While searches the date range, start date (date_reg) and end date (date_reg)
, the mysql result should be have each main_table rows contains latest return, received, balance of each products.
E.g.: Between 10-01-2014 and 10-05-2014, should retrieve values of each product within the date
Client Id | Return | Received | Balance
| prod 1 prod 2 | prod 1 prod 2 | prod 1 prod 2
--------------------------------------------------------------
1 | 2 [3] 2 [7] | 5 5 | 8 5
2 | 1 [5] 0 [8] | 5 5 | 9 3
3 | 0 [6] 1 [10]| 5 5 | 7 6
[id], where id is the primary key of sub_table
I have tried mysql query
SELECT p.product_name, ipd.id as ipd_id, i.id as i_id, ipd.*, i.*
FROM main_table i
LEFT JOIN sub_table ipd ON ipd.main_table_id=i.id AND ipd.product_id IN (1,2)
LEFT JOIN product p ON ipd.product_id=p.id
WHERE ipd.date_reg IN (SELECT MAX(ipd1.date_reg)
FROM sub_table ipd1
WHERE ipd1.main_table_id=i.id AND
date_reg BETWEEN '10-01-2014' AND '10-05-2014')
ORDER BY cl.id ASC LIMIT 0, 20
it only return single product of return, received and balance of each client
When you use the subquery WHERE 'ipd.date_reg IN 'SELECT MAX...' you're only going to get 1 entry based on your data - 10-04-2014. Working correctly.
Try use GROUP BY in the sub query
Also GROUP_CONCAT(expr); helps to do many-to-many info's which can be be used to concatenate column values into a single string.
I got the output. Thanks everyone for the helps.
I have used GROUP_CANCAT to concatenate the results into one string with comma seperated
SELECT p.product_name, ipd.id as ipd_id, i.id as i_id, ipd.*, i.*,
GROUP_CONCAT(product_id SEPARATOR ',') as group_product_id,
GROUP_CONCAT(ipd.return SEPARATOR ',') as group_return,
GROUP_CONCAT(ipd.received SEPARATOR ',') as group_received,
GROUP_CONCAT(ipd.balance SEPARATOR ',') as group_balance
FROM main_table i
LEFT JOIN sub_table ipd ON ipd.main_table_id=i.id AND ipd.product_id IN (1,2)
LEFT JOIN product p ON ipd.product_id=p.id
WHERE ipd.date_reg IN (SELECT MAX(ipd1.date_reg)
FROM sub_table ipd1
WHERE ipd1.main_table_id=i.id AND
date_reg BETWEEN '10-01-2014' AND '10-05-2014'
GROUP BY ipd1.product_id)
ORDER BY cl.id ASC LIMIT 0, 20
The Result
Client Id | group_product_id | group_return | group_received | group_balance
--------------------------------------------------------------------------
1 | 1, 2 | 2, 2 | 5,5 | 8,5
2 | 1, 2 | 1, 0 | 5,5 | 9,3
3 | 1, 2 | 0, 1 | 5,5 | 7,6
Then the strings can be exploded into an array.
Related
help please make sql select to database. There are such data.
My table is:
id news_id season seria date_update
---|------|---------|-----|--------------------
1 | 4 | 1 | 7 | 2017-04-14 16:38:10
2 | 4 | 1 | 7 | 2017-04-14 17:38:10
5 | 4 | 1 | 7 | 2017-04-14 16:38:10
3 | 4 | 1 | 7 | 2017-04-14 16:38:10
4 | 4 | 1 | 7 | 2017-04-14 16:38:10
6 | 4 | 1 | 7 | 2017-04-14 16:38:10
7 | 4 | 1 | 7 | 2017-04-14 16:38:10
8 | 1 | 1 | 25 | 2017-04-23 18:42:00
Need to get all cells grouped by max season and seria and date and sorted by date_update DESC.
In result i need next rows
id news_id season seria date_update
---|------|---------|-----|--------------------
8 | 1 | 1 | 25 | 2017-04-23 18:42:00
2 | 4 | 1 | 7 | 2017-04-14 17:38:10
Because this rows have highest season and seria and date_update per One news_id. I.e i need to select data wich have highest season and seria and date_update grouped by news_id and also sorted by date_update DESC
I tried so, but the data is not always correct, and it does not always for some reason cover all the cells that fit the condition.
