Question

我想检索具有相同国家/地区名称的所有用户，但除了用户之外，我正在与其他用户进行比较。此查询给出了一个错误，指出子查询返回多行。在这方面，任何人都可以帮助我。

#if defined(_WIN32) || defined(_WIN64)
#  include <gl/glew.h>
#  include <GL/gl.h>
#  include <GL/glu.h>
#elif __APPLE__
#  include "TargetConditionals.h"
#  if (TARGET_OS_IPHONE && TARGET_IPHONE_SIMULATOR) || TARGET_OS_IPHONE
#    include <OpenGLES/ES2/gl.h>
#    include <OpenGLES/ES2/glext.h>
#  else
#    include <OpenGL/gl.h>
#    include <OpenGL/glu.h>
#    include <OpenGL/glext.h>
#  endif
#elif defined(__ANDROID__) || defined(ANDROID)
#  include <GLES2/gl2.h>
#  include <GLES2/gl2ext.h>
#elif defined(__linux__) || defined(__unix__) || defined(__posix__)
#  include <GL/gl.h>
#  include <GL/glu.h>
#  include <GL/glext.h>
#else
#  error platform not supported.
#endif

Answer 1

select *
from users u, address a
where a.email=u.email
and a.country = (select country from address where email = 'ta@gmail.com')
and u.email <> 'ta@gmail.com'

OR

select *
from users u, address a,
(select country from address where email = 'ta@gmail.com') c
where a.email=u.email
and a.country = c.country
and u.email <> 'ta@gmail.com'

Answer 2

我之前没有理解你的问题。试试这个，

SELECT * FROM users s
JOIN address a on a.email = u.email 
WHERE a.country in(SELECT country from address where email = 'ta@gmail.com')
AND a.email != 'ta@gmail.com'

Answer 3

试试这个：

SELECT * FROM users u, address a 
  WHERE a.email=u.email 
  AND u.email != 'ta@gmail.com'
  AND a.country IN (SELECT country FROM address WHERE email ='ta@gmail.com');

检索除具有子查询的元组之外的所有元组

3 个答案: