Php 2个表的左联接查询失败

我正在尝试获取两个表的左外部联接，但我无法执行此查询，phpmyadmin在运行此查询时在第12行出现1064错误：

SELECT 
pt.id as planid,
pt.trip_name,
pt.description,
cor.latitude,
cor.longitude,
bb.id as bookmarkid,
bb.num_of_persons as persons
FROM 
planned_trips as pt,
coordinates as cor,
LEFT JOIN Bookmarkedby as bb,Users as user
ON 
user.id = 1 AND
user.id = bb.user_id AND
bb.plannedtrips_id = pt.id AND
pt.coordinates_id = cor.id'

我挣扎了一个小时，我错过了什么？？ 我的数据库架构如下所示：

我目前正在准备我的查询，我需要在codeIgnitor上运行此查询。

每个表都需要单独的左联接

SELECT 
    pt.id as planid,
    pt.trip_name,
    pt.description,
    cor.latitude,
    cor.longitude,
    bb.id as bookmarkid,
    bb.num_of_persons as persons
FROM planned_trips as pt
INNER JOIN coordinates as cor ON pt.coordinates_id = cor.id
LEFT JOIN Bookmarkedby as bb ON bb.plannedtrips_id = pt.id
LEFT JOIN Users as user ON user.id = bb.user_id AND user.id = 1

每个表都需要单独的左联接

SELECT 
    pt.id as planid,
    pt.trip_name,
    pt.description,
    cor.latitude,
    cor.longitude,
    bb.id as bookmarkid,
    bb.num_of_persons as persons
FROM planned_trips as pt
INNER JOIN coordinates as cor ON pt.coordinates_id = cor.id
LEFT JOIN Bookmarkedby as bb ON bb.plannedtrips_id = pt.id
LEFT JOIN Users as user ON user.id = bb.user_id AND user.id = 1

问题是在左边的行中加入Bookmarkedby作为bb，用户作为user。考虑将其改为

FROM 
planned_trips as pt
JOIN coordinates as cor ON pt.coordinates_id = cor.id
LEFT JOIN Bookmarkedby as bb ON bb.plannedtrips_id = pt.id
LEFT JOIN Users as user ON user.id = bb.user_id 
AND user.id = 1 

问题是在左边的行中加入Bookmarkedby作为bb，用户作为user。考虑将其改为

FROM 
planned_trips as pt
JOIN coordinates as cor ON pt.coordinates_id = cor.id
LEFT JOIN Bookmarkedby as bb ON bb.plannedtrips_id = pt.id
LEFT JOIN Users as user ON user.id = bb.user_id 
AND user.id = 1 

---

不能在左联接后放置两个表名。对于要连接的每个表，您都需要一个单独的左连接子句。对于其他表，您还应该使用显式的内部连接子句。问题是在哪里合并，内部连接查询与左连接查询为什么您要与用户连接，但在SELECT子句中不使用该表中的任何内容？@Drew这真的是一个很好的副本吗？这个问题的问题是缺少一组人。他的左连接语法已经正确了。标题与此匹配，但实际问题与此不匹配。不能将两个表名放在LEFT JOIN之后。对于要连接的每个表，您都需要一个单独的左连接子句。对于其他表，您还应该使用显式的内部连接子句。问题是在哪里合并，内部连接查询与左连接查询为什么您要与用户连接，但在SELECT子句中不使用该表中的任何内容？@Drew这真的是一个很好的副本吗？这个问题的问题是缺少一组人。他的左连接语法已经正确了。标题与此匹配，但实际问题与此不匹配。因此，我们可以重复使用单独的左键连接，这对我来说是新的谢谢我们可以重复使用单独的左键连接，这对我来说是新的谢谢