1前端解决方案:

https://juejin.im/post/5c51526fe51d455047338a2a 这位大神的方法 获取到jsonlint.js然后导入本地vue项目

本地使用

import jsonlint from "@/utils/jsonlint.js"


// 创建axios实例
const service = axios.create({
 在这里重写方法
  transformResponse: [function (data) {
    // Do whatever you want to transform the data
    if (typeof data === 'string') {
      try {
        data = jsonlint.parse(data);
      } catch (e) { /* Ignore */ }
    }
    return data;
  }]
})

2后端解决方案:

1.0

在JavaBean上之间加上下面的注解  (spring boot默认使用Jackson类库)，对象序列化成JSON时，将Long转成String

import com.fasterxml.jackson.databind.annotation.JsonSerialize; 
 
import com.fasterxml.jackson.databind.ser.std.ToStringSerializer; 
 
@JsonSerialize(using = ToStringSerializer.class) 

 2.0

全局处理 Springboot 2.X配置

import com.fasterxml.jackson.annotation.JsonInclude;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.module.SimpleModule;
import com.fasterxml.jackson.databind.ser.std.ToStringSerializer;
import org.springframework.boot.jackson.JsonComponent;
import org.springframework.context.annotation.Bean;
import org.springframework.http.converter.json.Jackson2ObjectMapperBuilder;

@JsonComponent
public class JsonSerializerManage {

    @Bean
    public ObjectMapper jacksonObjectMapper(Jackson2ObjectMapperBuilder builder) {
        ObjectMapper objectMapper = builder.createXmlMapper(false).build();
        //忽略value为null 时 key的输出
        objectMapper.setSerializationInclusion(JsonInclude.Include.NON_NULL);

        /**
         * 序列换成json时,将所有的long变成string
         * 因为js中得数字类型不能包含所有的java long值
         */
        SimpleModule module = new SimpleModule();
        module.addSerializer(Long.class, ToStringSerializer.instance);
        module.addSerializer(Long.TYPE, ToStringSerializer.instance);
        objectMapper.registerModule(module);
        return objectMapper;
    }

}