【发布时间】:2017-10-30 14:59:42
【问题描述】:
其实我有这个 Rest Template 请求:
@RequestMapping(value = "/uploadProperties", method = RequestMethod.POST)
public @ResponseBody RessourceMetadata uploadProperties(
@RequestParam(value = "group", required = true) String group,
@RequestParam(value = "id", required = true) String id,
@RequestParam(value = "version", required = true) String version,
@RequestParam(value = "env", required = true) String env) {
try {
Ressource ressource = new Ressource(content, group, id, version, env, PropertiesFileUtils.getPropertiesFilename());
getRessourceService().save(ressource);
return ressource.getMetadata();
} catch (RuntimeException e) {
log.error("Error while uploading.", e);
throw e;
} catch (Exception e) {
log.error("Error while uploading.", e);
throw new RuntimeException(e);
}
}
我想像这样添加@RequestBody:
@RequestBody @RequestParam(value = "content", required = true) ????? content
这个新的content 可以包含任何内容。
如何正确传递content参数?
【问题讨论】:
-
“可以包含任何东西”是什么意思?你想把组、版本、id、env等所有字段都封装到内容中,对吗?
-
@pvpkiran 不,是一个新字段,我可以在其中放置代码或键/值行,例如...我不知道是否可以按照我的要求做?
-
某事要么是
@RequestBody要么是@RequestParam不是两者都...
标签: spring rest spring-boot swagger