我有这段代码,我想在找不到资源的情况下抛出异常
Menu menu = menuService.findById(addMenuAmount.getMenuId()) .orElseThrow(com.tdk.web.exception.ResourceNotFoundException(“menu " + addMenuAmount.getMenuId() + " not found "));
但我得到一个编译错误:
com.tdk.web.exception cannot be resolved to a type
肥皂起泡泡
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