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Describe the feature
前端如果需要跳转到任意楼层,需要加载整个洞,这对于高回复数的洞是不可取的。
Describe how to implement it
使用 floor.ranking 字段,加载对应楼的上下各 size 个楼,默认 size = 10;返回值中将包含最多 2 * size + 1 个楼。由于一个洞可能很小,一个楼的前后可能都不足 size 个楼,需要前端来判断加载位置。
select hole, ranking from floor where id = ?; select * from floor where hold_id = ? and ranking between ? - size and ? + size;
What does this feature intend to do
对于高楼层的查询优化
Additional context
无
The text was updated successfully, but these errors were encountered:
该功能是否必要?实测对于1400层的楼可以在1秒内返回,且需要前端修改请求新floor的UI。
Sorry, something went wrong.
有必要的,1秒和10ms还是有质的区别。后端这边会先做出来,前端可以自己考虑是否上线,我建议使用这个新API,在压力非常高的场合,比如高搜索压力的时候可能可以用到。
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Describe the feature
前端如果需要跳转到任意楼层,需要加载整个洞,这对于高回复数的洞是不可取的。
Describe how to implement it
使用 floor.ranking 字段,加载对应楼的上下各 size 个楼,默认 size = 10;返回值中将包含最多 2 * size + 1 个楼。由于一个洞可能很小,一个楼的前后可能都不足 size 个楼,需要前端来判断加载位置。
What does this feature intend to do
对于高楼层的查询优化
Additional context
无
The text was updated successfully, but these errors were encountered: