如何将参数传递给子查询

Question

我只是使用 findAllregions 参数的 userIdx 来查找 FavoriteStore

这是我的查询

@Select("SELECT * FROM region")
@Results(value = {
        @Result(property = "regionIdx", column = "regionIdx"),
        @Result(property = "stores", javaType = Store.class, column = "storeIdx",
                many = @Many(select = "com.travely.travely.mapper.StoreMapper.findFavoriteStore", fetchType = FetchType.LAZY))
})
List<Region> findAllRegions(@Param("userIdx") final Long useridx);



@Select("SELECT s.* FROM store as s NATURAL JOIN favorite as f WHERE f.userIdx=#{userIdx} AND f.isFavorite = 1")
    List<Store> findFavoriteStore(@Param("userIdx") final long userIdx);

适用于 'findAllRegions' 的 select 区域但不适用于 'findFavoriteStore'

的 select 商店

Answer 1

查询失败，因为 column 属性配置不正确。

这是列属性的相关部分 documentation:

The column name from the database, or the aliased column label that holds the value that will be passed to the nested statement as an input parameter.

如您所见，只能使用主查询结果中的列。

这意味着您要么需要将人工列包含到查询中，然后像这样在 @Result.column 中使用它：

@Select("SELECT #{userIdx} userIdx, r.* FROM region r")
@Results(value = {
     @Result(property = "regionIdx", column = "regionIdx"),
     @Result(
        property = "stores", javaType = Store.class, column = "userIdx",
        many = @Many(
                 select = "com.travely.travely.mapper.StoreMapper.findFavoriteStore",
                 fetchType = FetchType.LAZY))
})
List<Region> findAllRegions(@Param("userIdx") final Long useridx);

或者，如果使用 java 8+，您可以使用默认接口方法来获取 associations/collections，如下所示：

interfact MyMapper {

  @Select("SELECT * FROM region")
  @Results(value = {
        @Result(property = "regionIdx", column = "regionIdx")
  })
  List<Region> findAllRegions();

  default List<Region> findAllRegions(Long userIdx) {
      List<Region> regions = findAllRegions();
      List<Strore> favoriteStores = findFavoriteStore(userIdx);
      for(Region region:regions) {
          region.setStores(favoriteStores);
      }
      return regions;
  }

  @Select("SELECT s.* FROM store as s NATURAL JOIN favorite as f WHERE f.userIdx=#{userIdx} AND f.isFavorite = 1")
  List<Store> findFavoriteStore(@Param("userIdx") final long userIdx);

}

请注意，这不使用延迟获取最喜欢的商店。似乎这不是必需的，因为在不需要最喜欢的商店的情况下，没有 userIdx 的查询可以（并且应该使用。

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