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in Technique[技术] by (71.8m points)

sql server - Each GROUP BY expression must contain at least one column that is not an outer reference

What am I doing wrong here? I am getting this error on:

SELECT LEFT(SUBSTRING(batchinfo.datapath, PATINDEX('%[0-9][0-9][0-9]%', batchinfo.datapath), 8000), 
            PATINDEX('%[^0-9]%', SUBSTRING(batchinfo.datapath, PATINDEX('%[0-9][0-9][0-9]%', 
            batchinfo.datapath), 8000))-1),
            qvalues.name,
            qvalues.compound,
            qvalues.rid
FROM batchinfo JOIN qvalues ON batchinfo.rowid=qvalues.rowid
WHERE LEN(datapath)>4
GROUP BY 1,2,3
HAVING rid!=MAX(rid)

I would like to group by the first, second, and third columns having the max rid.

It works fine without the group by and having.

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by (71.8m points)

To start with you can't do this:

having rid!=MAX(rid)

The HAVING clause can only contain things which are attributes of the aggregate groups.

In addition, 1, 2, 3 is not valid in GROUP BY in SQL Server - I think that's only valid in ORDER BY.

Can you explain why this isn't what you are looking for:

select 
LEFT(SUBSTRING(batchinfo.datapath, PATINDEX('%[0-9][0-9][0-9]%', batchinfo.datapath), 8000), PATINDEX('%[^0-9]%', SUBSTRING(batchinfo.datapath, PATINDEX('%[0-9][0-9][0-9]%', batchinfo.datapath), 8000))-1),
qvalues.name,
qvalues.compound,
MAX(qvalues.rid)
 from batchinfo join qvalues on batchinfo.rowid=qvalues.rowid
where LEN(datapath)>4
group by LEFT(SUBSTRING(batchinfo.datapath, PATINDEX('%[0-9][0-9][0-9]%', batchinfo.datapath), 8000), PATINDEX('%[^0-9]%', SUBSTRING(batchinfo.datapath, PATINDEX('%[0-9][0-9][0-9]%', batchinfo.datapath), 8000))-1),
qvalues.name,
qvalues.compound

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