2016-08-11 57 views
1

我试图用主表中的值创建一些表。专家在PHP中,但不在MySQL中。 主表中有此列:MySQL,选择内部选择独特返回多个值

Table Places 
ISO 
Country 
Language 
Region2 (is the estate) 
Region4 (is ths city council) 
ID (id for locality) 
Locality 

获得国家也并不困难:

CREATE TABLE countries (id integer(11) UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY, 
iso varchar(2) NOT NULL, language varchar(2) NOT NULL, 
name varchar(50) NOT NULL) ENGINE=MyISAM DEFAULT CHARSET=utf8; 

INSERT INTO countries (iso , language, name) 
SELECT DISTINCT ISO AS iso, Language as language, Country AS name FROM Places WHERE 1; 

现在我要创建国家,议会和城市,我一直在与各国试图两天像这样(我试过一些不同的代码):

CREATE TABLE states (id integer(11) UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY, 
country_id integer(11) UNSIGNED NOT NULL, country_iso varchar(2) NOT NULL, 
name varchar(80) NOT NULL) ENGINE=MyISAM DEFAULT CHARSET=utf8; 

INSERT INTO states (country_id, country_iso, name) 
SELECT DISTINCT 
    (SELECT countries.id AS country_id from countries WHERE countries.iso = Places.ISO), 
    ISO as country_iso , 
    Region2 AS name FROM Places WHERE 1; 

但是对于country_id这个选择返回所有的国家ID。 我只需要在country.iso国家代码中找到国家/地区id。

在这个表格之后,我必须创建委员会,从Places中获取价值,再次选择Disposect,并且再次尝试从各个国家获得州ID,也许还需要de国家ID。

请问,任何人都可以让我以这种选择嵌套的方式? 谢谢。

回答

1

看来你需要一个连接

INSERT INTO states (country_id, country_iso, name) 
SELECT DISTINCT countries.countries.id , ISO, Places.Region2 
from countries 
inner join Placesc on countries.iso = Places.ISO