Question

对于第一个1265的ID，我只想要一个这样的记录（见下图），现在我有三个不同的记录用于相同的ID并且不好。你可以看到在ID = 1265的第一个记录中，我希望将字符串连接成一个用'，'分隔的字符串：

1265 MARK, LISA, MAY

这是我的代码示例：

select 
    vup.UgovorId as ID, 
    concat(p.Naziv, p.Naziv) as NUMBER 
from 
    [TEST_Ugovori_Prod].dbo.VezaUgovorPartner as vup 
inner join 
    [TEST_MaticniPodaci2].[dbo].[Partner] as p on vup.PartnerId = p.PartnerID 
group by 
    vup.UgovorId, p.Naziv
order by 
    vup.UgovorId desc

以下是我的结果：

1265    MARK  
1265    LISA
1265    MAY
1264    LINDA  
1263    MARTINA
1262    MARKO                                                                                                                                                                                
1261    VIDAKS
1260    PEEKS 
1259    MARCUS
1258    MARKO                                                                                                                                                                                
1257    MATIA                                                                                                                                                                                 
1256    POPOVIC

Answer 1

SELECT ID, NUMBER = 
    STUFF((SELECT DISTINCT ', ' + NUMBER
           FROM example t2 
           WHERE t2.ID = t1.ID
          FOR XML PATH('')), 1, 2, '')
FROM example t1
GROUP BY ID
ORDER BY ID DESC

[DEMO HERE]

如果您在使用上述查询时遇到问题，这应该会有所帮助：

SELECT vup.UgovorId as ID, 
     STUFF((SELECT DISTINCT ', ' + p2.Naziv  
           FROM [TEST_Ugovori_Prod].dbo.VezaUgovorPartner vup2 
           INNER JOIN [TEST_MaticniPodaci2].[dbo].[Partner] p2 ON vup2.PartnerId = p2.PartnerID 
           WHERE vup2.UgovorId = vup.UgovorId   
           FOR XML PATH('')), 1, 2, '') NUMBER
FROM [TEST_Ugovori_Prod].dbo.VezaUgovorPartner vup 
INNER JOIN [TEST_MaticniPodaci2].[dbo].[Partner] p on vup.PartnerId = p.PartnerID 
GROUP BY vup.UgovorId
ORDER BY vup.UgovorId DESC

[Test blend]

如果同一ID

1 个答案: