如何将Java异常信息返回给jQuery.ajax REST调用？

import com.fasterxml.jackson.databind.ObjectMapper;

// 定义错误响应对象
class ErrorResponse {
    private int errorCode;
    private String errorMessage;

    public ErrorResponse(int errorCode, String errorMessage) {
        this.errorCode = errorCode;
        this.errorMessage = errorMessage;
    }

    // 省略getter和setter方法
}

// REST接口
@RestController
public class MyController {

    @GetMapping("/api/data")
    public ResponseEntity<String> getData() {
        try {
            // 业务逻辑代码
            // 如果发生异常，捕获并封装到ErrorResponse对象中
            throw new Exception("Something went wrong");
        } catch (Exception e) {
            // 创建ErrorResponse对象
            ErrorResponse errorResponse = new ErrorResponse(500, e.getMessage());

            // 将ErrorResponse对象转换为JSON字符串
            ObjectMapper objectMapper = new ObjectMapper();
            String json = objectMapper.writeValueAsString(errorResponse);

            // 返回JSON字符串作为响应体
            return ResponseEntity.status(HttpStatus.INTERNAL_SERVER_ERROR).body(json);
        }
    }
}

$.ajax({
    url: "/api/data",
    type: "GET",
    success: function(response) {
        // 解析JSON字符串
        var errorResponse = JSON.parse(response);
        
        // 提取错误信息字段
        var errorCode = errorResponse.errorCode;
        var errorMessage = errorResponse.errorMessage;
        
        // 处理错误信息
        console.log("Error code: " + errorCode);
        console.log("Error message: " + errorMessage);
    },
    error: function(xhr, status, error) {
        // 处理请求错误
        console.log("Request error: " + error);
    }
});