I have a database that store transaction logs, I would like to count all the logs for that day and group them based on prod_id
MySQL table structure:
Table name = products
+------+---------+------------+--------+
| ID | PROD_ID | DATE | PERSON |
+------+---------+------------+--------+
| 1 | 2 | 1400137633 | 1 |
| 2 | 2 | 1400137666 | 1 |
| 3 | 3 | 1400137125 | 2 |
| 4 | 4 | 1400137563 | 1 |
| 5 | 2 | 1400137425 | 2 |
| 6 | 3 | 1400137336 | 1 |
+------+---------+------------+--------+
MYSQL CODE:
$q = 'SELECT count(ID) as count
FROM PRODUCTS
WHERE PERSON ='.$db->qstr($person).'
AND DATE(FROM_UNIXTIME(DATE)) = DATE(NOW())';
so what I get is the number of items for the given date. Since the date is the same as all other entries. however I would like to group the items by prod_id, I tried GROUP BY PROD_ID but that did not give me what I want. I would like it to group if the PROD_ID is multiple and the date is the same display as one entry while still count the others
so here I should get an output ($Person = 1).... 2+2+2=1 +3 +4 so total should be 3
any suggestions?
Use DISTINCT with COUNT on PROD_ID.
Example:
SELECT count( distinct PROD_ID ) as count
FROM PRODUCTS
WHERE PERSON = 1 -- <---- change this with relevant variable
AND DATE( FROM_UNIXTIME (DATE ) ) = curdate();
And I suggest you to use Prepared Statement to bind values.
Related
Having a hard time wrapping my mind around what seems should be a simply query.
So let's say we have a table that keeps track of amount of widgets/balloons in each store by date. How would you get a list of stores and their latest widget/balloons count?
i.e.
mysql> SELECT * FROM inventory;
+----+------------+-------+---------+---------+
| id | invDate | store | widgets | balloons|
+----+------------+-------+---------+---------+
| 1 | 2011-01-01 | 3 | 50 | 35 |
| 2 | 2011-01-04 | 2 | 50 | 35 |
| 3 | 2013-07-04 | 3 | 12 | 78 |
| 4 | 2020-07-04 | 2 | 47 | 18 |
| 5 | 2020-08-06 | 2 | 16 | NULL |
+----+------------+-------+---------+---------+
5 rows in set (0.00 sec)
Would like the result table to list all stores and their latest inventory of widgets/baloons
store, latest widgets, latest balloons
+-------+-----------+---------+
| store | widgets | baloons |
+-------+-----------+---------+
| 2 | 16 | NULL |
| 3 | 12 | 78 |
+-------+-----------+---------+
or grab latest non NULL value for balloons.
This works for all versions of MySQL
select i.*
from inventory i
join
(
select store, max(invDate) as maxDate
from inventory
group by store
) tmp on tmp.store = i.store
and tmp.maxDate = i.invDate
With MySQL 8+ you can do window functions:
with cte as
(
select store, widgets, balloons,
ROW_NUMBER() OVER(PARTITION BY store ORDER BY invDate desc) AS rn
from inventory
)
select * from cte where rn = 1
You can use a correlated sub query to get latest record for each store
select i.*
from inventory i
where i.invDate = (
select max(invDate)
from inventory
where i.store = store
)
order by i.store
DEMO
I have a example table below. I am trying to create a SQL query that gets all user_ids besides user_id of the current user and then orders by number of matches to the row with the current user_id
For example, if the user has a user_id of '1', I want to get all of the user_ids corresponding with the rows of id 2-8, and then order the user_ids from most matches to the row of the current user to least matches with the row of the current user
Let's say var current_user = 1
Something like this:
SELECT user_id
FROM assets
WHERE user_id <> `current_user` and
ORDER BY most matches to `current_user`"
The output should get 7,8,3,9,2
I would appreciate anyone's input on how I can effectively achieve this.
