1前端解決方案:
https://juejin.im/post/5c51526fe51d455047338a2a 這位大神的方法 獲取到jsonlint.js然後導入本地vue項目
本地使用
import jsonlint from "@/utils/jsonlint.js"
// 創建axios實例
const service = axios.create({
在這裏重寫方法
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後端解決方案:
1.0
在JavaBean上之間加上下面的註解 (spring boot默認使用Jackson類庫),對象序列化成JSON時,將Long轉成String
import com.fasterxml.jackson.databind.annotation.JsonSerialize;
import com.fasterxml.jackson.databind.ser.std.ToStringSerializer;
@JsonSerialize(using = ToStringSerializer.class)
2.0
全局處理 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爲null 時 key的輸出
objectMapper.setSerializationInclusion(JsonInclude.Include.NON_NULL);
/**
* 序列換成json時,將所有的long變成string
* 因爲js中得數字類型不能包含所有的java long值
*/
SimpleModule module = new SimpleModule();
module.addSerializer(Long.class, ToStringSerializer.instance);
module.addSerializer(Long.TYPE, ToStringSerializer.instance);
objectMapper.registerModule(module);
return objectMapper;
}
}