SQL コード:
SELECT id, album_date AS timestamp, CONCAT((SELECT detail_value
FROM people_db.user_details_tbl WHERE detail_field = 'first_name' AND user_id = pictures_db.albums.owner), ' uploaded pictures!') AS title_html
FROM pictures_db.albums
WHERE id IN
(SELECT DISTINCT(album_id)
FROM pictures_db.album_pics
WHERE pic_id IN
(SELECT DISTINCT(picture_id)
FROM pictures_db.picture_access_tbl
WHERE grantee_group_id IN
(SELECT group_id
FROM people_db.group_membership_tbl
WHERE member_id = '2'
)
)
);
PHP コード:
$albums_sql = mysql_query("SELECT id, album_date AS timestamp, CONCAT((SELECT detail_value
FROM people_db.user_details_tbl
WHERE detail_field = 'first_name' AND user_id = pictures_db.albums.owner), ' uploaded pictures!') AS title_html
FROM pictures_db.albums
WHERE id IN (
SELECT DISTINCT(album_id)
FROM pictures_db.album_pics
WHERE pic_id IN (
SELECT DISTINCT(picture_id)
FROM pictures_db.picture_access_tbl
WHERE grantee_group_id IN (
SELECT group_id
FROM people_db.group_membership_tbl
WHERE member_id = '2'
)
)
)") or die(mysql_error());
PHP を実行すると、次のエラーが表示されます: テーブル 'pictures_db.albums' が存在しません
同じユーザーとして実行を試み、すべての権限を再付与し、権限をフラッシュしました。PHP ではなく、シェルで動作します。何か案は?