我把它变成Group By hostId，然后它只会显示第一行的值， 如:

A   B  C
1  10
2      3

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

我把它变成Group By hostId，然后它只会显示第一行的值， 如:

A   B  C
1  10
2      3

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;

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

利用Matt Fenwick的想法帮助我解决了这个问题(非常感谢)，让我们把它简化为一个问题:

select
    history.*,
    coalesce(sum(case when itemname = "A" then itemvalue end), 0) as A,
    coalesce(sum(case when itemname = "B" then itemvalue end), 0) as B,
    coalesce(sum(case when itemname = "C" then itemvalue end), 0) as C
from history
group by hostid