mysql：计算每组中的值数

Question

我有一个问题：

SELECT 
    count(session_id_open) as opens,
    count(session_id_visit) as visits,
    date_intervals_open,
    group_concat(date_intervals_visit)
FROM
    bla
GROUP BY date_intervals_open

我得到下表。我需要的是按百分比查找出现在 group_concat 中的每个值的出现次数。所以，基本上，我需要计算每组（data_intervals_open）中值的数量（date_intervals_visit）

opens   visits  date_intervals_open group_concat        
213    5        day (12-16)             evening (17-21),evening (17-21),day (12-16),day (12-16),day (12-16)
113    0        evening (17-21)         NULL
11     0        late evening (22-00)    NULL
396    12       morning (5-11)          morning (5-11),morning (5-11),morning (5-11),morning (5-11),morning (5-11),morning (5-11),morning (5-11),morning (5-11),morning (5-11),morning (5-11),morning (5-11),morning (5-11)
9      0        night (1-4)              NULL

这大约是我需要得到的表。在 First record night 中有 40 次，因为“晚上 (17-21)”出现了两次，所有出现的次数为 5。2/5*100=40

opens   visits  date_intervals_open evening(17-21)  day(12-16)  morning (5-11)  
213    5        day (12-16)             40          60        0
113    0        evening (17-21)         NULL        NULL      NULL
11     0        late evening (22-00)    NULL        NULL     NULL
396    12       morning (5-11)          0        0     100
9      0        night (1-4)             NULL

PS：我使用 group_concat 只是为了可视化我在那里的价值。我不必使用它，因为之后解析它需要额外的努力。

Answer 1

您基本上需要一个支点，并进行一些进一步的计算。我的回答的基础来自pivoting records in MySQL 上的以下优秀 SO 主题。我假设您有固定数量的date_intervals_visit 值，因为这些值似乎涵盖了一整天，因此我使用具有固定计数的条件计数方法。我将在示例代码中添加 2 个类别，您可以将其扩展为涵盖所有 date_intervals_visit 值。

SELECT 
    count(session_id_open) as opens,
    count(session_id_visit) as visits,
    date_intervals_open,
    round(sum(if(date_intervals_visit='morning (5-11)',1,0)) / count(session_id_visit) * 100,2) as `morning (5-11)`,
    round(sum(if(date_intervals_visit='day (12-16)',1,0)) / count(session_id_visit) * 100,2) as `day (12-16)`
FROM
    bla
GROUP BY date_intervals_open

如果date_intervals_open值可能有0次访问，那么你需要在表达式中检查0：

if(count(session_id_visit)=0, 0, <above formula>)

Answer 2

SELECT 
    count(session_id_open) as opens,
    @visits := count(session_id_visit) as visits,
    date_intervals_open,
    ROUND(100 * SUM(date_intervals_visit = 'evening(17-21)') / @visits) AS 'evening(17-21)',
    ROUND(100 * SUM(date_intervals_visit = 'day (12-16)') / @visits) AS 'day (12-16)',
    ROUND(100 * SUM(date_intervals_visit = 'morning (5-11)') / @visits)'morning (5-11)',
FROM
    bla
GROUP BY date_intervals_open

Answer 3

使用这样的函数：

CREATE FUNCTION [dbo].[fn_SplitString](
    @InputStr   varchar(Max),
    @Seperator  varchar(10))
RETURNS @OutStrings TABLE (ItemNo int identity(1,1), Item varchar(256))

AS
BEGIN

    DECLARE @Str varchar(2000),
            @Poz int, @cnt int

    --DECLARE @OutStrings TABLE (Item varchar(2000))

    SELECT @Poz = CHARINDEX (@Seperator, @InputStr), @cnt = 0
    WHILE @Poz > 0 AND @cnt <= 10000
    BEGIN
        SELECT @Str = SubString(@InputStr, 1, @Poz - 1)
        INSERT INTO @OutStrings(Item) VALUES(@Str)

        SELECT @InputStr = Right(@Inputstr, Len(@InputStr) - (len(@Str) + len(@Seperator)))
        SELECT @Poz = CHARINDEX (@Seperator, @InputStr), @cnt = @cnt + 1
    END
    IF @InputStr <> ''
    BEGIN
        INSERT INTO @OutStrings(Item) VALUES(@InputStr)
    END

    RETURN
END

通过以下方式：

SELECT  opens,
        visits,
        date_intervals_open,
        [evening(17-21)]/[All]*100 AS [evening(17-21)],
        [day(12-16)]/[All]*100 AS [day(12-16)],
        [morning (5-11)]/[All]*100 AS [morning (5-11)]
FROM
    (   
    SELECT 
        count(session_id_open) as opens,
        count(session_id_visit) as visits,
        date_intervals_open,
        (SELECT COUNT(Item) FROM [dbo].[fn_SplitString](LTRIM(RTRIM(group_concat(date_intervals_visit))), ',') WHERE item  = 'evening(17-21)') AS [evening(17-21)],
        (SELECT COUNT(Item) FROM [dbo].[fn_SplitString](LTRIM(RTRIM(group_concat(date_intervals_visit))), ',') WHERE item = 'day(12-16)') AS [day(12-16)],
        (SELECT COUNT(Item) FROM [dbo].[fn_SplitString](LTRIM(RTRIM(group_concat(date_intervals_visit))), ',') WHERE item = 'morning (5-11)') AS [morning (5-11)],
        (SELECT COUNT(Item) FROM [dbo].[fn_SplitString](LTRIM(RTRIM(group_concat(date_intervals_visit))), ',')) AS [All]
    FROM
        bla
    GROUP BY date_intervals_open
    )blabla