三种解决方法:Cannot deserialize value of type `java.util.Date` from String
作者:互联网
一、改前端
加入格式化: value-format="yyyy-MM-dd HH:mm:ss"
<el-date-picker
v-model="formValidate.pastDueTime"
value-format="yyyy-MM-dd HH:mm:ss"
type="datetime"
placeholder="选择日期时间"
/>
二、改后端实体类
注释掉 @JsonFormat 注解
// @JsonFormat(pattern = "yyyy-MM-dd HH:mm:ss",timezone="GMT+8")
private Date pastDueTime;
三、新增配置类
这种没试过,能用请评论留言
import java.util.ArrayList;
import java.util.List;
import org.springframework.context.annotation.Bean;
import org.springframework.context.annotation.Configuration;
import org.springframework.http.MediaType;
import org.springframework.http.converter.json.MappingJackson2HttpMessageConverter;
import com.fasterxml.jackson.databind.ObjectMapper;
@Configuration
public class WebConfig {
@Bean
public MappingJackson2HttpMessageConverter getMappingJackson2HttpMessageConverter() {
MappingJackson2HttpMessageConverter mappingJackson2HttpMessageConverter = new MappingJackson2HttpMessageConverter();
//设置日期格式
ObjectMapper objectMapper = new ObjectMapper();
mappingJackson2HttpMessageConverter.setObjectMapper(objectMapper);
//设置中文编码格式
List<MediaType> list = new ArrayList<MediaType>();
list.add(MediaType.APPLICATION_JSON_UTF8);
mappingJackson2HttpMessageConverter.setSupportedMediaTypes(list);
return mappingJackson2HttpMessageConverter;
}
}
============== 没有了
标签:java,String,deserialize,springframework,mappingJackson2HttpMessageConverter,org, 来源: https://blog.csdn.net/qq_16946803/article/details/121876177