i have two tables: users and user_visits.
For example i would get the user_id and name of users that have 0 visited where date > '2014-05-06' and date < '2014-05-11'.
USERS TABLE USERS_VISITS TABLE
USER_ID | NAME USERS_VISIT_ID | USER_ID | DATE
1 | John 1 | 1 | 2014-05-07
2 | Mark 2 | 1 | 2014-05-08
3 | Mike 3 | 4 | 2014-05-10
4 | Steven
The result must be:
USER_ID | NAME | COUNT(*) |
2 | Mark | 0 |
3 | Mike | 0 |
My query is:
SELECT COUNT(uv.user_id) AS count_visited,
FROM users u
LEFT JOIN users_visited uv ON (u.user_id=uv.user_id)
WHERE uv.date >= '2014-05-06'
AND uv.date <= '22014-05-11'
GROUP BY u.user_id
HAVING count_visited = 0;
This is how you can do it using left join
select
u.user_id,
u.name
from user u
left join user_visits uv on uv.user_id = u.user_id
AND uv.date >= '2014-05-06'
AND uv.date <= '22014-05-11'
where uv.user_id is NULL
Related
I have 3 tables created forum groups and group_members I want to get groups only those which are not in group members along with user id currently it is getting group which group id and user id is not present in the group members table if data is not in the table only if 1 group member exist it pulls up the record . In simple words I want to show show groups which user have not joined here is my table schema for both 3 tables
Groups
+----+----------+
| id | name |
+----+----------+
| 1 | group 1 |
| 2 | group 2 |
| 3 | group 3 |
| 4 | group 4 |
+----+----------+
forums
+------------------+-------------+
| id | title | group_id |
+------------------+-------------+
| 1 | test 1 | 2 |
| 2 | test 2 | 3 |
| 3 | test 3 | 2 |
| 4 | test 4 | 3 |
| 5 | test 5 | 2 |
| 6 | test 6 | 4 |
+------------------+-------------+
Group_members
+-----------------+-------------+
| id | user_id | group_id |
+-----------------+-------------+
| 1 | 107 | 2 |
| 2 | 106 | 3 |
+-----------------+-------------+
Here is my sql I have written
<?php
$sql_grp_chk = $this->db->query("SELECT * FROM groups WHERE NOT EXISTS (SELECT * FROM group_members WHERE groups.id == group_members.group_id)");
foreach($sql_grp_chk->result() as $data_ct):
$sql_gr_coun = $this->db->query("SELECT groups.*, (SELECT count(group_id) FROM forums WHERE groups.id = forums.group_id) as forumcount FROM groups WHERE groups.id != '".$data_ct->id."' ORDER BY forumcount DESC LIMIT 5");
foreach($sql_gr_coun->result() as $data_count):
$sql_follow = $this->db->get_where('group_members', array('group_id' => $data_count->id));
var_dump($data_count);
?>
<?php endforeach; ?>
<?php endforeach; ?>
Not sure why forums is there, but to select all groups that are not linked to a user you can do left join:
select g.* from groups g
left join group_members m on m.group_id = g.id and m.user_id = :userId
where m.id is null;
EDIT:
Select top 5 groups, by number of forums linked:
select g.*, count(nullif(f.id, 1)) as cnt from groups g
inner join forums f on f.group_id = g.id
group by g.id
order by cnt desc
limit 5;
Both queries together - top 5 groups, by number of forums linked, which user has not joined yet:
select g.*, count(nullif(f.id, 1)) as cnt from groups g
left join group_members m on m.group_id = g.id and m.user_id = :userId
left join forums f on f.group_id = g.id
where m.id is null
group by g.id
order by cnt desc
limit 5;
I have 2 table as below:
Table 1: common_member
+-----------+-----------+
| uid | username |
+-----------+-----------+
| 1 | haha |
| 2 | walao |
| 3 | alamak |
| 4 | hero |
| 5 | theone |
| 6 | nobody |
+-----------+-----------+
Table 2: labour_slog
+--------------+-------------+--------------+-------------+
| uid | slaveid | masterid | bytime |
+--------------+-------------+--------------+-------------+
| 1 | 2 | 3 | 123456 |
| 4 | 5 | 6 | 456789 |
+--------------+-------------+--------------+-------------+
I fetch the data as script below:
$queryLabourSlog = DB::query("SELECT * FROM ".DB::table('labour_slog')." ORDER BY id desc");
while($rowLabourSlog = DB::fetch($queryLabourSlog)) {
$user_list_slog[] = $rowLabourSlog;
};
array_multisort($idss, SORT_DESC, $user_list_slog);
In my html, I use
<!--{loop $user_list_slog $value}-->{$value[uid]} on {$value[bytime]} forced hire {$value[masterid]}'s employee {$value[slaveid]}.<!--{/loop}-->
The html will display:
1 on 123456 forced hire 3's employee 2.
