[英]How to disable swagger-codegen java.io.InputStream convertation
I got rests that return:我得到了回报:
java.io.InputStream
, java.io.InputStream
,
com.fasterxml.jackson.databind.JsonNode
, com.fasterxml.jackson.databind.JsonNode
,
org.springframework.core.io.InputStreamRecource
But但
In result swagger-codegen
generated api, i got rests, that return:在结果
swagger-codegen
生成的 api 中,我得到了休息,返回:
productName.client.model.InputStream
, productName.client.model.InputStream
,
ErrorModelNamenamespacecomFasterxmlJacksonDatabindNameJsonNode
, ErrorModelNamenamespacecomFasterxmlJacksonDatabindNameJsonNode
,
productName.client.model.InputStreamResource
I cant ignore them by .swagger-codegen-ignore
我
.swagger-codegen-ignore
ignore 忽略它们
And i think that replacing java.io.InputStream
by byte[]
is not good (how it described here ).而且我认为用
byte[]
替换java.io.InputStream
并不好( 这里是如何描述的)。 Cause stream has some pluses, such as exclusion of memory leaks.原因流有一些优点,例如排除内存泄漏。
What interesting, Swagger understand, that classes from springframework.http
must not be converted.有趣的是,Swagger 明白,来自
springframework.http
类不能被转换。 I got them all as original in generated api.我在生成的 api 中将它们全部作为原始版本。
Is it possiple to save java.io.InputStream
and org.springframework.core.io.InputStreamRecource
?是否可以保存
java.io.InputStream
和org.springframework.core.io.InputStreamRecource
?
Done by完成人
java -jar swagger-codegen-cli-3.0.27.jar generate --import-mappings InputStreamResource=org.springframework.core.io.InputStreamResource,JsonNode=com.fasterxml.jackson.databind.JsonNode,InputStream=java.io.InputStream -DhideGenerationTimestamp=true -i "http://$CONTAINER_IP:8080/app_name/rest/v3/api-docs" -l java --library resttemplate -o app_name-api-client
or through或通过
.swagger-codegen-ignore
file by add /src/main/java/app_name/client/InputStreamResource
.swagger-codegen-ignore
文件通过添加/src/main/java/app_name/client/InputStreamResource
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