MYSQL get row rank - php

I have a mysql table and I need to get random row and get the rank of total view
+--------+------------+---------+
| id | name |totalview|
+--------+------------+---------+
| 1 | ex1 | 20 |
| 2 | ex2 | 100 |
| 3 | ex3 | 30 |
| 4 | ex4 | 40 |
+--------+------------+---------+
for example :
SELECT * FROM `table` WHERE `id` = '$rand';
$rand may be 1 or 2 etc ..
I need to get rank of this row by totalview
thank's

SELECT *,
(SELECT COUNT(*) FROM table t2 WHERE totalview > t1.totalview ) + 1 cnt
FROM table t1
WHERE id = '$rand';

SELECT SUM(ref.totalview < t.totalview) FROM t1 CROSS JOIN t1 ref WHERE t1.id = '$rand'

Related

How to compare columns values with sum function in SQL?

I have three tables :
mls_category
points_martix
mls_entry
My first table (mls_category) is like below:
*--------------------------------*
| cat_no | store_id | cat_value |
*--------------------------------*
| 10 | 101 | 1 |
| 11 | 101 | 4 |
*--------------------------------*
My second table (points_martix) is like below:
*----------------------------------------------------*
| pm_no | store_id | value_per_point | max_distance |
*----------------------------------------------------*
| 1 | 101 | 1 | 10 |
| 2 | 101 | 2 | 50 |
| 3 | 101 | 3 | 80 |
*----------------------------------------------------*
My third table (mls_entry) is like below:
*-------------------------------------------*
| user_id | category | distance | status |
*-------------------------------------------*
| 1 | 10 | 20 | approved |
| 1 | 10 | 30 | approved |
| 1 | 11 | 40 | approved |
*-------------------------------------------*
I am using the following query to show the sum of distance with some condition:
SELECT SUM(t1.totald/c.cat_value)
AS total_distance
FROM mls_category c
JOIN
(SELECT SUM(distance) totald, user_id, category
FROM mls_entry
WHERE user_id = 1
AND status = 'approved'
GROUP BY user_id, category) t1
ON c.cat_no = t1.category
This gives me sum 60 as total_distance, that is correct which I wanted.
Now, I want to include the third table (points_matrix) and want to compare my sum(60) is less than or equal to 80(max_distance) then my new value would be 60*3=180.
So, suppose my sum comes 10 then my new value will be 10*1=10 and if my sum comes 25 then my new value will be according to point matrix 25*2=50.
Yon can using MIN() to calculate what value_per_point you need, and the whole sql is like this:
SELECT MIN(b.value_per_point) * d.total_distance FROM points_matrix b
JOIN
(
SELECT store_id, sum(t1.totald/c.cat_value) as total_distance FROM mls_category c
JOIN
(
SELECT SUM(distance) totald, user_id, category FROM mls_entry
WHERE user_id= 1 AND status = 'approved' GROUP BY user_id, category
) t1 ON c.cat_no = t1.category
) d ON b.store_id = d.store_id AND b.max_distance >= d.total_distance
Use Correlated Subquery:
SELECT
dt.total_distance * dt.max_points
FROM (
SELECT SUM(t1.totald/c.cat_value) AS total_distance,
(
SELECT value_per_point
FROM points_martix
WHERE SUM(t1.totald/c.cat_value) >= max_distance
ORDER BY max_distance ASC LIMIT 1
) AS max_points
FROM mls_category AS c
JOIN (
SELECT SUM(distance) AS totald,
user_id,
category
FROM mls_entry
WHERE user_id= 1 AND
status = 'approved'
GROUP BY user_id, category
) AS t1 on c.cat_no = t1.category
) AS dt

MySQL, Merge selects in order of one record from each select

I have a table that contains too many records and each bunch of records belong to someone:
---------------------
id | data | username
---------------------
1 | 10 | ali
2 | 11 | ali
3 | 12 | ali
4 | 20 | omid
5 | 21 | omid
6 | 30 | reza
now I want to create a query to result me like this:
1-10-ali
4-20-omid
6-30-reza
2-11-ali
5-21-omid
3-12-ali
Is there anyway to create a query to result me one record per each username and then one from another, and another to the end?
Unfortunately MySQL doesn't have a ranking system so you can use UDV (user defined variables) to rank your records like so.
SELECT id, `data`, name
FROM
( SELECT
id, `data`, name,
#rank := if(#name = name, #rank + 1, 1) as rank,
#name := name
FROM test
CROSS JOIN (SELECT #rank := 1, #name := '') temp
ORDER BY name, `data`
) t
ORDER BY t.rank, t.name, t.data
Sql Fiddle to play with
Output:
+---------------------+
| id | data | name |
+-----+------+--------+
| 1 | 10 | ali |
+---------------------+
| 4 | 20 | omid |
+---------------------+
| 6 | 30 | reza |
+---------------------+
| 2 | 11 | ali |
+---------------------+
| 5 | 21 | omid |
+---------------------+
| 3 | 12 | ali |
+---------------------+
The classic SQL approach is a self join and grouping that lets you determine a row's ranking position by counting the number of rows that come before it. As this is probably slower I doubt I could talk you out of the proprietary method but I mention it to give you an alternative.
select t.id, min(t.`data`), min(t.username)
from test t inner join test t2
on t2.username = t.username and t2.id <= t.id
group by t.id
order by count(*), min(t.username)
Your example would work with
SELECT id, `data`, name
FROM tbl
ORDER BY `data` % 10,
username
`data`;
If data and username do not have the desired pattern, then improve on the example.

