when(candidateService.findById(1)).thenReturn(new Candidate());
我想为任何Integer(不一定是1)
扩展此行为如果我啰嗦
when(candidateService.findById( any(Integer.class) )).thenReturn(new Candidate());
我有编译错误
CandidateService类型中的方法findById(Integer)不是 适用于论证(Matcher)
更新
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;
答案 0 :(得分:33)
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;