生成递归树
电脑版发表于:2021/11/9 16:45
注意:在生成树时最好从子集开始生产。
//Dto
public class FormsTest2Dto <E>{
@ApiModelProperty("框架id")
private Long Id;//子级id
@ApiModelProperty(value = "父级id") //example = "0代表顶级或无" 举例说明
private Long p_id; //父级id 0代表顶级或无
@ApiModelProperty("部门名称")
private String form_Name;//--部门名称
@ApiModelProperty(value = "人数")
private Integer people_Number;//人数(管理多少人)
private List<E> ChildrenE;
public FormsTest2Dto(Long id, String form_Name) {
Id = id;
this.form_Name = form_Name;
}
}
//子集树 方法
public List<FormsTest2Dto> getTree2(Long P_id)
{
List<FormsTest2Dto> matters = iFormsMapper.queryFormAllDto(new Forms(null, P_id, null));
List<FormsTest2Dto> listreturn=new ArrayList<>();
for (int i = 0; i < matters.size(); i++) {
FormsTest2Dto dto= new FormsTest2Dto(matters.get(i).getId(),matters.get(i).getForm_Name());
dto.setChildrenE(getTree2(matters.get(i).getId()));
listreturn.add(dto);
}
return listreturn;
}
案例
1
2.1
2.2
