Mysql show different columns for different where clauses - mysql

I want to show a output table that counts all the users found in a table.
Basically I want the output to look like:
+-----------+-----------+
| user1 | user2 |
+-----------+-----------+
| 5 | 2 |
+-----------+-----------+
I'm just using a dummy table to test this. My query looks like this:
(
select
name as user1
from
users
where
name = 'root'
) UNION (
select
name as user2
from
users
where
name = 'not_root'
)
Which only outputs something like this:
+-----------+
| user1 |
+-----------+
| 5 |
| 2 |
+-----------+

EDITED
The main idea of the approach it's treat a table as two different virtual tables in subquery. We can make nested select statement e.g. (select count(*) as c from users where name = 'root') u1 MySql treats it as particular table named u1 with one row and one column named c.
select u1.c as root_cnt, u2.c as not_root_cnt
from (select count(*) as c from users where name = 'root') u1,
(select count(*) as c from users where name = 'not_root') u2
or
Moreover if you have select statement that returns only one row you can put nested selects directly in fields list
select (select count(*) as c from users where name = 'root') as root_cnt, (select count(*) as c from users where name = 'not_root') as not_root_cnt
Definite disadvantage of such approach it's extra subqueries. Method based on using case when construction free from such disadvantage.

Try this
SELECT
SUM(CASE WHEN Name = 'root' THEN 1 ELSE 0 END) user1,
SUM(CASE WHEN Name = 'not_root' THEN 1 ELSE 0 END) user2
FROM Users

You can use a case statement inside of count to get the counts in separate columns
select
count(case when name = 'root' then 1 end) user1,
count(case when name = 'not_root' then 1 end) user2
from users
where name in ('root','not_root')

It seems your query is wrong..
union won't combine the results of two queries in the way you have described above.
union would combine the result of two or more select statements but wouldn't "join" it.
You might want to use joins for this cause. still you wouldn't be able to put 5|2 in same row as it basically suggests --> fetch user 1 and user 2 values for one particular type

i think group by is a much better approach:
select user, count(user) from users group by user

Related

Query: I have 4 rows, need to add the results from 3 rows into one, and leave the last row untouched

I have a kind of tricky question for this query. First the code:
SELECT user_type.user_type_description,COUNT(incident.user_id) as Quantity
FROM incident
INNER JOIN user ON incident.user_id=user.user_id
INNER JOIN user_type ON user.user_type=user_type.user_type
WHERE incident.code=2
GROUP BY user.user_type
What Am I doing?
For example, I am counting police reports of robbery, made from different kind of users. In my example, "admin" users reported 6 incidents of code "2" (robbery) and so on, as is showed in 'where' clause (incident must be robbery, also code 2).
this brings the following result:
+-----------------------+----------+
| user_type_description | Quantity |
+-----------------------+----------+
| Admin | 6 |
| Moderator | 8 |
| Fully_registered_user | 8 |
| anonymous_user | 9 |
+-----------------------+----------+
Basically Admin,Moderator and Fully_registered_user are appropriately registered users. I need to add them in a result where it shows like:
+--------------+------------+
| Proper_users | Anonymous |
+--------------+------------+
| 22 | 9 |
+--------------+------------+
I am not good with sql. Any help is appreciated. Thanks.
You can try to use condition aggregate function base on your current result set.
SUM with CASE WHEN expression.
SELECT SUM(CASE WHEN user_type_description IN ('Admin','Moderator','Fully_registered_user') THEN Quantity END) Proper_users,
SUM(CASE WHEN user_type_description = 'anonymous_user' THEN Quantity END) Anonymous
FROM (
SELECT user_type.user_type_description,COUNT(incident.user_id) as Quantity
FROM incident
INNER JOIN user ON incident.user_id=user.user_id
INNER JOIN user_type ON user.user_type=user_type.user_type
WHERE incident.code=2
GROUP BY user.user_type
) t1
You just need conditional aggregation:
SELECT SUM( ut.user_type_description IN ('Admin', 'Moderator', 'Fully_registered_user') ) as Proper_users,
SUM( ut.user_type_description IN ('anonymous_user') as anonymous
FROM incident i INNER JOIN
user u
ON i.user_id = u.user_id INNER JOIN
user_type ut
ON u.user_type = ut.user_type
WHERE i.code = 2;
Notes:
Table aliases make the query easier to write and to read.
This uses a MySQL shortcut for adding values -- just just adding the booelean expressions.
I would solve it with a CTE, but it would be better to have this association in a table.
WITH
user_type_categories
AS
(
SELECT 'Admin' AS [user_type_description] , 'Proper_users' AS [user_type_category]
UNION SELECT 'Moderator' AS [user_type_description] , 'Proper_users' AS [user_type_category]
UNION SELECT 'Fully_registered_user' AS [user_type_description] , 'Proper_users' AS [user_type_category]
UNION SELECT 'anonymous_user' AS [user_type_description] , 'Anonymous' AS [user_type_category]
)
SELECT
CASE WHEN utc.[user_type_category] = 'Proper_users' THEN
SUM(incident.user_id)
END AS [Proper_Users_Quantity]
, CASE WHEN utc.[user_type_category] = 'Anonymous' THEN
SUM(incident.user_id)
END AS [Anonymous_Quantity]
FROM
[incident]
INNER JOIN [user] ON [incident].[user_id] = [user].[user_id]
INNER JOIN [user_type] ON [user].[user_type] = [user_type].[user_type]
LEFT JOIN user_type_categories AS utc ON utc.[user_type_description] = [user_type].[user_type_description]
WHERE
[incident].[code] = 2

Join two subqueries and have a field: division of the results of two subqueries

I have a table like this:
userid | trackid | path
123 70000 ad
123 NULL abc.com
123 NULL Apply
345 70001 Apply
345 70001 Apply
345 NULL Direct
345 NULL abc.com
345 NULL cdf.com
And I want a query like this. When path='abc.com', num_website +1; when path='Apply', num_apply +1
userid | num_website | num_Apply | num_website/num_Apply
123 1 1 1
345 1 2 0.5
My syntax looks like this:
select * from
(select userid,count(path) as is_CWS
from TABLE
where path='abc.com'
group by userid
having count(path)>1) a1
JOIN
(select userid,count(userid) as Apply_num from
where trackid is not NULL
group by userid) a2
on a1.userid=a2.userid
My question is
1. how to have the field num_website/num_apply in term of my syntax above?
2. is there any other easier way to get the result I want?
Any spots shared will appreciate.
The simplest way to do it would be to change the select line:
SELECT a1.userid, a1.is_CWS, a2.Apply_num, a1.is_CWS/a2.Apply_num FROM
(select userid,count(path) as is_CWS
from TABLE
where path='abc.com'
group by userid
having count(path)>1) a1
JOIN
(select userid,count(userid) as Apply_num
from TABLE
where trackid is not NULL
group by userid) a2
on a1.userid=a2.userid
and then continue with the rest of your query as you have it. The star means "select everything." If you wanted to select only a few things, you would just list those things in place of the star, and if you wanted to select some other values based on those things, you would put those in the stars as well. In this case a1.is_CWS/a2.Apply_num is an expression, and MySql knows how to evaluate it based on the values of a1.is_CWS and a2.Apply_num.
In the same vein, you can do a lot of what those subqueries are doing in a single expression instead of a subquery. objectNotFound has the right idea. Instead of doing a subquery to retrieve the number of rows with a certain attribute, you can select SUM(path="abc.com") as Apply_num and you don't have to join anymore. Making that change gives us:
SELECT a1.userid,
SUM(path="abc.com") as is_CWS,
a2.Apply_num,
is_CWS/a2.Apply_num FROM
TABLE
JOIN
(select userid,count(userid) as Apply_num
FROM TABLE
where trackid is not NULL
group by userid) a2
on a1.userid=a2.userid
GROUP BY userid
Notice I moved the GROUP BY to the end of the query. Also notice instead of referencing a1.is_CWS I now reference just is_CWS (it's no longer inside the a1 subtable so we can just reference it)
You can do the same thing to the other subquery then they can share the GROUP BY clause and you won't need the join anymore.
to get you started ... you can build on top of this :
select
userid,
SUM(CASE WHEN path='abc.com'then 1 else 0 end ) as num_website,
SUM(CASE WHEN path='Apply' and trackid is not NULL then 1 else 0 end ) as Apply_Num
from TABLE
WHERE path='abc.com' or path='Apply' -- may not need this ... play with it
group by userid

