Question

我正在寻找一种易于阅读的智能SQL查询（SQLite引擎）来聚合列中的数据。通过一个例子更容易解释：

数据表：

*** Settings ***
Library       Collections

*** Test Cases ***
Test AAA
    ${aaaaa}    My Keyword    A=aaa    B=bbb    C=ccc
    ${bbbbb}    My Keyword    A=aaa    B=bbb    C=ccc    D=ddd

*** Keywords ***
My Keyword
    [Arguments]    @{args}
    ${resp}    Create Dictionary    ${args}
    [Return]    ${resp}

预期结果集：httpcode列，按时间编码。在这个例子中，时间聚集是0.2秒（但它可以在一秒钟或10秒内聚集）。我只对某些预期的http_code感兴趣：

 id  elapsedtime httpcode
 1          0.0      200
 2          0.1      200
 3          0.3      301
 4          0.6      404
 5          1.0      200
 6          1.1      404
 7          1.2      500

＆＃34; time＆＃34;并非强制性的。要连续的。在前面的示例中，可以删除时间为0.6和0.6的行。

目前，我可以通过执行4个不同的请求（一个通过http代码）来完成此操作，并在我的开发应用程序中重新聚合结果：

 time code_200 code_404 code_500 code_other
 0.0        2        0        0          0
 0.2        0        0        0          1
 0.4        0        1        0          0
 0.6        0        0        0          0
 0.8        0        0        0          0
 1.0        1        1        1          0

但我非常确定我可以用一个查询来实现这一目标。不幸的是，我没有掌握UNION关键字......

有没有办法在单个SELECT中获取这些数据？

参见SQLFiddle：http://sqlfiddle.com/#!5/2081f/3/1

Answer 1

找到一个比我原来的帖子更好的解决方案，我会留下来，以防你好奇。这是更好的解决方案：

with t1 as (
    select
    0.2 * cast (elapsedtime/ 0.2 as int) as time,
    case httpcode when 200 then 1 else 0 end code_200,
    case httpcode when 404 then 1 else 0 end code_404,
    case httpcode when 500 then 1 else 0 end code_500,
    case when httpcode not in (200, 404, 500) then 1 else 0 end code_other
    from test
)

select time,
sum(code_200) as count_200,
sum(code_404) as count_404,
sum(code_500) as count_500,
sum(code_other) as count_other
from t1
group by time;

旧解决方案：

这在眼睛上可能不太容易，但它或多或少都有效（只有你想要的输出和我得到的结果之间的差异是没有值的时间分组（在你的例子中是0.6和0.8）被省略：

with 

t_all as (select
0.2 * cast (elapsedtime/ 0.2 as int) as time, count(id) as total
from test
group by time
),

t_200 as (select
0.2 * cast (elapsedtime/ 0.2 as int) as time, count(id) as code_200
from test
where (httpcode=200)
group by time),

t_404 as (select
0.2 * cast (elapsedtime/ 0.2 as int) as time, count(id) as code_404
from test
where (httpcode=404)
group by time),

t_500 as (select
0.2 * cast (elapsedtime/ 0.2 as int) as time, count(id) as code_500
from test
where (httpcode=500)
group by time),

t_other as (select
0.2 * cast (elapsedtime/ 0.2 as int) as time, count(id) as code_other
from test
where (httpcode not in (200, 404, 500))
group by time)

select 
t_all.time,
total,
ifnull(code_200,0) as count_200,
ifnull(code_404,0) as count_404,
ifnull(code_500,0) as count_500,
ifnull(code_other,0) as count_other
from t_all
left join t_200 on t_all.time = t_200.time
left join t_404 on t_all.time = t_404.time
left join t_500 on t_all.time = t_500.time
left join t_other on t_all.time = t_other.time;

Answer 2

它可能对你有帮助

select
    0.2 * cast (elapsedtime/ 0.2 as int) as time, count(id) as code_200,
    0 as code_404,
    0 as code_500,
    0 as code_other
from
    test
where (httpcode=200)
group by time
union
select
    0.2 * cast (elapsedtime/ 0.2 as int) as time,0 as code_200,
    count(id) as code_404,
    0 as code_500,
    0 as code_other
from
    test
where (httpcode=404)
group by time
union
select
    0.2 * cast (elapsedtime/ 0.2 as int) as time,0 as code_200,
    0 as code_400,
    count(id) as code_500,
    0 as code_other
from
    test
where (httpcode=500)
group by time
union
select
     0.2 * cast (elapsedtime/ 0.2 as int) as time,0 as code_200,
     0 as code_400,
     0 as code_500,
     count(id) as code_other
from
  test
where (httpcode<>200 and httpcode <> 404 and httpcode <> 500)
group by time

按时间计算和聚集数据

2 个答案: