多個加入與和，並返回錯誤的結果

Question

SELECT `apt`.`id` as `aprtid`, 
    `apt`.`apt_no` as `aptno`, 
    `apt`.`core` as `aptcore`, 
    `apt`.`floor` as `aptfloor`, 
    `apt`.`bd_code` as `aptbuld`, 
    `bd`.`building_name` as `buldname`, 
    `rs`.`resident_name` as `rsdntname`, 
    `rs`.`resident_email` as `rsdntemail`, 
    `rs`.`resident_phone` as `rsdntphone`, 
    `rs`.`resident_pic` as `rsdntpic` 
FROM `apartments` as `apt` 
INNER JOIN `buildings` as `bd` 
    ON `bd`.`bd_code` = `apt`.`bd_code` 
INNER JOIN `residents` as `rs` 
    ON `rs`.`apartment_no` = `apt`.`apt_no` 
WHERE `apt`.`id` = '8' 
ORDER BY `apt`.`id` DESC 

當我使用此查詢它返回某種程度上acurate但與居民和公寓的情況下加入與它不匹配。 因爲我有匹配的居民和公寓我有相同的公寓沒有，但都有不同的建築物，但作爲一個輸出其返回我的結果如何可能我在這個查詢中做錯了。爲什麼我得到的結果我認爲和條件不起作用，但。

Answer 1

推測，您的join條件不完整。我猜想：

FROM `apartments` as `apt` INNER JOIN 
    `buildings` as `bd` 
    ON `bd`.`bd_code` = `apt`.`bd_code` INNER JOIN 
    `residents` as `rs` 
    ON `rs`.`apartment_no` = `apt`.`apt_no` AND 
     rs.bd_code = apt.bd_code 

或者：

FROM `apartments` as `apt` INNER JOIN 
    `buildings` as `bd` 
    ON `bd`.`bd_code` = `apt`.`bd_code` INNER JOIN 
    `residents` as `rs` 
    ON `rs`.`apt_id` = `apt`.`apt_id`