如何从表中检索每条第 N 条记录? [英] How do I retrieve every Nth record from a table?

问题描述

查询保持不变，但偏移量不同.

The query remains constant, but the offset varies.

SELECT NAME from tbl WHERE alphabet_index='A' limit 880,1;
SELECT NAME from tbl WHERE alphabet_index='A' limit 1760,1;
SELECT NAME from tbl WHERE alphabet_index='A' limit 2640,1;
....
SELECT NAME from tbl WHERE alphabet_index='A' limit 34320,1;

有没有更好的方法来做到这一点，而不必运行具有不同偏移量的 N 个查询?

Is there a better way to do this without having to run N queries with different offsets?

更新:我正在尝试将名称范围构建为向下钻取的索引.因此，如果一个alphabet_index 有50000 条记录并且我想显示40 个范围，我正在计算offset = total/rangeCount.我想要的最终结果是

Update: I am trying to build name ranges as an index for a drill down. So if an alphabet_index has 50000 records and I would like to display 40 ranges, I am calculating offset = total/rangeCount. My desired end result would be

AA...-AA...
AA...-AB...
AB...-AC...
AC...-AC...
AD...-AF...
...
...
AW...-AZ...

推荐答案

您可以像这样在单个查询中执行不同的偏移量查询

You can do a varying offset query in a single query like this

select NAME from 
   (select @row:=@row+1 as row, t.NAME from 
    tbl t, (select @row := 0) y 
    where alphabet_index='A' order by alphabet_index) z 
where row % 880 = 1;

这将通过@row 变量为每一行添加一个唯一的整数 id.然后它将通过模运算符和该变量每隔 880 选择一行.需要 order by 子句才能获得可重复的行为，否则结果实际上是随机的.

This will add a unique integer id to each row via the @row variable. Then it will select a row every other 880 via the modulo operator and that variable. An order by clause is required to get repeatable behavior, else the result would be effectively random.

这篇关于如何从表中检索每条第 N 条记录?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！