MYSQL select from table where field is in the field

UPDATED
id | id_list
1 | 2,3,5,7
2 | 1,4,5,6
3 | 1,4,6,7
4 | 2,3,5,8
5 | 1,2,4,8
6 | 2,3,7,8
7 | 1,3,6,9
8 | 4,5,6,9
9 | 7,8
let's say I'm up to the content of id=1
I wanted to select all the rows where id is in id_list of id=1 PLUS the row where id=1
so the result would be
rows with id = 1,2,3,5,7
How to do this query guys?
You can also use a self join
Using IN()
select * from atable a
join atable b on (a.id = b.id )
where 1 IN (a.id_list) or b.id =1
Fiddle with IN()
Using FIND_IN_SET()
select * from atable a
join atable b on (a.id = b.id )
where FIND_IN_SET('1', a.id_list) or b.id =1
Fiddle with FIND_IN_SET()
Using UNION ALL
select * from atable
where id =1
UNION ALL
select * from atable
where 1 IN (id_list)
Fiddle with UNION ALL
Your database design is broken; id_list should be represented as a join table instead of as a string. However, to solve your immediate problem:
select * from table where id=1
or id_list like '1%'
or id_list like '%,1,%'
or id_list like '%,1'
Adjust as needed for PreparedStatement. You have to provide all three cases because if you just did
or id_list like '%1%'
and id_list contained the value 11, you'd get an incorrect match
Try this (see SQL-Fiddle):
SELECT * FROM tbl
WHERE id = 1
OR INSTR((SELECT id_list FROM tbl WHERE id = '1'), id) > 0;
Tested with MySQL 5.5.30
try this one
select *
from tbl
where id=1
or id_list like '%1%'
This appears to call for a union of two sets. The one set would be the single row whose id matches the specified value:
SELECT
id
FROM
atable
WHERE
id = #id
The other set would be the result of this self-join:
SELECT
item.id
FROM
atable AS item
INNER JOIN
atable AS list
ON
FIND_IN_SET(item.id, list.id_list)
WHERE
list.id = #id
That is, the row with the specified id is matched against every row in the table on the condition that the other row's id is found in the specified row's id_list.
You can try the complete query at SQL Fiddle.
Please note that lists aren't a very good design feature. In your situation, it might be better to use a many-to-many table as suggested by #Othman. Only I would probably use a slightly different query to get the desired output, because his doesn't include the specified row itself:
SELECT
id
FROM
manytomany
WHERE
id = #id
UNION
SELECT
linked_id
FROM
manytomany
WHERE
id = #id
;
While the entries in manytomany are assumed to be unique, the query uses the UNION DISTINCT operator because of the potential duplicates returned by the first subquery, although it is possible to move the application of DISTINCT to the first subquery only like this:
SELECT DISTINCT
id
FROM
manytomany
WHERE
id = #id
UNION ALL
SELECT
linked_id
FROM
manytomany
WHERE
id = #id
;
That first subquery could actually be rewritten simply as SELECT #id AS id, but the rewrite would only make sense if the passed value was guaranteed to be valid, i.e. that it would definitely be found in manytomany.id.
Here's a demo for the other approach too (all three variations, including the SELECT #id AS id one).
I don't have an answer for your question but I encourage you to redesign your table like this I think this called many to many relation
id | friend
1 | 2
1 | 3
1 | 5
1 | 7
2 | 1
2 | 4
2 | 5
2 | 6
3 | 1
3 | 4
3 | 6
3 | 7
And then your query will be like this
SELECT DISTINCT(friend) FROM `new` WHERE id = 1
I am assuming you are using php..
My suggestion is to grab the id_list for id 1.
Explode that id_list on the comma, and then do another mysql query to grab the remaining results for 5|3|6|8
ex) $idarray = explode(",", $result);
select * from your_table where id in ('5','3','6','8')
OPTION 2:
SELECT * FROM your_table
WHERE id = '1'
OR id IN ('\''+(SELECT REPLACE(id_list,',','\',\'') FROM your_table WHERE id = '1')+'\'')
EDIT: Oops, sorry, that should be an OR instead.

