MySQL的：我怎样写选取ID不是另一个表

Question

 
I have a simple structure of 2 tables: contacts and group_contacts. 

A contact can belong to one, many or no groups. 

I'm trying to write a select statement that will give me all the contacts 
that don't belong to group_id '123'. The negative, don't, has me confused. 

CREATE TABLE IF NOT EXISTS `contacts` (
    `contact_id` bigint(20) unsigned NOT NULL AUTO_INCREMENT, 
    `name` varchar(50) NOT NULL) 

CREATE TABLE IF NOT EXISTS `group_contacts` (
    `contact_id` bigint(20) unsigned NOT NULL, 
    `group_id` int(11) unsigned NOT NULL) 



thanks 

Answer 1

您将有两个宏步进行的：

1. 首先，您需要在联系人和group_contacts表之间留下外部联结，以便选择所有在group_contacts表中具有且没有任何相关联系的联系人
2. 然后在group_contacts的where子句中应该加载123的group_id。

Answer 2

试试这个

select * from contacts a left join group_contacts b 
on a.contact_id = b.contact_id 
where b.group_id !=123 

Answer 3

select * from contacts as ct 
left join group_contacts as gc on ct.contact_id=gc.contact_id 
where gc.group_id!=123 

Answer 4

select a.contact_id from contacts a, group_contacts b where b.group_id<>123 and b.contact_id=a.contact_id;