关于mybatis 传入基本参数类型（Integer）使用if test 判断出错的问题

        SELECT p1.id,p1.name,p1.manager_id,p1.is_scrum,u1.real_name as subManagerName,p1.parent_id,p2.name as parentProName,u2.real_name as parManagerName,p2.manager_id as parManagerId from
        project p1
        left join project p2
        on p1.parent_id=p2.id
        left join user u1
        on p1.manager_id=u1.id
        left join user u2
        on p2.manager_id=u2.id
        where p1.id in (
        SELECT p1.id from
        (SELECT case when parent_id=0 then id else parent_id end as parent_id,id,manager_id,is_scrum from project where is_scrum=1 ) p1
        left join project_user p2 on p1.id=p2.project_id
        left join project p3 on p1.manager_id=p3.id
        
            where p1.manager_id=#{userId} or p2.user_id=#{userId} or p3.manager_id=#{userId} group by p1.id
        
        )
    

此时会报错没有在integer中找到userid的get方法的错误。假如这里userid用的是String类型则不会报这个错误。

解决办法
1、使用一个包装类将userid包装起来。
2、需要使用_parameter 当做入参
例：

        SELECT p1.id,p1.name,p1.manager_id,p1.is_scrum,u1.real_name as subManagerName,p1.parent_id,p2.name as parentProName,u2.real_name as parManagerName,p2.manager_id as parManagerId from
        project p1
        left join project p2
        on p1.parent_id=p2.id
        left join user u1
        on p1.manager_id=u1.id
        left join user u2
        on p2.manager_id=u2.id
        where p1.id in (
        SELECT p1.id from
        (SELECT case when parent_id=0 then id else parent_id end as parent_id,id,manager_id,is_scrum from project where is_scrum=1 ) p1
        left join project_user p2 on p1.id=p2.project_id
        left join project p3 on p1.manager_id=p3.id
        
            where p1.manager_id=#{userId} or p2.user_id=#{userId} or p3.manager_id=#{userId} group by p1.id
        
        )
    

        select
        *
        from
       reason op
        where
        1=1
        
        and
        id = #{_parameter, jdbcType=BIGINT}