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rlfprog

Does Conversation Exist?

Jan 4th, 2015
260
0
Never
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SQL 0.30 KB | None | 0 0
  1. SELECT cu.id
  2. FROM conversation_user cu
  3. WHERE cu.conversation_id = 1
  4.   AND EXISTS
  5.     (SELECT *
  6.      FROM conversation_user
  7.      WHERE conversation_id = 1
  8.        AND user_id = 32)
  9.   AND EXISTS
  10.     (SELECT *
  11.      FROM conversation_user
  12.      WHERE conversation_id = 1
  13.        AND user_id = 3);
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