Full Outer Join in ACC | Adobe Higher Education
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pradyumns539618
Level 2
February 25, 2020
解決済み

Full Outer Join in ACC

  • February 25, 2020
  • 2 の返信
  • 7195 ビュー

Hi ,

 

Does Adobe Campaign support Full Outer Join ?

My requirement is : We have two table A(Customer) and B(Restaurant) ,Table A has column (X coordinates  and Y coordinates ) geographic location of Customer and Table B has column ((X coordinates  and Y coordinates ) geographic location of Restaurant .

I would like create 1 recordset of data from both table having the all four columns . I tried with Union but it failed , as there is no link between both tables .So we can't do it using reconciliation . Please let us know if is there any workaround for it .

 

Please let me know if you need more detail on it .

 

Thanks!

 

@jonathon_wodnicki @jean-serge_biro @venu_reddy  @kapilkochar 

このトピックへの返信は締め切られました。
ベストアンサー Jonathon_wodnicki

Hi,

 

Cartesian join? Every customer maps to every restaurant?

This can be done with either schema definition, join key is 1=1, or enrichment with simple join (any cardinality is fine), 1=1.

 

Thanks,

-Jon

2 の返信

Adhiyan
Adobe Employee
AdhiyanAdobe Employee
Adobe Employee
February 25, 2020

Hi Pradyumn,

 

Did you try enrichment in this scenario. It should be possible to add the table B as a link and extract the columns.

You should query the records of table A , then add a enrichment activity and add the data as a link for table B.

 

In the enrichment , select add data and then select A link :

 

 

In the next step , choose the table B and it would extract the columns from that table.

 

Regards,
Adhiyan

 

P
pradyumns539618作成者
Level 2
March 17, 2020
Hi Adhiyan , Sorry for late response ! I can't see any option here to mark as answered .
Jonathon_wodnicki
Community Advisor
Jonathon_wodnickiCommunity Advisor回答
Community Advisor
February 25, 2020

Hi,

 

Cartesian join? Every customer maps to every restaurant?

This can be done with either schema definition, join key is 1=1, or enrichment with simple join (any cardinality is fine), 1=1.

 

Thanks,

-Jon

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