sql-server - T-SQL STRING_AGG 问题不知道是写得不好还是不工作

Question

为什么我在第一部分没有看到我的 i.inventoryId 的分离？但是当我在另一个选择中选择它们时，我会得到更多的结果。还有“绝地”==（PlayerAvatarId 6）

https://i.ibb.co/XW4NJC7/image.png

SELECT p.FirstName [Spelers Voornaam]
    ,p.LastName [Spelers Achternaam]
    ,pa.AvatarName [Spelers Avatarnaam]
    ,pa.FamilyName [Familie's Groeps Naam]
    ,Av.Type [Avatar's Type]
    ,string_agg (i.InventoryId, ',') as [In Inventory]

FROM Player AS p
LEFT JOIN PlayerAvatar AS pa ON p.PlayerId = pa.PlayerId
LEFT JOIN Avatar AS Av ON pa.AvatarId = Av.AvatarId
LEFT JOIN Avatar AS a ON pa.AvatarId = a.AvatarId
LEFT JOIN Inventory as i on  i.InventoryId = pa.InventoryId
LEFT JOIN Item as it on it.ItemId = i.ItemId
WHERE pa.AvatarName = 'Jedi'
GROUP BY p.FirstName, p.LastName, pa.AvatarName, pa.FamilyName, av.Type, i.InventoryId

SELECT *
from Inventory
Where PlayerAvatarId = 6

Answer 1

如果这是您要连接的内容，我认为您不想按 InventoryId 分组...试试这个：

编辑，您需要删除逐行不同的列。

SELECT p.FirstName [Spelers Voornaam]
    ,p.LastName [Spelers Achternaam]
    ,pa.FamilyName [Familie's Groeps Naam]
    ,string_agg (i.InventoryId, ',') as [In Inventory]
FROM Player AS p
LEFT JOIN PlayerAvatar AS pa ON p.PlayerId = pa.PlayerId
LEFT JOIN Avatar AS Av ON pa.AvatarId = Av.AvatarId
LEFT JOIN Avatar AS a ON pa.AvatarId = a.AvatarId
LEFT JOIN Inventory as i on  i.InventoryId = pa.InventoryId
LEFT JOIN Item as it on it.ItemId = i.ItemId
GROUP BY p.FirstName, p.LastName, pa.AvatarName, pa.FamilyName, av.Type

或者您也可以聚合这些列。

SELECT p.FirstName [Spelers Voornaam]
    ,p.LastName [Spelers Achternaam]
    ,string_agg(pa.AvatarName,',') [Spelers Avatarnaam]
    ,pa.FamilyName [Familie's Groeps Naam]
    ,string_agg(Av.Type,',') [Avatar's Type]
    ,string_agg (i.InventoryId, ',') as [In Inventory]

FROM Player AS p
LEFT JOIN PlayerAvatar AS pa ON p.PlayerId = pa.PlayerId
LEFT JOIN Avatar AS Av ON pa.AvatarId = Av.AvatarId
LEFT JOIN Avatar AS a ON pa.AvatarId = a.AvatarId
LEFT JOIN Inventory as i on  i.InventoryId = pa.InventoryId
LEFT JOIN Item as it on it.ItemId = i.ItemId
GROUP BY p.FirstName, p.LastName, pa.FamilyName,