I have a table with two columns:
id
num
1
2
2
8
1
7
7
3
I want to get as an answer to my query only ids that have more than 1 nums.
For example in my table I would want to get as a result:
id
1
How should I express my query?
Thanks in advance.
You might need something like this:
SELECT id
FROM your_table_name
GROUP BY id
HAVING count(DISTINCT num) > 1;
Google 'Aggregate functions'. Here the aggregate function is count() and it works always coupled with a GROUP BY clause. Pretty fun.
I could use some help with some SQL. Part of a website that I need to do is a place where you can add/remove stores from users. Now the query to get the stores that have already been assigned to a certain user is simple, however I can't seem to figure out how to do the opposite. Meaning getting all that haven't been added to the user yet.
So here's a template of 1 of the tables.
user_id store_id
1 11
1 12
1 14
2 15
4 16
If I run this (that 1 will be php variable): SELECT store_id FROM store_user WHERE user_id= 1;
I get the result like that:
store_id
11
12
14
And the rest would be something like: SELECT * FROM store except where id = 11,12,14.
I would appreciate if anyone could help me with that last part and preferrably put it all in to 1 query.
Edit: for those who stumble here with a similar problem, this is what worked in the end:
SELECT * FROM store where id NOT IN (select store_id from store_user where user_id = 1)
Thanks for the help everyone!
You can use IN or NOT IN
select store_id from store where id in (1,2....)
select store_id from store where id not in (1,2....)
use NOT IN
SELECT * FROM store where
id NOT IN (select store_id from store_user where user_id = 1)
I have a SQL problem. I have a table where a user gets a row for every experience they complete. The schema looks similar to this fiddle: http://sqlfiddle.com/#!2/5d6a87/4
I am trying to write a query that lists every user that has expid 1-5. So in my example it would list userids: 1,2, and 4. Since userid 3 does not have 5 rows, one for each experience that user shouldn't be listed.
What you would like to use is "group by"
the query would look like it:
select userid
from some_table
group by userid
having count(*)>=5
you can also be creative and force 5 different expeids by
select userid
from some_table
group by userid
having count(distinct expid)>=5
light reading about group by:
http://www.w3schools.com/sql/sql_groupby.asp
Good luck!
Try this:
SELECT id,userid,expid FROM
some_table
GROUP BY userid
HAVING count(*)>= 5
Result:
ID USERID EXPID
1 1 1
6 2 1
13 4 1
See result in SQL Fiddle.
You just need to add the having clause. Read more here.
Here
SELECT * FROM some_table GROUP BY UserID having count(*)>= 5;
They way I see it should be like this but your question is not 100% clear, since you said 5 rows for every ExpID.
SELECT * FROM some_table GROUP BY UserID, ExpID having count(*)>= 5;
I have a table like this
id | user_id | code | type | time
-----------------------------------
2 2 fdsa r 1358300000
3 2 barf r 1358311000
4 2 yack r 1358311220
5 3 surf r 1358311000
6 3 yooo r 1358300000
7 4 poot r 1358311220
I want to get the concatenated 'code' column for user 2 and user 3 for each matching time.
I want to receive a result set like this:
code | time
-------------------------------
fdsayooo 1358300000
barfsurf 1358311000
Please note that there is no yackpoot code because the query was not looking for user 4.
You can use GROUP_CONCAT function. Try this:
SELECT GROUP_CONCAT(code SEPARATOR '') code, time
FROM tbl
WHERE user_id in (2, 3)
GROUP BY time
HAVING COUNT(time) = 2;
SQL FIDDLE DEMO
What you are looking for is GROUP_CONCAT, but you are missing a lot of details in your question to provide a good example. This should get you started:
SELECT GROUP_CONCAT(code), time
FROM myTable
WHERE user_id in (2, 3)
GROUP BY time;
Missing details are:
Is there an order required? Not sure how ordering would be done useing grouping, would need to test if critical
Need other fields? If so you will likely end up needing to do a sub-select or secondary query.
