皮皮虾的倔强
于 2022-05-23 13:25:09 发布
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文章标签: java json 开发语言
版权
一、准备工作
定义三个json字符串用于以下测试:
//json字符串-简单对象
String jsonStr = "{\"studentName\":\"张三\",\"studentAge\":18}";
//json字符串-数组类型
String jsonArrStr = "[{\"studentName\":\"张三\",\"studentAge\":18},{\"studentName\":\"李四\",\"studentAge\":17}]";
//json字符串-复杂对象
String complexJsonStr= "{\"teacherName\":\"李寻欢\",\"teacherAge\":30,\"course\":{\"courseName\":\"武术\",\"code\":110},
\"students\":[{\"studentName\":\"张三\",\"studentAge\":18},{\"studentName\":\"李四\",\"studentAge\":19}]}";
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二、json字符串、json对象、java对象的转换方法
1.JSON字符串到JSON对象的转换
(1)json字符串-简单对象与JSONObject之间的转换
JSONObject jsonObj = JSON.parseObject(jsonStr);
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(2)json字符串-数组类型与JSONArray之间的转换
JSONArray jsonArray = JSON.parseArray(jsonArrStr);
//遍历JSONArray方法1
for(int i = 0; i < jsonArray.size(); i++){
JSONObject jsonObj = jsonArray.getJSONObject(i);
}
//遍历JSONArray方法2
for(Object obj : jsonArray){
JSONObject jsonObject = (JSONObject) obj;
}
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(3)json字符串-复杂对象与JSONObject之间的转换
JSONObject jsonObj = JSON.parseObject(complexJsonStr);
//取出复杂对象中各项内容
String teacherName = jsonObj.getString("teacherName");
Integer teacherAge = jsonObj.getInteger("teacherAge");
JSONObject course = jsonObj.getJSONObject("course");
JSONArray students = jsonObj.getJSONArray("students");
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2.JSON对象到JSON字符串的转换
JSONObject jsonObj = new JSONObject();
//JSONObject到JSON字符串的转换
String jsonStr = jsonObj.toJSONString();
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3.JSON字符串到Java对象的转换
JSON字符串与JavaBean之间的转换建议使用TypeReference类
(1)json字符串-简单对象与Java对象之间的转换
// 方法1
Student student = JSON.parseObject(jsonStr , new TypeReference() {});
// 方法2
Student student = JSON.parseObject(jsonStr , Student.class);
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(2)json字符串-数组与Java对象之间的转换
ArrayListstudents = JSON.parseObject(jsonArrStr, new TypeReference() {});
for (Student student : students) {
System.out.println(student.getStudentName()+":"+student.getStudentAge());
}
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(3)json字符串-复杂对象与Java对象之间的转换
Teacher teacher = JSON.parseObject(complexJsonStr, new TypeReference() {});
//获取teacher中的内容
String teacherName = teacher.getTeacherName();
Integer teacherAge = teacher.getTeacherAge();
Course course = teacher.getCourse();
Liststudents = teacher.getStudents();
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4.Java对象到JSON字符串的转换
Teacher teacher = new Teacher();
String jsonStr = JSON.toJSONString(teacher);
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5.Java对象到JSON对象的转换
String jsonStr= JSON.toJSONString(student);
JSONObject jsonObj = JSON.parseObject(jsonStr);
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6.JSON对象到Java对象的转换
# 方法1,先转换为json字符串,再使用parseObject
String jsonStr = jsonObj.toJSONString();
Student stu = JSON.parseObject(jsonStr,new TypeReference() {});
# 方法2,直接使用toJavaObject
Student stu = JSON.toJavaObject(jsonObj, Student.class);
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例子
1.json字符串转化为java实体类 (parseObject)
ApprovalVo approvalVo = JSON.parseObject(str, ApprovalVo.class);
// str == json字符串
// ApprovalVo == 实体类
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2.json字符串转化为list对象 (parseArray)
String str2 = "[{'password':'123123','username':'zhangsan'},{'password':'321321','username':'lisi'}]";
Listusers = JSON.parseArray(jsonStr2, User.class);
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3.json字符串转化为复杂java对象 (parseObject)
// 复杂对象->>>>对象中嵌套对象的
String str3 = "{'name':'userGroup','users':[{'password':'123123','username':'zhangsan'},{'password':'321321','username':'lisi'}]}";
UserGroup userGroup = JSON.parseObject(jsonStr3, UserGroup.class);
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把实体类转化成json字符串
String str = JSON.toJSONString(ApprovalVo);
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把json字符串转化成JSONObject
String jsonStr = "{\"school\":\"商职\",\"sex\":\"男\",\"name\":\"wjw\",\"age\":22}";
JSONObject jsonObject = JSONObject.parseObject(jsonStr);
System.out.println(jsonObject.getString("name"));
System.out.println(jsonObject.getInteger("age"));
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参考:
https://zhuanlan.zhihu.com/p/476747417
https://blog.51cto.com/u_15281317/3008066
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版权声明:本文为CSDN博主「皮皮虾的倔强」的原创文章,遵循CC 4.0 BY-SA版权协议,转载请附上原文出处链接及本声明。
原文链接:https://blog.csdn.net/Wolves_howl/article/details/124921713
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