SELECT serial.*
FROM serial as serial
INNER JOIN (SELECT id, MAX(season) AS maxseason, MAX(seria) AS maxseria FROM serial GROUP BY news_id) as one_serial
ON serial.id = one_serial.id
WHERE serial.season = one_serial.maxseason AND serial.seria = one_serial.maxseria
ORDER BY serial.date_update
Please, help. Thank.
The specification is unclear.
But we do know that the GROUP BY news_id clause is going collapse all of the rows with a common value of news_id into a single row. (Other databases would throw an error with this syntax; we can get MySQL to throw a similar error if we include ONLY_FULL_GROUP_BY in the sql_mode.)
My suggestion would be to remove the GROUP BY news_id clause from the end of the query.
But that's just a guess. It's not at all clear what you are trying to achieve.
EDIT
SELECT t.*
FROM (
SELECT r.news_id
, r.season
, r.seria
, MAX(r.date_update) AS max_date_update
FROM (
SELECT p.news_id
, p.season
, MAX(p.seria) AS max_seria
FROM (
SELECT n.news_id
, MAX(n.season) AS max_season
FROM serial n
GROUP BY n.news_id
) o
JOIN serial p
ON p.news_id = o.news_id
AND p.season = o.max_season
) q
JOIN serial r
ON r.news_id = q.news_id
AND r.season = q.season
AND r.seria = q.max_seria
) s
JOIN serial t
ON t.news_id = s.news_id
AND t.season = s.season
AND t.seria = s.seria
AND t.date_update = s.max_date_update
GROUP BY t.news_id
ORDER BY t.news_id
Or, an alternate approach making use of MySQL user-defined variables...
SELECT s.id
, s.season
, s.seria
, s.date_update
FROM (
SELECT IF(q.news_id = #p_news_id,0,1) AS is_max
, q.id
, #p_news_id := q.news_id AS news_id
, q.season
, q.seria
, q.date_update
FROM ( SELECT #p_news_id := NULL ) r
CROSS
JOIN serial q
ORDER
BY q.news_id DESC
, q.season DESC
, q.seria DESC
, q.date_update DESC
) s
WHERE s.is_max
ORDER BY s.news_id
The subquery selects the maximum season and the maximum seria per news_id. How many records exist for the news_id that match both the maximum season and the maximum seria we don't know. It can be, one or two or thousand or zero.
So with the join you get an unknown number of records per news_id. Then you group by news_id. This gets you one result row per news_id. How then can you select serial.*? * means all columns from a row, but which row,when there can be many for a news_id? MySQL usually picks values arbitrarily in this case (usually all from the same row, but even that is not guaranteed). So you end up with random rows which you order by date_update.
This doesn't make much sense. So the question is: what do you really want to achieve? Maybe my explanation suffices and you are able now to fix your query yourself.
I have mysql table with hospitals and treatments(associated with sub treatments) that they provide. I need to make mysql query on the table which returns hospitals providing all treatment/sub_treatment given in a list. For example:
From table below I need hospitals providing all treatments in list: (tretament_id, sub_treatment_id) = (1-1, 1-2). So result must be hospitals with id 1 and 8.
hospital_id | treatment_id | sub_treatment_id
-------------------------------------------------
1 | 1 | 1
1 | 1 | 2
1 | 1 | 3
_________________________________________________
4 | 1 | 1
4 | 2 | 1
_________________________________________________
8 | 1 | 1
8 | 1 | 2
_________________________________________________
7 | 2 | 1
I tried WHERE IN but it works like OR so returns hospital 4 which satisfies only (1,1). How can I write an sql query like WHERE IN but which works like AND?
Try this:
SELECT hospital_id
FROM mytable
WHERE (treatment_id, sub_treatment_id) IN ((1, 1), (1, 2))
GROUP BY hospital_id
HAVING COUNT(CASE
WHEN (treatment_id, sub_treatment_id) IN ((1, 1), (1, 2))
THEN 1
END) = 2
Demo here
You can do this using group by and having:
select hospital_id
from t
where treatment_id = 1 and sub_treatment_id in (1, 2)
group by hospital_id
having count(*) = 2;
Note: This assumes that there are no duplicates in the table. That is easy enough to fix using count(distinct), but probably not necessary.