Do you only want results with multiple times?
Do you really want no separator between values in the results column (specify the delimiter with SEPARATOR '' in the GROUP_CONCAT
Notes:
You can add more fields to the GROUP BY if you want to do it by something else (like user_id and time).
Is it possible to sort in MySQL by "order by" using a predefined set of column values (ID) like order by (ID=1,5,4,3) so I would get records 1, 5, 4, 3 in that order out?
UPDATE: Why I need this...
I want my records to change sort randomly every 5 minutes. I have a cron task to update the table to put different, random sort order in it.
There is just one problem! PAGINATION.
I will have visitors who come to my page, and I will give them the first 20 results. They will wait 6 minutes, go to page 2 and have the wrong results as the sort order has already changed.
So I thought that if I put all the IDs into a session on page 2, we get the correct records even if the sorting had already changed.
Is there any other better way to do this?
You can use ORDER BY and FIELD function.
See http://lists.mysql.com/mysql/209784
SELECT * FROM table ORDER BY FIELD(ID,1,5,4,3)
It uses Field() function, Which "Returns the index (position) of str in the str1, str2, str3, ... list. Returns 0 if str is not found" according to the documentation. So actually you sort the result set by the return value of this function which is the index of the field value in the given set.
You should be able to use CASE for this:
ORDER BY CASE id
WHEN 1 THEN 1
WHEN 5 THEN 2
WHEN 4 THEN 3
WHEN 3 THEN 4
ELSE 5
END
On the official documentation for mysql about ORDER BY, someone has posted that you can use FIELD for this matter, like this:
SELECT * FROM table ORDER BY FIELD(id,1,5,4,3)
This is untested code that in theory should work.
SELECT * FROM table ORDER BY id='8' DESC, id='5' DESC, id='4' DESC, id='3' DESC
If I had 10 registries for example, this way the ID 1, 5, 4 and 3 will appears first, the others registries will appears next.
Normal exibition
1
2
3
4
5
6
7
8
9
10
With this way
8
5
4
3
1
2
6
7
9
10
There's another way to solve this. Add a separate table, something like this:
CREATE TABLE `new_order` (
`my_order` BIGINT(20) UNSIGNED NOT NULL,
`my_number` BIGINT(20) NOT NULL,
PRIMARY KEY (`my_order`),
UNIQUE KEY `my_number` (`my_number`)
) ENGINE=INNODB;
This table will now be used to define your own order mechanism.
Add your values in there:
my_order | my_number
---------+----------
1 | 1
2 | 5
3 | 4
4 | 3
...and then modify your SQL statement while joining this new table.
SELECT *
FROM your_table AS T1
INNER JOIN new_order AS T2 on T1.id = T2.my_number
WHERE ....whatever...
ORDER BY T2.my_order;
This solution is slightly more complex than other solutions, but using this you don't have to change your SELECT-statement whenever your order criteriums change - just change the data in the order table.
If you need to order a single id first in the result, use the id.
select id,name
from products
order by case when id=5 then -1 else id end
If you need to start with a sequence of multiple ids, specify a collection, similar to what you would use with an IN statement.
select id,name
from products
order by case when id in (30,20,10) then -1 else id end,id
If you want to order a single id last in the result, use the order by the case. (Eg: you want "other" option in last and all city list show in alphabetical order.)
select id,city
from city
order by case
when id = 2 then city else -1
end, city ASC
If i had 5 city for example, i want to show the city in alphabetical order with "other" option display last in the dropdown then we can use this query.
see example other are showing in my table at second id(id:2) so i am using "when id = 2" in above query.
record in DB table:
Bangalore - id:1
Other - id:2
Mumbai - id:3
Pune - id:4
Ambala - id:5
my output:
Ambala
Bangalore
Mumbai
Pune
Other
SELECT * FROM TABLE ORDER BY (columnname,1,2) ASC OR DESC