Question

我有一个assigns表，其中包含以下列：

id - int
id_lead - int
id_source - int
date_assigned - int (this represents a unix timestamp)

现在，假设我在此表中有以下数据：

id id_lead id_source date_assigned
1  20      5         1462544612
2  20      6         1462544624
3  22      6         1462544615
4  22      5         1462544626
5  22      7         1462544632
6  25      6         1462544614
7  25      8         1462544621

现在，假设我想得到id_source为6的行的计数，并且是每个潜在客户的第一个条目（按date_assigned asc排序）。

所以在这种情况下，count会= 2，因为有2个引导（id_lead 22和25），其第一个id_source是6。

我如何编写此查询以使其快速并且可以作为子查询选择正常工作？我在想这样的事情是行不通的：

select count(*) from `assigns` where `id_source`=6 order by `date_assigned` asc limit 1

我不知道如何以最佳方式编写此查询。任何帮助将不胜感激。

Answer 1

简单查询

点击http://sqlfiddle.com/#!9/8666e0/7

select count(*) from 
(select * from assigns group by id_lead )t 
where t.id_source=6

Answer 2

伪代码：

select rows 
with a.id_source = 6
but only if
    there do not exist any row
        with same id_lead
        and smaller date_assigned

将其翻译为SQL

select *                                     -- select rows 
from assigns a
where a.id_source = 6                        -- with a.id_source = 6
  and not exists (                           -- but only if there do not exist any row
    select 1
    from assigns a1
    where a1.id_lead = a.id_lead             -- with same id_lead
      and a1.date_assigned < a.date_assigned -- and smaller date_assigned
  )

现在将select *替换为select count(*)，您就会得到结果。

http://sqlfiddle.com/#!9/3dc0f5/7

<强>更新

可以将NOT-EXIST查询重写为排除LEFT JOIN查询：

select count(*)
from assigns a
left join assigns a1
    on  a1.id_lead = a.id_lead
    and a1.date_assigned < a.date_assigned
where a.id_source = 6
  and a1.id_lead is null

如果您想获取id_source的所有值的计数，则下面的查询可能是最快的：

select a.id_source, count(1)
from (
    select a1.id_lead, min(a1.date_assigned) date_assigned
    from assigns a1
    group by a1.id_lead
) a1
join assigns a
    on  a.id_lead = a1.id_lead
    and a.date_assigned = a1.date_assigned
group by a.id_source

您仍然可以将group by a.id_source替换为where a.id_source = 6。

查询需要assigns(id_source)和assigns(id_lead, date_assigned)上的索引。

使用另一列排序的列值计算第一次出现

2 个答案: