2022年12月7日
idea创建代码模版
原文:【精品】IntelliJ 文件模板 创建 通用Controller #yyds干货盘点#_51CTO博客_intellij idea创建jsp文件
我搞了两个简单的
xxController
#set($NameNoController = $NAME.length() - 10)
#set($NameLowerFirst = ${NAME.substring(0,1).toLowerCase()} + $NAME.substring(1,$NameNoController))
#set($NameUpperFirst = $NAME.substring(0,$NameNoController))
#if (${PACKAGE_NAME} && ${PACKAGE_NAME} != "")package ${PACKAGE_NAME};#end
#parse("File Header.java")
@Slf4j
@Api(tags = "")
@RestController
@RequestMapping("/")
public class ${NAME} extends AdminBaseController {
}
xxApi
#set($NameNoController = $NAME.length() - 3)
#set($NameLowerFirst = ${NAME.substring(0,1).toLowerCase()} + $NAME.substring(1,$NameNoController))
#set($NameUpperFirst = $NAME.substring(0,$NameNoController))
#if (${PACKAGE_NAME} && ${PACKAGE_NAME} != "")package ${PACKAGE_NAME};#end
#parse("File Header.java")
@Slf4j
@RestController
@RequestMapping("/")
@Api(value = "", tags = "")
public class ${NAME} extends WxBaseApi {
@GetMapping("/")
public AjaxResult get() {
return AjaxResult.success();
}
}