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Not an alternative, but introductory work

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This was not intended to be an alternative, but for explanation on how to begin such a query. This one only returns all required friendships including those already existing.

Then better delete it completely.
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Stefan Koch 2013-07-30 13:03:38 +02:00
parent 3bb436e81a
commit 5a93365187
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documents/musterloesung-db-2012-09-24/d2c1.2.sql
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SELECT f1.person2, f2.person2
FROM FriendshipSymmetric f1
JOIN FriendshipSymmetric f2 ON f1.person1 = f2.person1
WHERE f1.person2 != f2.person2
AND f1.person1 = <id>;