Delete all rows, except last 10 for each client that has related row(s) in the table in one query?

So my situation is this:
Clients table - has client data etc, not too exciting
Recently Viewed table - table that has recently viewed things for the client(s), And has structure like this:
( id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
, client_id INT NOT NULL
, cookie_user_id INT NOT NULL
, hotel_id INT NOT NULL
, added DATETIME NOT NULL
, comment TEXT
,status TINYINT NOT NULL DEFAULE 1
);
I currently have a partially working SQL to delete rows in the recently viewed table that right now globally limits number of latest remaining undeleted records in it. This is how it looks like now
DELETE FROM `recently_viewed`
WHERE `recently_viewed`.`id` NOT IN (
SELECT id
FROM (
SELECT `id`
FROM `recently_viewed`
WHERE `client_id` IN (SELECT `id` FROM `klijenti`)
ORDER BY `id` DESC
LIMIT 5
) x
)
AND `client_id` <> 0
"LIMIT 5" part should limit to the N records to remain in recently viewed table on a "per client" basis. Right now it limits records in recently viewed table to 5 no matter how many clients actually have records there. So if I have 10 clients, each of them has 8 records in that table, I would like this query to delete as many oldest records as needed to leave only 5 newest recently viewed items for EACH client and not just leave 5 overall in the table, ignoring the "per each client" logic. Hope that makes sense to you :)
Currently, this query would be ok if I would first fetch all clients in the app and then do a foreach loop to make another query for each client and leave 5 of his latest recently viewed items, but would like to do this in one SQL query instead.
How could this be done ? Thank you
You can do it like this:
DELETE FROM `recently_viewed`
WHERE `recently_viewed`.`id` NOT IN (
SELECT id
FROM (
SELECT t.`id`,count(*) as rnk
FROM `recently_viewed` t
INNER JOIN `recently_viewed` s
ON(t.`client_id` = s.`client_id` and t.added <= s.added)
WHERE t.`client_id` IN (SELECT `id` FROM `klijenti`)
GROUP BY t.`ID`
) x
WHERE rnk <= 5
)
AND `client_id` <> 0
You can use vartiables to in order to count the 5 more recent records per client_id:
DELETE FROM `recently_viewed`
WHERE `recently_viewed`.`id` NOT IN
(
SELECT id
FROM (
SELECT `id`,
#rn := IF(#cid = `client_id`, #rn + 1,
IF(#cid := `client_id`, 1, 1)) AS rn
FROM `recently_viewed`
CROSS JOIN (SELECT #rn := 0, #cid := 0) AS vars
WHERE `client_id` IN (SELECT `id` FROM `klijenti`)
ORDER BY `client_id`, `id` DESC) x
WHERE x.rn <= 5
)
Giorgos's answer is faster, but here's another method...
Consider the following...
SELECT * FROM my_table ORDER BY x,i;
+---+------+
| i | x |
+---+------+
| 2 | A |
| 3 | A |
| 6 | A |
| 8 | A |
| 1 | B |
| 5 | B |
| 4 | C |
| 7 | C |
| 9 | C |
+---+------+
Let's say we want to select the two latest i for each x. Here's one way to do that...
SELECT m.* FROM my_table m JOIN my_table n ON n.x = m.x AND n.i >= m.i GROUP BY m.i HAVING COUNT(*) <= 2;
+---+------+
| i | x |
+---+------+
| 1 | B |
| 5 | B |
| 6 | A |
| 7 | C |
| 8 | A |
| 9 | C |
+---+------+
The inverse of this set can be found as follows....
SELECT m.* FROM my_table m JOIN my_table n ON n.x = m.x AND n.i >= m.i GROUP BY m.i HAVING COUNT(*) > 2;
+---+------+
| i | x |
+---+------+
| 2 | A |
| 3 | A |
| 4 | C |
+---+------+
...which in turn can be incorporated in a DELETE. Here's a crude method for doing that...
DELETE a FROM my_table a
JOIN
( SELECT m.* FROM my_table m JOIN my_table n ON n.x = m.x AND n.i >= m.i GROUP BY m.i HAVING COUNT(*) > 2 ) b
ON b.i = a.i;
Query OK, 3 rows affected (0.03 sec)
SELECT * FROM my_table ORDER BY x,i;
+---+------+
| i | x |
+---+------+
| 6 | A |
| 8 | A |
| 1 | B |
| 5 | B |
| 7 | C |
| 9 | C |
+---+------+
As I say, if performance is critical, then look at a solution along the lines that Giorgos has provided.

