Question

我几乎在那里，但没有完全返回我需要的结果。

我有3张桌子。 program_location，hcp_data，guest_data，都有正确的主键。

hcp_data中的每一行都包含一个引用（通过外键）到1个程序位置。 guest_data中的每一行都包含对hcp_data中一行的引用。可能有多个guest虚拟机绑定到一个hcp_data条目。将这些表格视为聚会地点，被邀请者以及受邀者选择将他们带到同一地点的人员。因此，通过他们的关系，我可以通过他们链接到的hcp的位置来确定访客的位置。

所以我要做的是从hcp_data和additional_guest表中选择所有行，并从program_locations表中连接几行。本质上，目的是转储数据，使其作为单个平面文件有意义。

这就是我所拥有的：

SELECT lname, fname, degree, email, job_title, office_addr_line_1, office_addr_line_2, hcp_data.city, hcp_data.state, hcp_data.zip, phone, timestamp, program_location.restaurant_name, program_location.city AS restaurant_city, program_location.state AS restaurant_state
FROM hcp_data 
LEFT JOIN program_location
ON hcp_data.program_location_idprogram_location = program_location.idprogram_location
UNION 
SELECT lname, fname, degree, email, job_title, office_addr_line_1, office_addr_line_2, additional_guest.city, additional_guest.state, additional_guest.zip, phone, timestamp, program_location.restaurant_name, program_location.city AS restaurant_city, program_location.state AS restaurant_state
FROM additional_guest
LEFT JOIN program_location
ON

目前最后一个ON是空白的。我相信这是我被困的地方。我想我需要为其中一方做复合选择，但我不确定。

请帮忙！谢谢！

编辑，BTW，每个表的主键只是前面附加“id”的表名，即“idprogram_location” 外键是它们来自的表，下划线，它们来自的表中列的名称：即“program_location_idprogram_location”

Answer 1

所以我已经弄清楚了。如果有更简单或更好的方式（我不怀疑），请分享！

SELECT lname, fname, degree, email, job_title, office_addr_line_1, office_addr_line_2, hcp_data.city, hcp_data.state, hcp_data.zip, phone, timestamp, program_location.restaurant_name, program_location.city AS restaurant_city, program_location.state AS restaurant_state
FROM hcp_data 
LEFT JOIN program_location
ON program_location.idprogram_location = hcp_data.program_location_idprogram_location
UNION 
SELECT lname, fname, degree, email, job_title, office_addr_line_1, office_addr_line_2, additional_guest.city, additional_guest.state, additional_guest.zip, phone, timestamp, program_location.restaurant_name, program_location.city AS restaurant_city, program_location.state AS restaurant_state
FROM additional_guest
LEFT JOIN program_location
ON program_location.idprogram_location = 
(SELECT idprogram_location 
FROM program_location 
WHERE idprogram_location = 
(SELECT program_location_idprogram_location 
FROM hcp_data 
WHERE idhcp_data = 
(SELECT hcp_data_idhcp_data 
FROM additional_guest LIMIT 1)));

具有多个连接和复合选择的MySQL查询

1 个答案: