MYSQL查询使用所有其他参与者的数量来获取用户的参与地点

作者: 电动少女
发布时间: 2024-04-14 09:58:35 (24天前)
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⑩Ⅵ嵗Ｄ夨憶 | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <pre> <code> SELECT p.place_id, p.place_name, c.city, placeCount.Cnt FROM places p INNER JOIN user_place up ON p.place_id = up.place_id INNER JOIN fa_user u ON up.user_id = u.id INNER JOIN city c ON p.city_id = c.id LEFT JOIN ( SELECT place_id, COUNT( DISTINCT user_id ) Cnt FROM user_place WHERE user_id <>53 GROUP BY place_id )placeCount ON p.place_id = placeCount.place_id WHERE up.user_id =53 LIMIT 0 , 30 </code> </pre> </DIV>

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v-star*위위 | 2019-08-31 10-32

<div class =“post-text”itemprop =“text”> <P> 你可以看看这样的东西 </p> <pre> <code> SELECT u.user_id, u.NAME, p.place_id, p.place_name, c.city_id, c.city_name, placeCount.Cnt FROM places p INNER JOIN user_place up ON p.place_id = up.place_id INNER JOIN users u ON up.user_id = u.user_id INNER JOIN city c ON p.city_id = c.city_id LEFT JOIN ( SELECT place_id, COUNT(DISTINCT user_id) Cnt FROM user_place WHERE user_id <> 43 GROUP BY place_id ) placeCount ON p.place_id = placeCount.place_id WHERE p.type_id = 2 AND up.user_id = 43 </code> </pre> </DIV>

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