2013-07-25 18 views
-4

profile_table我有兩個表格配置文件和profil_calculation。配置表看起來像

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 

我想導致像.....

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

請幫助.... ......

+1

是什麼在profil_description輪廓FK?似乎沒有。 –

+0

profile_description表中的calc_id是profile_table中profile_id的外鍵? –

+0

@AyyappanSekar你怎麼能這麼說? –

回答

1

這將是這樣的:

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; 

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; 
0
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 
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