2016-04-24 45 views
0

我有两个表一个是random_questions,另一个是personal_questions,II想在personal_questions其中personal_questions.question_to = current_session_userpersonal_question.copy_id = random_question.id ....如何准备查询从另一个表检查,然后从另一个表中提取

检查copy_id列这里是我的代码:

$session_user = $_SESSION['user_id']; 

$sql = mysqli_query($con, "SELECT DISTINCT random_questions.id AS id, random_questions.question_by_id AS question_by_id, random_questions.question AS question, random_questions.total_answers AS answers FROM random_questions INNER JOIN personal_questions WHERE (random_questions.total_answers != '8') OR (personal_questions.question_to != '$session_user' AND personal_questions.copy_id != random_questions.id)"); 

我没有得到的结果正确,如果personal_questions的copy_id是等于任何random_questions的id,然后自动产生重复。我希望你明白。请帮助。

回答

0
Select * 
from random_questions p, 
     personal_questions a 
where a.question_to = current_session_user 
and a.copy_id  = p.id;