how to find duplicate count without counting original

I need to count the number of duplicate emails in a mysql database, but without counting the first one (considered the original). In this table, the query result should be the single value "3" (2 duplicate x#q.com plus 1 duplicate f#q.com).
TABLE
ID | Name | Email
1 | Mike | x#q.com
2 | Peter | p#q.com
3 | Mike | x#q.com
4 | Mike | x#q.com
5 | Frank | f#q.com
6 | Jim | f#q.com
My current query produces not one number, but multiple rows, one per email address regardless of how many duplicates of this email are in the table:
SELECT value, count(lds1.leadid) FROM leads_form_element lds1 LEFT JOIN leads lds2 ON lds1.leadID = lds2.leadID
WHERE lds2.typesID = "31" AND lds1.formElementID = '97'
GROUP BY lds1.value HAVING ( COUNT(lds1.value) > 1 )
It's not one query so I'm not sure if it would work in your case, but you could do one query to select the total number of rows, a second query to select distinct email addresses, and subtract the two. This would give you the total number of duplicates...
select count(*) from someTable;
select count(distinct Email) from someTable;
In fact, I don't know if this will work, but you could try doing it all in one query:
select (count(*)-(count(distinct Email))) from someTable
Like I said, untested, but let me know if it works for you.
Try doing a group by in a sub query and then summing up. Something like:
select sum(tot)
from
(
select email, count(1)-1 as tot
from table
group by email
having count(1) > 1
)

Count 2 values in 1 query

I need help with writing a query (mySQL) to
Total number of users in a project, and
Number of users in that project that have been "disabled"
Now, I can do it separately, like this:
SELECT count(users) FROM t1 WHERE project = 3
SELECT count(users) FROM t1 WHERE project = 3 AND status = "disabled"
But there's gotta be a simple way of combining the two in one query...
You can use UNION:
mysql> SELECT count(users) AS result FROM t1 WHERE project = 3 UNION
SELECT count(users) AS result FROM t1 WHERE project = 3 AND status = "disabled"
+--------+
| result |
+--------+
| 2 |
| 1 |
+--------+
Other solution are subqueries:
mysql> SELECT (SELECT COUNT(ID) FROM t1 WHERE project = 3) AS total_users,
(SELECT COUNT(ID) FROM t1 WHERE project = 3 AND status = 'disabled') AS disabled_users;
+-------------+----------------+
| total_users | disabled_users |
+-------------+----------------+
| 2 | 0 |
+-------------+----------------+
Anyway, the truth about those solutions is that two separate queries will do just fine. Only on very large systems this would matter. When we look at database performance, those two will be slower than two separate queries. The only reason we would want to do that is overhead from connection to the database.
Select
Project,
Count(Users),
Sum(Case When Status = 'Disabled' Then 1 Else 0 End)
From
T1
Group By Project
While I was typing stu gave an answer already. I had basically the same one with slightly different syntax:
select count(users), sum(if(status='disabled',1,0))
from t1
where project = 3 ;
Well you could do something like
SELECT status, count(users) FROM t1 **group by status** WHERE project = 3
to get a resultset of status and count of how many records in that status. Add all counts together for the total, remember the count for "disabled" for the disabled count.
SELECT status, count(users) FROM t1 group by status WHERE project = 3 AND status = "disabled"