此问题与JsonFormat无法处理JSON字符串中的复杂日期-时间格式有关。要解决此问题，您可以为ZonedDateTime类型创建自定义验证程序。
下面是一个示例：

import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonDeserializer;
import java.io.IOException;
import java.time.ZonedDateTime;
import java.time.format.DateTimeFormatter;

public class ZonedDateTimeDeserializer extends JsonDeserializer<ZonedDateTime> {

    private static final DateTimeFormatter FORMATTER = new DateTimeFormatterBuilder()
            .parseCaseInsensitive()
            .appendPattern("dd-MMM-yy hh.mm.ss.SSSSSSSSS a z")
            .toFormatter();

    @Override
    public ZonedDateTime deserialize(JsonParser jsonParser, DeserializationContext deserializationContext) throws IOException {
        String dateString = jsonParser.getText();
        return ZonedDateTime.parse(dateString, FORMATTER);
    }
}

字符串
你的财产

public class Test {

    @JsonProperty("createdTs")
    @JsonDeserialize(using = ZonedDateTimeDeserializer.class)
    private ZonedDateTime createdTs;
//Getter & Settere
}

型
测试

public static void main(String[] args) throws JsonProcessingException {
        String s = """
                {
                "createdTs": "29-OCT-21 12.00.00.000000000 AM UTC"
                }
                """;

        Test t = JsonMapper.builder().findAndAddModules().build().readValue(s, Test.class);

        System.out.println(t);
    }

型
输出Test(createdTs=2021-10-29T00:00Z[UTC])
试试看，然后告诉我