MySql PDO - How to fetch "newest rows from group" when using GROUP BY, ORDER BY and INNER JOIN?

products_table: | p_id | name |
| 1 | name1 |
| 2 | name2 |
| 3 | name3 |
favourites_table: | id | p_id | deleted | group_id |
fetch-> | 1 | 1 | 0 | 11 |
| 2 | 1 | 0 | 11 |
fetch-> | 3 | 2 | 0 | 22 |
| 4 | 2 | 0 | 22 |
fetch-> | 5 | 3 | 0 | 33 |
| 6 | 3 | 0 | 33 |
$sth = $db->prepare(' SELECT a.p_id, b.name
FROM favourites_table AS a
INNER JOIN products_table AS b
ON a.p_id = b.p_id
WHERE a.deleted=0
GROUP BY a.group_id
ORDER BY a.id ASC
LIMIT 0, 10;');
$sth->execute();
while(($query_data = $sth->fetch()) !== false) {
echo $query_data['p_id'] . ':' . $query_data['name'] . '<br>';
}
This query fetches rows 1, 3, 5 from 'favourites_table'.
How to change it so it fetches "newest rows" (2, 4, 6) ?
Do I have to change the whole query or am I missing something?
You're confused by the pernicious misfeature in MySQL called the GROUP BY extension. Read this. http://dev.mysql.com/doc/refman/5.6/en/group-by-extensions.html
You want the rows you define as latest for each value of group_id. These rows are in fact the undeleted ones with the highest id values.
So, first you need to use a subquery -- a virtual table -- to find those rows, as follows:
SELECT MAX(id) AS id, group_id FROM favourites_table WHERE deleted = 0 GROUP BY group_id
Then, you need to use that resultset to find the right rows in your main query. You would do this like so:
SELECT a.p_id, b.name
FROM favourites_table AS a
INNER JOIN products_table AS b ON a.p_id = b.p_id
INNER JOIN (
SELECT MAX(id) AS id, group_id FROM favourites_table WHERE deleted = 0 GROUP BY group_id
) AS c ON a.id = c.id
GROUP BY a.group_id
ORDER BY a.id ASC
LIMIT 0, 10
This should get your results.
Question: Why order them oldest (lowest id value) first? Why only show the oldest ten results? Is that what you want?

MySQL - Sum values width same ID in same table

I want sum values into my database by the same ID in the same table.
Table in database:
| ID | Value_o | Value_t | Value_tt |
| 1 | 40 | 20 | 10 |
query:
SELECT SUM(Value_o) AS Value_o, SUM(Value_t) AS Value_t, SUM(Value_tt) AS Value_TT
WHERE ID IN(1, 1)
And now the output id:
| Value_o | Value_t | Value_tt |
| 40 | 20 | 10 |
but I want:
| Value_o | Value_t | Value_tt |
| 80 | 40 | 20 |
I want get this output without JOIN.
Thanks!
PS. Sorry for my bad eng :/
Maybe this is what you are looking for:
SELECT
SUM(Value_o) AS Value_o,
SUM(Value_t) AS Value_t,
SUM(Value_tt) AS Value_TT
FROM
(
SELECT ID, Value_o, Value_t, Value_tt FROM Table1
UNION ALL
SELECT ID, Value_o, Value_t, Value_tt FROM Table1
) Table2
WHERE ID IN(1, 1);
Demo
The MySQL in operator doesn't work this way. Even if you have a value multiple times in the set, it doesn't duplicate the rows of your result.
If you want to have all the rows multiple times, you must use union all and sum over that
SELECT SUM(Value_o) AS Value_o, SUM(Value_t) AS Value_t, SUM(Value_tt) AS Value_TT
from (select * from mytable union all select * from mytable) t
WHERE ID IN (1)
Try this:
SELECT SUM(Value_o) AS Value_o, SUM(Value_t) AS Value_t, SUM(Value_tt) AS Value_TT
FROM TABLE
GROUP BY ID
HAVING ID = 1

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