Re: tabellenvergleich
Posted by:
Hauke Betz
Date: November 22, 2006 01:03AM
Hallo,
Das folgende Statement würde alle Datensätze aus Tab2 ausgeben, deren Id nicht in der ersten vorhanden ist. Habe es zwar auf nem 5er-Server getestet aber es sollte auch unter 4 keine Probleme machen:
select tab2.* from tab2 left join tab1 on tab2.id = tab1.id where tab1.id is null;
Gruß,
Hauke
Edited 2 time(s). Last edit at 11/22/2006 01:10AM by Hauke Betz.
Subject
Views
Written By
Posted
5697
November 21, 2006 10:15AM
Re: tabellenvergleich
3004
November 22, 2006 01:03AM
2794
November 22, 2006 03:20AM
2609
November 23, 2006 01:06AM
Sorry, you can't reply to this topic. It has been closed.
Content reproduced on this site is the property of the respective copyright holders.
It is not reviewed in advance by Oracle and does not necessarily represent the opinion
of Oracle or any other party.