Mysql in查询 结果集 乱序
SQL: select * from table where id IN (3,6,9,1,2,5,8,7);
这样的情况取出来后，其实，id还是按1,2,3,4,5,6,7,8,9,排序的，但如果我们真要按IN里面的顺序排序怎么办？SQL能不能完成？是否需要取回来后再foreach一下？其实mysql就有这个方法
sql: select * from table where id IN (3,6,9,1,2,5,8,7) order by field(id,3,6,9,1,2,5,8,7);
出来的顺序就是指定的顺序了。。。。
写法二: select * from table where id in ( 3,6,9,1,2,5,8,7 ) order by find_in_set(id,' 3,6,9,1,2,5,8,7 ')；
但是这么写 explain select * from table where id IN (3,6,9,1,2,5,8,7) order by field(id,3,6,9,1,2,5,8,7);
会Using filesort 导致查询效率降低
所以可以这样 把结果集查询出来 自己sort
List<Employee> employees = new ArrayList<>();employees.add(new Employee(123, "Jack", "Johnson", LocalDate.of(1988, Month.APRIL, 12)));employees.add(new Employee(345, "Cindy", "Bower", LocalDate.of(2011, Month.DECEMBER, 15)));employees.add(new Employee(567, "Perry", "Node", LocalDate.of(2005, Month.JUNE, 07)));employees.add(new Employee(467, "Pam", "Krauss", LocalDate.of(2005, Month.JUNE, 07)));employees.add(new Employee(435, "Fred", "Shak", LocalDate.of(1988, Month.APRIL, 17)));employees.add(new Employee(678, "Ann", "Lee", LocalDate.of(2007, Month.APRIL, 12)));employees=employees.stream().sorted((e1, e2) -> e1.getHireDate().compareTo(e2.getHireDate())).collect(Collectors.toList());接着:
描述下问题的由来
存在条件:
1. 有一个合辑表 A, 有一个内容表 C. C表中有一个列a_id 关联两个表
2. A表中有3000条数据 C表中有2w条数据 合辑表中有的有内容 有的没有内容
需求: 1. 分页显示合辑列表, 每个合辑有个ContentSize字段 表示合辑的内容数量.
解析步骤:
分两次查询: 第一次查分页. 第二次查合辑的内容数量.
1.分页sql 忽略.
2.统计list<合辑id> 合辑id列表中的 每合辑对应的内容数量
当时想到了第一种方式 连接查询 count in group by order by id的顺序 (次方法有两点不好1. left join 连表 效率低 2. 结果集顺序 Using filesort 效率低)
@Query(value = "select count(c.id) from abm_album a left join abm_album_content c on a.id=c.album_id and c.is_deleted =:isDeleted where a.id in (:albumIds) group by a.id order by field(a.id,:albumIds) ",nativeQuery = true)public List<BigInteger> countByAlbumIdsAndIsDeleted(@Param("isDeleted")Integer isDeleted,@Param("albumIds")List<Integer> albumIds); ---------------------------方式一:解析SQL如下------------------------- select count(c.id)from abm_album a left join abm_album_content con a.id=c.album_id and c.is_deleted =0where a.id in (5138822,5160757,5000142,5160750,5159885)group by a.idorder by field(a.id,5138822,5160757,5000142,5160750,5159885) @Query(value = "select c.album_id ,count(c.id) from abm_album_content c where c.is_deleted =:isDeleted and c.album_id in (:albumIds) group by c.album_id ",nativeQuery = true)public List<Object[]> countMapByAlbumIdsAndIsDeleted(@Param("isDeleted")Integer isDeleted,@Param("albumIds")List<Integer> albumIds); ---------------------------方式二:解析SQL如下-------------------------select c.album_id , count( c.id) from abm_album_content cwhere c.is_deleted =0 and c.album_id in ( 5138822 , 5160757 , 5000142 , 5160750 , 5159885 ) group by c.album_id ---------------------------查询结果处理------------------------- List<Album> datas = albumRepository.find(album, page.getStart(), page.getPageSize(), sort);if(null!=datas && datas.size()>0){//设置稿件数量 try { List<Integer> ids= datas.stream().map(Album::getId).collect(Collectors.toList()); // 方式一 list查询 in带顺序 然后顺序set sql关联查询 执行效率低 /* List<BigInteger> bigIntegers = albumContentRepository.countByAlbumIdsAndIsDeleted(AlbumConstant.UNDELETED, ids); if(datas.size()==ids.size()&&datas.size()==bigIntegers.size()){ for (int i=0;i<datas.size();i++){ datas.get(i).setContentsCount(bigIntegers.get(i).intValue()); } } */ // 方式二 执行单表count 返回值list<Object[]> 循环设置 效率高 List<Object[]> maps = albumContentRepository.countMapByAlbumIdsAndIsDeleted(AlbumConstant.UNDELETED, ids); Map<Integer,BigInteger> countMap=new HashMap<>(); maps.forEach(map->countMap.put((Integer)map[0],(BigInteger)map[1])); datas.forEach(data-> {if(null!=countMap.get(data.getId())){ data.setContentsCount(countMap.get(data.getId()).intValue()); }else { data.setContentsCount(0); } }); }catch (Exception e){log.error("查询合辑下稿件数量出错",e); }}//set结果集page.setDatas(datas);总结:
1. in 查询 返回的结果集 如果有数据不存在的时候 返回结果集的个数 和合辑id_list的size 是不一样的.
2. 能单表查询 尽量不要链表查询
OK. 到此结束 小记...
转载于:https://www.cnblogs.com/wangdaijun/p/5893328.html
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