MySQL的两个表，从一个表中选择所有值，比较是否存在，然后返回一个随机值

Question

I'm new here as for posting questions, but I've been looking for answers since long time ago. Still what I'm trying to do now I didn't found for the right solution or I just didn't know how to use it if I did found it.

Here's my SQL query:

$sql = "SELECT * FROM table_links AS link_id WHERE link_points > '1' 
AND NOT EXISTS (SELECT * FROM table_surf AS slink_id WHERE user_id = '$user'
AND surf_time = '$time') ORDER by RAND() LIMIT 1";

I'm trying to select a random LINK id (wich has points) and check if that link was allready shown to a user today by looking into the 2nd table where all LINK ID's are saved as SLINK ID's with user ID's and time of the last visits.

If last visit was today ( $time ) then it should not return anything.

I'm new to PHP/mysql but I'm trying to learn, still this is a situation where I'm not sure if it is even possible to realise a complex query like this one.

UPDATE: Based on Akhil post this is what seems to work for me -

$sql = "SELECT * FROM (SELECT RAND() temp, link_id FROM table_links WHERE link_points > '1' AND link_status NOT IN ('spam') AND NOT EXISTS (SELECT slink_id FROM table_surf AS links WHERE user_id = '$user' AND surf_time = '$time' AND slink_id = '$surflink')) a ORDER by temp LIMIT 1";

Answer 1

Use a sub query

$sql = "select * from (SELECT RAND() temp, table_links.* FROM table_links AS link_id WHERE link_points > '1' 
AND NOT EXISTS (SELECT * FROM table_surf AS slink_id WHERE user_id = '$user'
AND surf_time = '$time')) a  ORDER by temp LIMIT 1";

See if it solves your problem

Answer 2

如果您的链接ID列在表table_links称为my_link_id，则使用NOT IN像这样

$sql = "SELECT * FROM table_links AS link_id WHERE link_points > '1' AND link_id.my_link_id NOT IN (SELECT * FROM table_surf AS slink_id WHERE user_id = '$user' AND surf_time = '$time') ORDER by RAND() LIMIT 1";

Answer 3

$sql = "SELECT * FROM table_link1 
        WHERE id IN (SELECT id FROM table_link2) ORDER BY RAND()" LIMIT 1;