Question

我有一个包含容器编号的表和一个带有时间戳的completed_on日期字段。它还有船只id和起重机id。我需要通过询问completed_on是否为空来计算已完成容器的数量并且未完成。

我写过类似的内容，但它不起作用。

select vessel,
  crane_no,
  count(container_no) tot_moves,
  case when completed_on is null then count(container_no) end as pending,
  case when completed_on is not null then count(container_no) end as completed,
min(completed_on) first_m,
max(completed_on) last_m
from
containers
group by vessel, crane_no, completed_on

有什么想法吗？

Answer 1

你非常接近...... case语句应该在你的聚合count函数中：

select 
  vessel,
  crane_no,
  count(container_no) tot_moves,
  count(case when completed_on is null then 1 end) as pending,
  count(case when completed_on is not null then 1 end) as completed,
  min(completed_on) first_m,
  max(completed_on) last_m
from
  containers
group by 
  vessel, 
  crane_no, 
  completed_on

Answer 2

试试这个：

    select 
      vessel,
      crane_no,
      count(container_no) tot_moves,
      count(case when completed_on is null then 1 end) as pending,
      count(case when completed_on is not null then 1 end) as completed,
      min(completed_on) first_m,
      max(completed_on) last_m
    from containers group by  vessel, crane_no, completed_on

Answer 3

还有另一种方法，不使用CASE。我不确定Oracle如何优化，但可能更快（您需要自己进行分析）。

select vessel,
       crane_no,
       count(container_no) tot_moves,
       count(*) - count(completed_on) as pending,
       count(completed_on) as completed,
       min(completed_on) first_m,
       max(completed_on) last_m
from containers
group by vessel, crane_no, completed_on

这很有效，因为几乎所有聚合（包括COUNT()）都会忽略空值。您可以使用总行数与已完成行数的差异来获取待处理的行数。此外，count(completed_on) 应缓存，并且不会运行两次（某些RDBMS允许在SELECT子句中重用列别名，但我不知道如果Oracle支持这个）。

Answer 4

类似的东西：

select vessel,
    crane_no,
    count(container_no) tot_moves,
    sum(case when completed_on is null then 1 else 0 end) as pending,
    sum(case when completed_on is not null then 1 else 0 end) as completed,
    min(completed_on) first_m,
    max(completed_on) last_m
from containers
group by vessel, crane_no, completed_on;

case决定是否应计算每一行，sum根据这些计算实际计数。

Answer 5

将count（）放在案例周围。

select vessel,
  crane_no,
  count(container_no) tot_moves,
  count(case when completed_on is null then container_no end) as pending,
  count(case when completed_on is not null then container_no end) as completed,
min(completed_on) first_m,
max(completed_on) last_m
from
containers
group by vessel, crane_no;

Answer 6

SELECT
  vessel,
  crane_no,
  COUNT(container_no)                                       AS "tot_moves",
  SUM(CASE WHEN completed_on IS NULL     THEN 1 ELSE 0 END) AS "pending",
  SUM(CASE WHEN completed_on IS NOT NULL THEN 1 ELSE 0 END) AS "completed",
  MIN(completed_on) first_m,
  MAX(completed_on) last_m
FROM containers
GROUP BY vessel, crane_no;

Answer 7

默认情况下，count函数不会计数null。您可以使用nvl2函数来操作查询，以帮助您只计算空值。例如：

select count(completed_on) as completed , 
       count(nvl2(completed_on,null,1)) as pending
from containers;

根据字段计数

7 个答案: