i think it is wrong

this query below must return LDM

BECAUSE employee no 407133 exist on two columns rean83 and rean84

DROP TABLE IF EXISTS #F6000059;

CREATE TABLE #F6000059

(

RequestNo INT NULL

,REAN82 INT NULL

,REAN83...

I prefere using check sum

my case have numbers only on feature id and feature value

also i working on sql server 2019

so are using CHECKSUM_AGG will working

or not working good

yes

but issue on performance

so can i write statement above with another way best

yes i put index for

CREATE UNIQUE CLUSTERED INDEX cdx ON dbo.PartsHaveBestDisplayOrder (CodeRulesId,PackageId,DisplayOrder,PartId);

but still slow because data is big

so are there are any solution for large amount of data

by rewrite query...

• This reply was modified 2 years, 1 month ago by  ahmed_elbarbary.2010.

thanks for support

thanks much for support

i need only to use stuff instead of string aggreagte and heck time and performance

i post counts so what i do after that

or what query best for achieve that quickly

SELECT f1_AllCount = COUNT(f1.PartId)
,f1_UnqCount = COUNT(DISTINCT f1.PartID)
FROM Technology.Receipe ft
JOIN ExtractReports.dbo.TechnologyPlPartsFeaturValuesOrg f1 ON ft.featureid = f1.featureid
WHERE ft.operatorid = 1;

result is

f1_AllCount         f1_UnqCount
124140   ...

so how to use cross apply as you write below

The optimizer attempts to SORT 65 million rows after a merge join.  Yeeeesh.   One way to re-write the query would be...

i found issue

os.path.isdir()  working local only

so use another thing working remote to check directory

so How to handle that

it remain without change

meaning after apply script

nothing change

but not append to existing file

an nothing change

the goal mfrom asking question is to get result above without string aggreagte or comma separated

suppose i have

p1 1,2,3

p2 2,2,2

both p1 and p2 both have same count and same sum

so...

yes exactly

thanks

if Feature name and Feature value not exist on table #partsfeature  then i will apply dense rank to take new number for technology id

if Feature name and Feature value exist...

after insert part id below

(3900,'grail','51V',NULL)

Null will be 2 because grail feature name and value 51V exist before with technologyid 2