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Mapper 接口如何传递多个参数?
阅读量:606 次
发布时间:2019-03-12

本文共 1722 字,大约阅读时间需要 5 分钟。

  • 方式一、接口中传多个参数,在 xml 中使用 #{param0}、#{param1}…
  • 方式二、使用 @param 注解指定名称,在 xml 中使用 #{名称}
  • 方式三、多个参数封装到 Java bean 中
  • 方式四、多个参数指定 key,put 到 Map 中
//方式一//javaSystem.out.println("------ selectUserByParamIndex ------");user = userMapper.selectUserByParamIndex(31, "ConstXiong1");System.out.println(user);//xml< select id="selectUserByParamIndex" resultType="constxiong.po.User">    select * from user where id = #{arg0} and name = #{arg1}< /select>//方式二//javaSystem.out.println("------ selectUserByAnnotation ------");user = userMapper.selectUserByAnnotation(31, "ConstXiong1");System.out.println(user);//xml< select id="selectUserByAnnotation" resultType="constxiong.po.User">    select * from user where id = #{id} and name = #{name}< /select>//方式三//javaSystem.out.println("------ selectUserByPo ------");user = userMapper.selectUserByPo(new User(31, "ConstXiong1"));System.out.println(user);//xml< select id="selectUserByPo" resultType="constxiong.po.User" parameterType="constxiong.po.User">    select * from user where id = #{id} and name = #{name}< /select>//方式四//javaSystem.out.println("------ selectUserByMap ------");Map
param = new HashMap<>();param.put("id", 31);param.put("name", "ConstXiong1");user = userMapper.selectUserByMap(param);System.out.println(user);//xml< select id="selectUserByMap" resultType="constxiong.po.User"> select * from user where id = #{id} and name = #{name}< /select>

 

打印结果

------ selectUserByParamIndex ------User{id=31, name='ConstXiong1', mc='null'}------ selectUserByAnnotation ------User{id=31, name='ConstXiong1', mc='null'}------ selectUserByPo ------User{id=31, name='ConstXiong1', mc='null'}------ selectUserByMap ------User{id=31, name='ConstXiong1', mc='null'}

 

 


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