Table assets
+----------+---------+-------+--------+-------+
| id | user_id | cars | houses | boats |
+----------+---------+-------+--------+-------+
| 1 | 1 | 3 | 2 | 3 |
| 2 | 8 | 3 | 2 | 5 |
| 3 | 3 | 3 | 2 | 2 |
| 4 | 2 | 5 | 1 | 5 |
| 5 | 9 | 5 | 7 | 3 |
| 8 | 7 | 3 | 2 | 3 |
+----------+---------+-------+--------+-------+
I think you can just do this:
select a.*
from assets a cross join
assets a1
where a1.user_id = 1 and a.user_id <> a1.user_id
order by ( (a.cars = a1.cars) + (a.houses = a1.houses) + (a.boats = a1.boats) ) desc;
In MySQL, a boolean expression is treated as an integer in a numeric context, with 1 for true and 0 for false.
If you want to be fancier, you could order by the total difference:
order by ( abs(a.cars - a1.cars) + abs(a.houses - a1.houses) + abs(a.boats - a1.boats) );
This is called Manhattan distance, and you would be implementing a version of a nearest neighbor model.
I have the following table:
+---------+--------------+----------+
| item_id | location_id | price |
+---------+--------------+----------+
| 1 | 1 | 100 |
| 1 | 1 | 250 |
| 1 | 2 | 50 |
| 2 | 1 | 250 |
| 2 | 1 | 1000 |
| 3 | 1 | 1000 |
| 3 | 2 | 100 |
+---------+--------------+----------+
I can reduce this down to the minimum values using this query
SELECT
item_id, location_id, MIN(price) AS Price
from
table
GROUP BY item_id , location_id
This gets me
+---------+--------------+----------+
| item_id | location_id | price |
+---------+--------------+----------+
| 1 | 1 | 100 |
| 1 | 2 | 50 |
| 2 | 1 | 250 |
| 3 | 1 | 1000 |
| 3 | 2 | 100 |
+---------+--------------+----------+
I want to reduce this further. I am using the rows with a location_id of 1 as a reference row. For each row that has an item_id matching the reference row's item_id but a different location id. I want to compare that row's price with the reference row's price. If the price is lower than the reference row's price, I want to filter that row out.
My final result should include the reference row for each item id and any rows that met the criteria of the price being lower than the reference row price.
I have a hunch that I can use the HAVING clause to do this but I am having trouble compiling the statement. How should I construct the HAVING statement?
Thanks in advance
Nah, having can't help you like this, having is for things like you need filter min() result for something
e.g:
select id,min(price) from table where date = '2016-3-18' group by id having min(price) = 50
it will show you the records that min(price)=50
let's back to your case, there are lots of way to do that,
1. left join
select a.item_id,a.location_id,a.price
from table a
left join table b
on a.location_id = b.location_id and a.price > b.price
where b.price is null
2. exists
select a.item_id,a.location_id,a.price
from table a
where exists(
select 1 from
(select location_id,min(price)as price from table group by location_id)b
where a.location_id = b.location_id and a.price = b.price
)
normally i ll recommand you use exists
I have two tables, one that store product information and one that stores reviews for the products.
I am now trying to get the number of reviews submitted for the products between two dates but for some reason I get the same results regardless of the dates i put.
