Java User 对象 JSON 解析错误：JSONArray 转换为 JSONObject 失败

{"title":"Java User 对象 JSON 解析错误：JSONArray 转换为 JSONObject 失败","description":"本文介绍了在使用 FastJSON 解析 JSON 字符串时，由于将 JSONArray 对象强制转换为 JSONObject 对象而导致的 ClassCastException 异常，并提供了解决方案。","keywords":"Java, JSON, FastJSON, JSONArray, JSONObject, ClassCastException, 解析错误, 解决方法","content":"public class User { private Integer id; private String name; private String address; private Date birth;

public User() {
}

public User(Integer id, String name, String address, Date birth) {
	this.id = id;
	this.name = name;
	this.address = address;
	this.birth = birth;
}
public Integer getId() {
	return id;
}

public void setId(Integer id) {
	this.id = id;
}

public String getName() {
	return name;
}

public void setName(String name) {
	this.name = name;
}

public String getAddress() {
	return address;
}

public void setAddress(String address) {
	this.address = address;
}

public Date getBirth() {
	return birth;
}

public void setBirth(Date birth) {
	this.birth = birth;
}


@Override
public String toString() {
	return "User{"
			+ "id="
			+ id
			+ ", name='"
			+ name
			+ '', address='"
			+ address
			+ '', birth="
			+ birth
			+ '}';

}

} String json="[{"address":"广州","birth":1691063608890,"id":101,"name":"黄某"},{"address":"广州","birth":1691063608890,"id":102,"name":"黄某"},{"address":"广州","birth":1691063608890,"id":103,"name":"黄某"},{"address":"广州","birth":1691063608890,"id":104,"name":"黄某"},{"address":"广州","birth":1691063608890,"id":105,"name":"黄某"}]"; JSONObject parse= (JSONObject) JSON.parse(json); System.out.println(json); System.out.println(parse.get("id")); System.out.println(parse.get("name")); System.out.println(parse.get("address")); System.out.println(parse.get("birth")); User user=JSON.parseObject(json,User.class); System.out.println(user); /*Exception in thread "main" java.lang.ClassCastException: com.alibaba.fastjson.JSONArray cannot be cast to com.alibaba.fastjson.JSONObject at COPY.配置文件.test12.main(test12.java:40) 什么原因导致内容：这个错误是由于将JSONArray对象强制转换为JSONObject对象引起的。在给定的JSON字符串中，方括号"[]"表示一个JSONArray，而不是单个的JSONObject。因此，需要使用JSONArray来解析JSON字符串，而不是JSONObject。

你可以尝试以下代码来解决这个问题：

String json = "[{"address":"广州","birth":1691063608890,"id":101,"name":"黄某"},{"address":"广州","birth":1691063608890,"id":102,"name":"黄某"},{"address":"广州","birth":1691063608890,"id":103,"name":"黄某"},{"address":"广州","birth":1691063608890,"id":104,"name":"黄某"},{"address":"广州","birth":1691063608890,"id":105,"name":"黄某"}]";
JSONArray jsonArray = JSON.parseArray(json);

for (int i = 0; i < jsonArray.size(); i++) {
	JSONObject jsonObject = jsonArray.getJSONObject(i);
	User user = JSON.parseObject(jsonObject.toJSONString(), User.class);
	System.out.println(user);
}

这段代码将JSON字符串解析为一个JSONArray对象，然后遍历JSONArray并将每个JSONObject转换为User对象。最后，打印每个User对象的信息。

"