(的父位置中删除子记录,因为这些子信息来自父子信息,这些记录不是父记录
{
"statusCode": 200,
"statusMessage": "Success",
"dataCount": 0,
"data": null,
"dataList": [
{
"id": 1051,
"parentId": null,
"questionInfo": {
"id": 1051,
"description": "--",
"question": "--",
"answersInfo": [
{
"id": 2024,
"description": "--",
"isCorrect": "--",
"name": "--",
"subQuestions": []
},
{
"id": 2023,
"description": "--",
"isCorrect": "--",
"name": "--",
"subQuestions": [
{
"id": 1050,
"parentId": 1051,
"questionInfo": {
"id": 1050,
"description": "--",
"question": "--",
"answersInfo": [
{
"id": 2041,
"description": "--",
"isCorrect": "--",
"name": "--",
"subQuestions": []
},
{
"id": 2040,
"description": "--",
"isCorrect": "--",
"name": "--",
"subQuestions": []
}
]
}
}
]
}
]
}
},
{
"-------": "-------"
}
],
"dataMap": null
}
我正在使用Spring MVC + Hibernate标准API + Jackson库那么,任何人都可以告诉我在序列化时从列表中排除对象(基于某些条件的完整对象(的最佳方法?
正如我们与@Developer讨论的那样,回答这个问题的最佳方法是使用某种工具序列化所有对象,就像从Jackson
ObjectMapper
一样,一旦构建了 List,迭代它以仅包含您想要保留的元素。
性能方面,这不是最好的事情,但这将确保您不会错过任何想要的物品