This is my query:
SELECT
productName,
COUNT(*) as `count`,
avg(rating) as `rating`
FROM `Reviews`
LEFT JOIN `Products` using(`productID`)
WHERE `date` BETWEEN '2015-07-20' AND '2015-07-30'
GROUP BY
`productName`
ORDER BY `count` DESC, `rating` DESC;
This returns:
+------------+---------------------+
| productName| count|rating |
+------------+------+--------------+
| productA | 23 | 4.3333333 |
| productB | 17 | 4.25 |
| productC | 10 | 3.5 |
+------------+---------------------+
Products table:
+---------+-------------+
|productID | productName|
+---------+-------------+
| 1 | productA |
| 2 | productB |
| 3 | productC |
+---------+-------------+
Reviews table
+---------+-----------+--------+---------------------+
|reviewID | productID | rating | date |
+---------+-----------+--------+---------------------+
| 1 | 1 | 4.5 | 2015-07-27 17:47:01|
| 2 | 1 | 3.5 | 2015-07-27 18:54:22|
| 3 | 3 | 2 | 2015-07-28 13:28:37|
| 4 | 1 | 5 | 2015-07-28 18:33:14|
| 5 | 2 | 1.5 | 2015-07-29 11:58:17|
| 6 | 2 | 3.5 | 2015-07-30 15:04:25|
| 7 | 2 | 2.5 | 2015-07-30 18:11:11|
| 8 | 1 | 3 | 2015-07-30 18:26:23|
| 9 | 1 | 3 | 2015-07-30 21:35:05|
| 10 | 1 | 4.5 | 2015-07-31 14:25:47|
| 11 | 3 | 0.5 | 2015-07-31 14:47:48|
+---------+-----------+--------+---------------------+
when I put two random dates that I do know for sure they not on the date column, I will still get the same results. Even when I want to retrieve records only on a certain day, I get the same results.
You should not use left join, because by doing so you retrieve all the data from one table. What you should use is something like :
select
productName,
count(*) as `count`,
avg(rating) as `rating`
from
products p,
reviews r
where
p.productID = r.productID
and `date` between '2015-07-20' and '2015-07-30'
group by productName
order by count desc, rating desc;
If the result, given your sample data, that you're looking for is:
| productName | count | rating |
|-------------|-------|--------|
| productA | 5 | 4 |
| productB | 3 | 3 |
| productC | 1 | 2 |
This is the count and average of reviews made on any date between 2015-07-20 and 2015-07-30 inclusive.
Then the there are two issues with your query. First, you need to change the join to a inner join instead of a left join, but more importantly you need to change the date condition as you are currently excluding reviews that fall on the last date on the range, but after midnight.
This happens because your between clause compares datetime values with date values so the comparison ends up being date between '2015-07-20 00:00:00' and '2015-07-30 00:00:00' which clearly excludes some dates at the end.
The fix is to either change the date condition so that the end is a day later:
where date >= '2015-07-20' and date < '2015-07-31'
or cast the date column to a date value, which will remove the time part:
where date(date) between '2015-07-20' and '2015-07-30'
Sample SQL Fiddle
You are using a LEFT JOIN between your reviews and your products tables. This will result in all the rows of reviews being shown with some rows having all product columns left empty.
You should use INNER JOIN, as this will filter only the wanted results.
(In the end I can only guess, since I don't even know which column belongs to which table ...)
The full query (very similar to Angelo Giannis's solution):
select
productName,
count(*) as `count`,
avg(rating) as `rating`
from
products INNER JOIN reviews USING(productId)
where date between '2015-07-20' and '2015-07-30'
group by productName
order by count desc, rating desc;
Here a fiddle with my and Angelo's solution (they both work).
I have a MySQL table like this.
| season_id | round_1 | names | score_round_1
| 5 | 10 | John1 | 5
| 5 | 10 | John2 | 3
| 5 | 11 | John3 | 2
| 5 | 11 | John4 | 5
I want to select the records with highest score_round_1 in each round_1(10,11) group .
In this case the first and last rows would be selected.
I tried using the GROUP BY round_1 but that only returns the first row from the two.
Any advice?
Zolka
This is simple
select max(score_round_1),
name
from score
group by round_1
SELECT *
FROM table p1
WHERE score_round_1 = (
SELECT MAX( p2.score_round_1 )
FROM table p2
WHERE p1.round_1 = p2.round_1 ) ANDround_1 !=0
Use aggregate function MAX
SELECT names, MAX(score_round_1) GROUP BY round_1