MySQL -行到列

我试着搜索帖子，但我只找到SQL Server/Access的解决方案。我需要一个解决方案在MySQL (5.X)。

我有一个表(称为历史)3列:hostid, itemname, itemvalue。如果我执行select (select * from history)，它会返回

   +--------+----------+-----------+
   | hostid | itemname | itemvalue |
   +--------+----------+-----------+
   |   1    |    A     |    10     |
   +--------+----------+-----------+
   |   1    |    B     |     3     |
   +--------+----------+-----------+
   |   2    |    A     |     9     |
   +--------+----------+-----------+
   |   2    |    C     |    40     |
   +--------+----------+-----------+

如何查询数据库以返回类似的内容

   +--------+------+-----+-----+
   | hostid |   A  |  B  |  C  |
   +--------+------+-----+-----+
   |   1    |  10  |  3  |  0  |
   +--------+------+-----+-----+
   |   2    |   9  |  0  |  40 |
   +--------+------+-----+-----+

当前回答

SELECT 
    hostid, 
    sum( if( itemname = 'A', itemvalue, 0 ) ) AS A,  
    sum( if( itemname = 'B', itemvalue, 0 ) ) AS B, 
    sum( if( itemname = 'C', itemvalue, 0 ) ) AS C 
FROM 
    bob 
GROUP BY 
    hostid;

2009-08-10 13:12:13

其他回答

这不是你正在寻找的确切答案，但这是一个解决方案，我需要在我的项目，希望这有助于某人。这将列出用逗号分隔的1到n行项目。Group_Concat使这在MySQL中成为可能。

select
cemetery.cemetery_id as "Cemetery_ID",
GROUP_CONCAT(distinct(names.name)) as "Cemetery_Name",
cemetery.latitude as Latitude,
cemetery.longitude as Longitude,
c.Contact_Info,
d.Direction_Type,
d.Directions

    from cemetery
    left join cemetery_names on cemetery.cemetery_id = cemetery_names.cemetery_id 
    left join names on cemetery_names.name_id = names.name_id 
    left join cemetery_contact on cemetery.cemetery_id = cemetery_contact.cemetery_id 

    left join 
    (
        select 
            cemetery_contact.cemetery_id as cID,
            group_concat(contacts.name, char(32), phone.number) as Contact_Info

                from cemetery_contact
                left join contacts on cemetery_contact.contact_id = contacts.contact_id 
                left join phone on cemetery_contact.contact_id = phone.contact_id 

            group by cID
    )
    as c on c.cID = cemetery.cemetery_id


    left join
    (
        select 
            cemetery_id as dID, 
            group_concat(direction_type.direction_type) as Direction_Type,
            group_concat(directions.value , char(13), char(9)) as Directions

                from directions
                left join direction_type on directions.type = direction_type.direction_type_id

            group by dID


    )
    as d on d.dID  = cemetery.cemetery_id

group by Cemetery_ID

这个墓地有两个公共名称，因此名称被列在不同的行中，由一个id连接，但有两个名称id，查询产生如下内容 CemeteryID Cemetery_Name 纬度 1 阿普尔顿,Sulpher弹簧35.4276242832293

2013-10-10 14:58:42

你可以使用几个LEFT join。请使用此代码

SELECT t.hostid,
       COALESCE(t1.itemvalue, 0) A,
       COALESCE(t2.itemvalue, 0) B,
       COALESCE(t3.itemvalue, 0) C 
FROM history t 
LEFT JOIN history t1 
    ON t1.hostid = t.hostid 
    AND t1.itemname = 'A' 
LEFT JOIN history t2 
    ON t2.hostid = t.hostid 
    AND t2.itemname = 'B' 
LEFT JOIN history t3 
    ON t3.hostid = t.hostid 
    AND t3.itemname = 'C' 
GROUP BY t.hostid

2022-02-02 03:25:03

我编辑阿贡Sagita的答案从子查询加入。我不确定这两种方式有多大区别，但只是作为另一个参考。

SELECT  hostid, T2.VALUE AS A, T3.VALUE AS B, T4.VALUE AS C
FROM TableTest AS T1
LEFT JOIN TableTest T2 ON T2.hostid=T1.hostid AND T2.ITEMNAME='A'
LEFT JOIN TableTest T3 ON T3.hostid=T1.hostid AND T3.ITEMNAME='B'
LEFT JOIN TableTest T4 ON T4.hostid=T1.hostid AND T4.ITEMNAME='C'

2012-12-19 07:22:31

SELECT 
    hostid, 
    sum( if( itemname = 'A', itemvalue, 0 ) ) AS A,  
    sum( if( itemname = 'B', itemvalue, 0 ) ) AS B, 
    sum( if( itemname = 'C', itemvalue, 0 ) ) AS C 
FROM 
    bob 
GROUP BY 
    hostid;

2009-08-10 13:12:13

如果你可以使用MariaDB，有一个非常非常简单的解决方案。

自MariaDB-10.02以来，添加了一个新的存储引擎CONNECT，可以帮助我们将另一个查询或表的结果转换为数据透视表，就像你想要的那样: 你可以看看这些文件。

首先安装connect存储引擎。

现在我们的表的主列是itemname，每一项的数据都位于itemvalue列中，所以我们可以使用这个查询得到结果数据透视表:

create table pivot_table
engine=connect table_type=pivot tabname=history
option_list='PivotCol=itemname,FncCol=itemvalue';

现在我们可以从数据透视表中选择我们想要的:

select * from pivot_table

详情请点击这里

2019-06-19 15:06:29

MySQL -行到列

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