Here is a solution using GROUP_CONCAT and JOIN:
select distinct t.hospital_id
from hospitals h and treatments t ON h.id = t.hospital_id
having GROUP_CONCAT(CONCAT(t.treatment_id, '-', t.sub_treatment_id)
ORDER BY t.treatment_id, t.sub_treatment_id)
= '1-1,1-2';
I have 2 tables that I am trying to join but I am not sure how to make it the most time efficient.
Tasks Table:
nid | created_by | claimed_by | urgent
1 | 11 | 22 | 1
2 | 22 | 33 | 1
3 | 33 | 11 | 1
1 | 11 | 43 | 0
1 | 11 | 44 | 1
Employee Table:
userid | name
11 | EmployeeA
22 | EmployeeB
33 | EmployeeC
Result I am trying to get:
userid | created_count | claimed_count | urgent_count
11 | 3 | 1 | 3
22 | 1 | 1 | 2
33 | 1 | 1 | 2
created_account column will show total # of tasks created by that user.
claimed_count column will show total # of tasks claimed by that user.
urgent_count column will show total # of urgent tasks (created or claimed) by that user.
Thanks in advance!
I would start by breaking this up into pieces and then putting them back together. You can get the created_count and claimed_count using simple aggregation like this:
SELECT created_by, COUNT(*) AS created_count
FROM myTable
GROUP BY created_by;
SELECT claimed_by, COUNT(*) AS claimed_count
FROM myTable
GROUP BY claimed_by;
To get the urgent count for each employee, I would join the two tables on the condition that the employee is either the created_by or claimed_by column, and group by employee. Instead of counting, however, I would use SUM(). I am doing this because it appears each row will be either 0 or 1, so SUM() will effectively count all non-zero rows:
SELECT e.userid, SUM(t.urgent)
FROM employee e
JOIN task t ON e.userid IN (t.created_by, t.claimed_by)
GROUP BY e.userid;
Now that you have all the bits of data you need, you can use an outer join to join all of those subqueries to the employees table to get their counts. You can use the COALESCE() function to replace any null counts with 0:
SELECT e.userid, COALESCE(u.urgent_count, 0) AS urgent_count, COALESCE(crt.created_count, 0) AS created_count, COALESCE(clm.claimed_count, 0) AS claimed_count
FROM employee e
LEFT JOIN(
SELECT e.userid, SUM(t.urgent) AS urgent_count
FROM employee e
JOIN task t ON e.userid IN (t.created_by, t.claimed_by)
GROUP BY e.userid) u ON u.userid = e.userid
LEFT JOIN(
SELECT claimed_by, COUNT(*) AS claimed_count
FROM task
GROUP BY claimed_by) clm ON clm.claimed_by = e.userid
LEFT JOIN(
SELECT created_by, COUNT(*) AS created_count
FROM task
GROUP BY created_by) crt ON crt.created_by = e.userid;
Here is an SQL Fiddle example.
hi i have 2 mysql table as follow:
items
id item_name user_id
1 test1 1
2 test2 1
3 test3 1
4 test4 1
project
id user_id items
1 1 1,3
2 1 2,4
how can write a join query that can return each items in a project?
project1 =>
item1=>
[id1] =>
[name1] =>
item3=>
[id3] =>
[name3] =>
Thanks.
UPDATE
item tbl
project tbl
First of all don't store strings of delimited values in your db. You're limiting your self with the means to normally maintain and query data. Normalize your data (in this case by introducing project_items table with project_id and item_id columns). It'll pay off big time in a long run.
In the mean time you can use FIND_IN_SET() to join your tables
SELECT p.id project_id, p.user_id, i.id item_id, i.item_name
FROM project p LEFT JOIN items i
ON FIND_IN_SET(i.id, p.items) > 0
AND p.user_id = i.user_id
ORDER BY p.id, i.id
Output:
| PROJECT_ID | USER_ID | ITEM_ID | ITEM_NAME |
----------------------------------------------
| 1 | 1 | 1 | test1 |
| 1 | 1 | 3 | test3 |
| 2 | 1 | 2 | test2 |
| 2 | 1 | 4 | test4 |
Here is SQLFiddle demo
UPDATE: Values of items should not contain spaces. Either remove them or use REPLACE() like this
ON FIND_IN_SET(i.id, REPLACE(p.items, ' ', '')) > 0
Here is SQLFiddle demo
Your approach is not good. You have to create a relational table between items and projects. Including the list of values followed by commas in a record is not a good idea.
You should create an additional table relational called: project_items
and you can use the following sentence to retrieve the items from a project
select project_items.id_project, items.item_name, items.user_id
from project_items
left join items on project_items.id_item = items.id
That's a better approach
I have the following table with messages:
+---------+---------+------------+----------+
| msg_id | user_id | m_date | m_time |
+-------------------+------------+----------+
| 1 | 1 | 2011-01-22 | 06:23:11 |
| 2 | 1 | 2011-01-23 | 16:17:03 |
| 3 | 1 | 2011-01-23 | 17:05:45 |
| 4 | 2 | 2011-01-22 | 23:58:13 |
| 5 | 2 | 2011-01-23 | 23:59:32 |
| 6 | 2 | 2011-01-24 | 21:02:41 |
| 7 | 3 | 2011-01-22 | 13:45:00 |
| 8 | 3 | 2011-01-23 | 13:22:34 |
| 9 | 3 | 2011-01-23 | 18:22:34 |
| 10 | 3 | 2011-01-24 | 02:22:22 |
| 11 | 3 | 2011-01-24 | 13:12:00 |
+---------+---------+------------+----------+
What I want is for each day, to see how many messages each user has sent BEFORE and AFTER 16:00:
SELECT
user_id,
m_date,
SUM(m_time <= '16:00') AS before16,
SUM(m_time > '16:00') AS after16
FROM messages
GROUP BY user_id, m_date
ORDER BY user_id, m_date ASC
This produces:
user_id m_date before16 after16
-------------------------------------
1 2011-01-22 1 0
1 2011-01-23 0 2
2 2011-01-22 0 1
2 2011-01-23 0 1
2 2011-01-24 0 1
3 2011-01-22 1 0
3 2011-01-23 1 1
3 2011-01-24 2 0
Because user 1 has written no messages on 2011-01-24, this date is not in the resultset. However, this is undesirable. I have a second table in my database, called "date_range":
+---------+------------+
| date_id | d_date |
+---------+------------+
| 1 | 2011-01-21 |
| 1 | 2011-01-22 |
| 1 | 2011-01-23 |
| 1 | 2011-01-24 |
+---------+------------+
I want to check the "messages" against this table. For each user, all these dates have to be in the resultset. As you can see, none of the users have written messages on 2011-01-21, and as said, user 1 has no messages on 2011-01-24. The desired output of the query would be:
user_id d_date before16 after16
-------------------------------------
1 2011-01-21 0 0
1 2011-01-22 1 0
1 2011-01-23 0 2
1 2011-01-24 0 0
2 2011-01-21 0 0
2 2011-01-22 0 1
2 2011-01-23 0 1
2 2011-01-24 0 1
3 2011-01-21 0 0
3 2011-01-22 1 0
3 2011-01-23 1 1
3 2011-01-24 2 0
How can I link the two tables so that the query result also holds rows with zero values for before16 and after16?
Edit: yes, I have a "users" table:
+---------+------------+
| user_id | user_date |
+---------+------------+
| 1 | foo |
| 2 | bar |
| 3 | foobar |
+---------+------------+
Test bed:
create table messages (msg_id integer, user_id integer, _date date, _time time);
create table date_range (date_id integer, _date date);
insert into messages values
(1,1,'2011-01-22','06:23:11'),
(2,1,'2011-01-23','16:17:03'),
(3,1,'2011-01-23','17:05:05');
insert into date_range values
(1, '2011-01-21'),
(1, '2011-01-22'),
(1, '2011-01-23'),
(1, '2011-01-24');
Query:
SELECT p._date, p.user_id,
coalesce(m.before16, 0) b16, coalesce(m.after16, 0) a16
FROM
(SELECT DISTINCT user_id, dr._date FROM messages m, date_range dr) p
LEFT JOIN
(SELECT user_id, _date,
SUM(_time <= '16:00') AS before16,
SUM(_time > '16:00') AS after16
FROM messages
GROUP BY user_id, _date
ORDER BY user_id, _date ASC) m
ON p.user_id = m.user_id AND p._date = m._date;
EDIT:
Your initial query is left as is, I hope it doesn't requires any explanations;
SELECT DISTINCT user_id, dr._date FROM messages m, date_range dr will return a cartesian or CROSS JOIN of two tables, which will give me all required date range for each user in subject. As I'm interested in each pair only once, I use DISTINCT clause. Try this query with and without it;
Then I use LEFT JOIN on two sub-selects.
This join means: first, INNER join is performed, i.e. all rows with matching fields in the ON condition are returned. Then, for each row in the left-side relation of the join that has no matches on the right side, return NULLs (thus the name, LEFT JOIN, i.e. left relation is always there and right is expected to have NULLs). This join will do what you expect — return user_id + date combinations even if there were no messages in the given date for a given user. Note that I use user_id + date sub-select first (on the left) and messages query second (on the right);
coalesce() is used to replace NULL with zero.
I hope this clarifies how this query works.
Give this a shot:
select u.user_id, u._date,
sum(_time <= '16:00') as before16,
sum(_time > '16:00') as after16
from (
select m.user_id, d._date
from messages m
cross join date_range d
group by m.user_id, d._date
) u
left join messages m on u.user_id=m.user_id
and u._date=m._date
group by u.user_id, u._date
The inner query is just building a set of all possible/desired user-date pairs. It would be more efficient to use a users table, but you didn't mention that you had one, so I won't assume. otherwise, you just need the left join to not remove the non-joined records.
EDIT
--More detailed explanation: taking the query apart.
Start with the innermost query; the goal is to get a list of all desired dates for every user. Since there's a table of users and a table of dates it can look like this:
select distinct u.user_id, d.d_date
from users u
cross join date_range d
The key here is the cross join, taking every row in the users table and associating it with every row in the date_range table. The distinct keyword is really just a shorthand for a group by on all columns, and is here just in case there's duplicated data.
Note that there are several other methods of getting this same result set (like in my original query), but this is probably the simplest from both a logical and computational standpoint.
Really, the only other steps are to add the left join (associating all of the rows we got above to all available data, and not removing anything that doesn't have any data) and the group by and select components which are basically the same as you had before. So, putting everything together it looks like this:
select t.user_id, t.d_date,
sum(m.m_time <= '16:00') as before16,
sum(m.m_time > '16:00') as after16
from (
select distinct u.user_id, d.d_date
from users u
cross join date_range d
) t
left join messages m on t.user_id = m.user_id
and t.d_date = m.m_date
group by t.user_id, t.d_date
Based on some other comments/questions, note the explicit use of prefixes for all uses of all tables and sub-queries (which is pretty straight forward since we're not using any table more than once anymore): u for the users table, d for the date_range table, t for the sub-query containing the dates to use for each user, and m for the message table. This is probably where my first explanation fell a little short, since I used the message table twice, both times with the same prefix. It works there because of the context of both uses (one was in a sub-query), but it probably isn't the best practice.
It is not neat. But if you have a user table. Then maybe something like this:
SELECT
user_id,
_date,
SUM(_time <= '16:00') AS before16,
SUM(_time > '16:00') AS after16
FROM messages
GROUP BY user_id, _date
UNION
SELECT
user_id,
date_range,
0 AS before16,
0 AS after16
FROM
users,
date_range
ORDER BY user_id, _date ASC
chezy525's solution works great, I ported it to postgresql and removed/renamed some aliases:
select users_and_dates.user_id, users_and_dates._date,
SUM(case when _time <= '16:00' then 1 else 0 end) as before16,
SUM(case when _time > '16:00' then 1 else 0 end) as after16
from (
select messages.user_id, date_range._date
from messages
cross join date_range
group by messages.user_id, date_range._date
) users_and_dates
left join messages on users_and_dates.user_id=messages.user_id
and users_and_dates._date=messages._date
group by users_and_dates.user_id, users_and_dates._date;
and ran on my machine, worked perfectly