Question

我的数据库中有三个表：评论，地标，用户，我希望从每个表中获取一些属性，我尝试使用以下查询，但没有重新启动，这是我的3个表

user (userid,firstName,lastName)
comment (userid,L_ID,review)
landmarks (L_ID,Title)

用户和对userid的评论和L_ID

上的地标评论

  $q = mysqli_query($con,"SELECT l.*,c.id,c.review,CONCAT(u.firstName,' ',u.lastName) name FROM landmarks l inner join comment c on l.L_ID on c.L_ID inner join users u on u.userid=c.userid WHERE ".(isset($userid) ? "c.userid=$userid" : (isset($landmarkid) ? "c.L_ID=$landmarkid" : "") )." ORDER BY c.id");

Answer 1

我相信你的＆＃34;内部联接评论c＆＃34;

的ON部分有错误

SELECT l.*,c.id,c.review,CONCAT(u.firstName,' ',u.lastName) name 
FROM landmarks l 
inner join comment c 
    on l.L_ID 
    on c.L_ID 
inner join users u 
    on u.userid=c.userid 
WHERE ".(isset($userid) ? "c.userid=$userid" : (isset($landmarkid) ? "c.L_ID=$landmarkid" : "") )." 
ORDER BY c.id

我相信你想要的是

SELECT l.*,c.id,c.review,CONCAT(u.firstName,' ',u.lastName) name 
FROM landmarks l 
inner join comment c 
    on c.L_ID = l.L_ID 
inner join users u 
    on u.userid=c.userid 
WHERE ".(isset($userid) ? "c.userid=$userid" : (isset($landmarkid) ? "c.L_ID=$landmarkid" : "") )." 
ORDER BY c.id

内连接有三个表

1 个答案: