php - 我有两个表配置文件和 profile_calculation。配置文件表看起来像

Question

profile_table

profile_id profile_name Profile_description
1          VINU         ADMIN

profile_description 表

calc_id proile_calc profile_result
1       20            45
2       30            43
3       42            82

我想要这样的结果......

profile_id profile_name Profile_description calc_id proile_calc profile_result
  1             VINU          ADMIN             1       20            45
                                                2       30            43
                                                3       42            82

请帮忙..........

Answer 1

它会是这样的：

SELECT pt.profile_id, pt.profile_name , pt.Profile_description, pd.calc_id , pd.proile_calc, pd.profile_result FROM profile_table as pt LEFT JOIN profile_description as pd ON pt.profile_id = pd.calc_id;

或者

SELECT profile_table.profile_id, profile_table.profile_name , profile_table.Profile_description, profile_description.calc_id , profile_description.proile_calc, profile_description.profile_result FROM profile_table LEFT JOIN profile_description ON profile_table.profile_id = profile_description.calc_id;

Answer 2

Table is poorly design. No FK in profile_description table. but if i suppose calc_id is FK 
in profile_description table  then 
this query will help you.

SELECT `profile_table`.*,`profile_description`.*
FROM profile_table
RIGHT JOIN profile_description
ON profile_table.profile_id=profile_description.calc_id