Question

我尝试在Entity类上编写if else条件但是当我运行tomcat时它不起作用。它有一个错误。我很想知道如何在spring roo项目中编写if else条件？我可以写哪个文件？

我尝试通过写:(在Student.java中）测试条件

  9 @RooJavaBean
 10 @RooToString
 11 @RooJpaActiveRecord
 12 public class Student {
 13     private long id;
 14     private String name; 
 15     private Integer age; 
 16     if(true){} //test if condition

当我运行tomcat时这是错误。

[ERROR] Failed to execute goal org.apache.maven.plugins:maven-compiler-plugin:2.3.2:compile (default-compile) on project tap: Compilation failure: Compilation failure:
[ERROR] Employee.java:[22,1] illegal start of type
[ERROR] Employee.java:[22,4] illegal start of type
[ERROR] -> [Help 1]
[ERROR] 
[ERROR] To see the full stack trace of the errors, re-run Maven with the -e switch.
[ERROR] Re-run Maven using the -X switch to enable full debug logging.

Answer 1

失败表示您有语法错误。

是的，第16行的语法错误。你不能在某个类的某个地方写一个if语句，它必须在一个方法中！

@RooJavaBean
@RooToString
@RooJpaActiveRecord
public class Student {
   private long id;
   private String name; 
   private Integer age; 

   public void someMethod() {
       if(1 == 1){
          // if 1 is 1 then do this.
       } else {
         // if 1 is different to 1 then do this.
       }                
   }

你应该注意到，这个Roo类是普通的Java类（由一些Roo生成的AspectJ文件增强），但这是真正的类，但没有配置文件。

春天roo项目如何使用if else条件？

1 个答案: