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(); } }