0

2 番目のテーブルに一致し、存在しない 3 つのエントリのグループを選択する次のクエリがあります。

以下のように、個々のエントリ ID ではなく、見つかったグループの数を「count_result」経由で返す方法を見つける必要があります。

どうすればこれを達成できますか？

SELECT COUNT(sub.entry_id) as count_result
FROM exp_submissions AS sub
LEFT JOIN exp_judging_portfolios AS jud1 ON sub.entry_id = jud1.entry_id_1
LEFT JOIN exp_judging_portfolios AS jud2 ON sub.entry_id = jud1.entry_id_2
LEFT JOIN exp_judging_portfolios AS jud3 ON sub.entry_id = jud1.entry_id_3
WHERE jud1.entry_id_1 IS NULL
AND jud2.entry_id_2 IS NULL
AND jud3.entry_id_3 IS NULL
AND sub.member_group = 6
AND sub.type_id = 1
GROUP BY sub.member_id, sub.portfolio_number
HAVING count(sub.portfolio_number) = 3
4

2 に答える 2

1

サブクエリは必要ありません。

SELECT COUNT(DISTINCT(sub.member_id, sub.portfolio_number)) as count_result
FROM exp_submissions AS sub
LEFT JOIN exp_judging_portfolios AS jud1 ON sub.entry_id = jud1.entry_id_1
LEFT JOIN exp_judging_portfolios AS jud2 ON sub.entry_id = jud1.entry_id_2
LEFT JOIN exp_judging_portfolios AS jud3 ON sub.entry_id = jud1.entry_id_3
WHERE jud1.entry_id_1 IS NULL
AND jud2.entry_id_2 IS NULL
AND jud3.entry_id_3 IS NULL
AND sub.member_group = 6
AND sub.type_id = 1
GROUP BY sub.member_id, sub.portfolio_number
HAVING count(sub.portfolio_number) = 3
于 2013-10-02T09:44:26.393 に答える
0

結果をカウントするだけの別のクエリのソースとしてクエリを使用します。

SELECT COUNT(*) FROM (
      -- YOUR QUERY GOES HERE --
) AS t
于 2013-10-02T09:43:53.180 に答える