4 on 456789 forced hire 6's employee 5.
How do I join the Table 1's username data to get the loop display as below?
haha on 123456 forced hire alamak's employee walao.
hero on 456789 forced hire nobody's employee theone.
Thanks.
SELECT table_slog.*, u1.`name`, u2.`name` FROM `table_slog` LEFT JOIN `common_member` AS u1 ON `table_slog`.`uid`=u1.`uid` LEFT JOIN `common_member` AS u2 ON `table_slog`.`slaveid`=u2.`uid` LEFT JOIN `common_member` AS u3 ON `table_slog`.`masterid`=u3.uid ORDER BY `id` DESC
Something like that:
SELECT `uid_main`.`username` AS `main_username`,
`uid_main`.`uid` AS `main_uid`,
`uid_master`.`username` AS `master_username`,
`uid_master`.`uid` AS `master_uid`,
`uid_slave`.`username` AS `slave_username`,
`uid_slave`.`uid` AS `slave_uid`,
`ls`.`bytime` AS `bytime`
FROM `labour_slog` AS `ls`
LEFT JOIN `common_member` AS `uid_main` ON (`ls`.`uid` = `cm`.uid)
LEFT JOIN `common_member` AS `uid_master` ON (`ls`.`master_id` = `cm`.uid)
LEFT JOIN `common_member` AS `uid_master` ON (`ls`.`slave_id` = `cm`.uid)
ORDER BY `cm`.`uid` DESC
I have 2 tables in database:
How can I get total user for each group. i.e: group 1: total are 2 users;
group2: total are 2 users;
group3: total is 1 user
You need normalization and never store comma-separated data.
Consider the following
mysql> select * from user_table ;
+---------+---------------+
| user_id | user_group_id |
+---------+---------------+
| 1 | 1,2 |
| 2 | 2 |
| 3 | 1,3 |
+---------+---------------+
3 rows in set (0.00 sec)
mysql> select * from group_table ;
+----------+------------+
| group_id | group_name |
+----------+------------+
| 1 | a |
| 2 | b |
| 3 | c |
+----------+------------+
3 rows in set (0.00 sec)
The above data is not normalized and to get the desired result out of these you need to use some in-efficient query as
select
g.group_id,
count(*) as total
from group_table g
left join user_table u on find_in_set(g.group_id,u.user_group_id) > 0
group by g.group_id ;
+----------+-------+
| group_id | total |
+----------+-------+
| 1 | 2 |
| 2 | 2 |
| 3 | 1 |
+----------+-------+
Now lets do normalization and store user-group data in a different table as
mysql> select * from user_to_group ;
+---------+----------+
| user_id | group_id |
+---------+----------+
| 1 | 1 |
| 1 | 2 |
| 2 | 2 |
| 3 | 1 |
| 3 | 3 |
+---------+----------+
You can easily write different queries from these tables now and here are some examples
select group_id,count(*) as tot from user_to_group group by group_id ;
+----------+-----+
| group_id | tot |
+----------+-----+
| 1 | 2 |
| 2 | 2 |
| 3 | 1 |
+----------+-----+
Joining the tables would even more easy
select
g.group_id,
g.group_name,
count(*) as tot
from user_to_group ug
join group_table g on g.group_id = ug.group_id
join user_table u on u.user_id = ug.user_id
group by g.group_id
+----------+------------+-----+
| group_id | group_name | tot |
+----------+------------+-----+
| 1 | a | 2 |
| 2 | b | 2 |
| 3 | c | 1 |
+----------+------------+-----+
SELECT group_name, COUNT(*) FROM user_table u, group_table g WHERE u.user_group_id LIKE %g.group_id% GROUP BY g.group_name;
this should work and give you a list of all groups and how many users are in them.
I will recommend you to create a third table which holds the information about which users are in which groups.
CREATE TABLE users_in_groups
(
user_id INT
, group_id INT
);
Then you can join like this:
SELECT
gt.group_id
, count(ut.user_id)
FROM
user_table AS ut
, INNER JOIN users_in_groups AS uig ON uig.user_id = ut.user_id
, INNER JOIN group_table AS gt ON gt.group_id = uig.group_id
GROUP BY
gt.group_id
;
To use the table you have now will you have to do something like this (in mysql):
SELECT
gt.group_id
, count(ut.user_id)
FROM
user_table AS ut
, INNER JOIN group_table AS gt ON LOCATE(gt.group_id, ut.user_group_id) > 0
GROUP BY
gt.group_id
Remember, when using group by, always locate what makes your group unique!
This is not an answer to your specific question but rather an alternative data structure proposal that might be better.
