sphinx для слейва отдельным индексом
Posted: Thu Jan 21, 2021 3:12 am
Можно ли такое сделать? Отдельный индекс отдельную дельту
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Code: Select all
sql_query_pre = UPDATE rot_settings SET value = (SELECT MAX(gallery_id) FROM rot_gallery_info) WHERE name = 'sphinx_max_gallery_id3'
sql_query = SELECT gi.gallery_id, UNIX_TIMESTAMP(gi.activation_date) as date, \
\
gd.alt, gd.description, gi.duration, gi.sponsor_id, gs.total_ctr, gi.content_type, rot_sponsors.sponsor_name, rot_sponsors.sponsor_site,\
\
(SELECT group_concat(tag_id) FROM rot_gal2tag g2t \
WHERE g2t.gallery_id = gi.gallery_id) as tag_ids, \
\
(SELECT group_concat(tag_name) FROM rot_gal2tag g2t \
LEFT JOIN rot_tags as t on t.tag_id = g2t.tag_id \
WHERE g2t.gallery_id = gi.gallery_id) as tags, \
\
(SELECT group_concat(name) FROM rot_groups \
WHERE rot_groups.id in (SELECT group_id FROM rot_gallery_stats3 \
WHERE rot_gallery_stats3.gallery_id = gi.gallery_id AND group_id != 0) \
) as group_names, \
\
(SELECT group_concat(model_name) FROM rot_models \
WHERE rot_models.model_id in (SELECT model_id FROM rot_gal2model \
WHERE rot_gal2model.gallery_id = gi.gallery_id) \
) as model_names, \
\
(SELECT group_concat(distinct(gss.group_id)) FROM rot_gallery_stats3 as gss \
WHERE gss.gallery_id = gi.gallery_id AND group_id != 0) as categories \
\
FROM rot_gallery_info AS gi \
JOIN rot_sponsors on rot_sponsors.sponsor_id = gi.sponsor_id \
JOIN rot_gallery_data3 AS gd ON gi.gallery_id = gd.gallery_id \
JOIN rot_gallery_stats3 AS gs ON gs.gallery_id = gi.gallery_id \
WHERE gallery_status = 'active' and gallery_type = 0 \
and gs.group_id = 0 \
AND gs.group_id NOT IN (SELECT group_id FROM rot_groups_data WHERE status != 1)