- when(candidateService.findById(1)).thenReturn(new Candidate());
我想扩展这种行为为任何整数(不一定为1)
如果我吃惊
- when(candidateService.findById( any(Integer.class) )).thenReturn(new Candidate());
我有编译错误
The method findById(Integer) in the type CandidateService is not
applicable for the arguments (Matcher)
UPDATE
进口:
- import static org.junit.Assert.assertEquals;
- import static org.mockito.Matchers.anyInt;
- import static org.mockito.Matchers.anyString;
- import static org.mockito.Mockito.mock;
- import static org.mockito.Mockito.times;
- import static org.mockito.Mockito.verify;
- import static org.mockito.Mockito.when;
- import static org.springframework.test.web.servlet.result.MockMvcResultMatchers.status;
- import java.util.ArrayList;
- import java.util.HashSet;
- import org.junit.Before;
- import org.junit.Test;
- import org.junit.runner.RunWith;
- import org.mockito.InjectMocks;
- import org.mockito.Mock;
- import org.mockito.MockitoAnnotations;
解决方法
尝试anyInt():
- when(candidateService.findById( anyInt())).thenReturn(new Candidate());
例如我的项目中有anyLong():
- when(dao.getAddress(anyLong())).thenReturn(Arrays.asList(dto));
编辑:
您必须导入:
- import static org.mockito.Matchers.anyInt;