Introduce a new table members that looks like
# members
user_id | group_id
1 | 1
1 | 2
2 | 2
3 | 1
3 | 3
Then you could SELECT group_id, count(*) FROM members GROUP BY group_id
+----------+----------+
| group_id | count(*) |
+----------+----------+
| 1 | 2 |
| 2 | 2 |
| 3 | 1 |
+----------+----------+
This structure might also make it easier for you to manage your memberships. user_id + group_id should be unique. And if supported let them be foreign keys.
I'm a bit stuck with a php/Mysql Query. I got 2 tables :
table_users table_ranks
---------------- ------------------------
| id | points | | name | points_needed |
---------------- ------------------------
| 1 | 2 | | lvl0 | 0 |
| 2 | 10 | | lvl1 | 10 |
| 3 | 21 | | lvl2 | 20 |
| 4 | 29 | | lvl3 | 30 |
---------------- ------------------------
I need an ouput like this :
User_1 = lvl0 (because user has 2 points)
User_2 = lvl1 (because user has just reached 10 points)
...
User_4 = lvl2 (because user has not yet reached 30 points)
Think you :)
Regards.
You can do it like this
SELECT
tu.id,
tr.name,
tu.points
FROM table_ranks as tr
LEFT JOIN (SELECT * FROM table_ranks LIMIT 1,69596585953484) as l
ON l.points_needed = (SELECT MIN(points_needed) FROM table_ranks WHERE points_needed > tr.points_needed limit 1)
LEFT OUTER JOIN table_users AS tu ON tu.points >= tr.points_needed AND tu.points < l.points_needed
WHERE tu.id IS NOT NULL
group by tu.id
Fiddle
Output
-------------------------
| id | points | name |
-------------------------
| 1 | lvl0 | 2 |
| 2 | lvl1 | 10 |
| 3 | lvl2 | 21 |
| 4 | lvl2 | 29 |
-------------------------
give this a try, a little bit messy due to table design,
SELECT u.*, r.name
FROM table_users u
INNER JOIN
(
SELECT x.name,
x.points_needed start_point,
COALESCE(y.points_needed - 1, 2000000) end_point
FROM
(
SELECT name, points_needed, #rank:=#rank+1 ranks
FROM table_ranks a, (SELECT #rank:=0) b
ORDER BY points_needed
) x
LEFT JOIN
(
SELECT name, points_needed, #rank1:=#rank1+1 ranks
FROM table_ranks a, (SELECT #rank1:=0) b
ORDER BY points_needed
) y ON x.ranks+1 = y.ranks
) r ON u.points BETWEEN r.start_point AND r.end_point
SQLFiddle Demo
Hi there coders around the world,
I'm working on a project where users can do certain things and gain points for it. To simplify this question let's say we got 2 tables user and points.
-- table user -- table points
+---------------+ +-----------------------------+
| id | name | | id | points | user_id |
+---------------+ +-----------------------------+
| 1 Tim | | 1 5 1 |
| 2 Tom | | 2 10 1 |
| 3 Marc | | 3 5 1 |
| 4 Tina | | 4 12 2 |
| 5 Lutz | | 5 2 2 |
+---------------+ | 6 7 1 |
| 7 40 3 |
| 8 100 1 |
+-----------------------------+
Now to get the complete highscore-list I use the following query
SELECT u.*, SUM( p.points ) AS sum_points
FROM user u
LEFT JOIN points p ON p.user_id = u.id
GROUP BY u.id
ORDER BY sum_points DESC
resulting in a fine highscore-list with all users from first to last
+------------------------------+
| id | name | sum_points |
+------------------------------+
| 1 Tim 127 |
| 3 Marc 40 |
| 2 Tom 14 |
| 4 Tina 0 |
| 5 Lutz 0 |
+------------------------------+
Alright back to the question itself. On the profile of a single user I'd like to show his ranking within the highscore-list.
Can this be done using a single query just showing that for example Tom (id=2) is ranked in place 3?
Thanks alot :-)
The idea is to ask, "how many players rank above #this_user":
select count(*) + 1 from
(
/* list of all users */
SELECT SUM( p.points ) AS sum_points
FROM user u
LEFT JOIN points p ON p.user_id = u.id
GROUP BY u.id
) x
/* just count the ones with higher sum_points */
where sum_points > (select sum(points) from points where user_id = #this_user)
Edited to make result 1-based instead of 0-based
SELECT q.*,
#r := #r + 1 AS rank
FROM (
SELECT #r := 0
) vars,
(
SELECT u.*,
SUM(p.points) AS sum_points
FROM
user u
LEFT JOIN
points p
ON p.user_id = u.id
GROUP BY
u.id
ORDER BY
sum_points DESC
) q