How to find only id that has different values in MySQL? - mysql

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.

Related

How to get exact rows count of particular column in MySQL table

I want to get exact row count of specified column
Example: Table
Name Id Age
_______________________________
Jon 1 30
Merry 2 40
William 50
David
There are 4 rows in table but i want to count ID column.
I am using below query to achieve it
select count(Id) from table;
But its returning 4 and I know why it is returning 4 but I want output as 2 because there are only two rows in Id column.
How can i achieve it?

Try this:
select count(Id) from table where id>0;

with the help of #blabla_bingo and #Edwin Dijk finally i have achieved it by below query
select count(Id) from table where Id!="";

SQL: Repeated records by grouping some columns

I have a data like,
ID Name ItemA ItemB ItemC
OXZ234 Adam 4 4 5
OXZ234 Adam 1 2 3
OXZ345 Tarzen 6 7 8
OXDER2 William 9 8 2
OXDER2 William 0 8 0
I need to find how much of food each person eats. For example by referring first two records I can say, Adam of ID OXZ234 ate ItemA-5, ItemB-6 and ItemC-8. But for small amount of data this kind of manual calculation is affordable. I have a million data records like this. So initially I need to find the records which is having same ID and name but only items count differing.
I have tried the query to find duplicate records by grouping all columns like below,
select ID,Name,ItemA,ItemB,ItemC, COUNT(*)
from DATA_REFRESH
group by ID,Name,ItemA,ItemB,ItemC
having COUNT(*) > 1
But Now I have to identify records having items columns differed.
So the expected output is like,
OXZ234 Adam 2
OXDER2 William 2
OXZ345 Tarzen 1
Any suggestion would be helpful!

You want SUM
select ID,
Name,
sum(ItemA) as ItA,
sum(ItemB) as ItB,
sum(ItemC) as ItC,
count(ID) as Occurrences -- Counts the number of entries per person
from DATA_REFRESH
group by ID,Name
having count(ID) >1 -- restricts this so only those with more than one entry appear

Hi, You can have a simple query without having clause,
select ID,Name,COUNT(*)
from DATA_REFRESH
group by ID,Name order by COUNT(*) desc ;

Simply try like this,
select ID,Name,COUNT(*)
from Sample_Check
group by ID,Name
having COUNT(*) > 1

mysql query that a user has a row for every experience id

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;

Retrieve unique data from MYSQL database

I have a table in my database which contains 5 rows. I am trying to write an sql statement that will retrieve all rows which only have 1 agency assigned to them.
case_id agency_ID
1 4
2 4
3 3
4 2
4 4
To clarify I would like to select the required rows (and any further rows) but only if the case_id is unique. Any rows with duplicates would be ommited.
I have tried to use DISTINCT(case_id), COUNT(*) to count all rows but it doesn't work and it's slowly sapping away my soul. It is probably an easy fix, but for the life of me I just can't see it.
Hope this is enough information to go on. Any help would be greatly appreciated.

SELECT * FROM your_table GROUP BY case_id HAVING COUNT(agency_ID) = 1

You can try
SELECT case_id,agency_ID,COUNT(case_id) as c
FROM yourTable
GROUP BY case_id
HAVING (c=1)

Returning query results in predefined order

Is it possible to do a SELECT statement with a predetermined order, ie. selecting IDs 7,2,5,9 and 8 and returning them in that order, based on nothing more than the ID field?
Both these statements return them in the same order:
SELECT id FROM table WHERE id in (7,2,5,9,8)
SELECT id FROM table WHERE id in (8,2,5,9,7)

I didn't think this was possible, but found a blog entry here that seems to do the type of thing you're after:
SELECT id FROM table WHERE id in (7,2,5,9,8)
ORDER BY FIND_IN_SET(id,"7,2,5,9,8");
will give different results to
SELECT id FROM table WHERE id in (7,2,5,9,8)
ORDER BY FIND_IN_SET(id,"8,2,5,9,7");
FIND_IN_SET returns the position of id in the second argument given to it, so for the first case above, id of 7 is at position 1 in the set, 2 at 2 and so on - mysql internally works out something like
id | FIND_IN_SET
---|-----------
7 | 1
2 | 2
5 | 3
then orders by the results of FIND_IN_SET.

Your best bet is:
ORDER BY FIELD(ID,7,2,4,5,8)
...but it's still ugly.

Could you include a case expression that maps your IDs 7,2,5,... to the ordinals 1,2,3,... and then order by that expression?

All ordering is done by the ORDER BY keywords, you can only however sort ascending and descending. If you are using a language such as PHP you can then sort them accordingly using some code but I do not believe it is possible with MySQL alone.

This works in Oracle. Can you do something similar in MySql?
SELECT ID_FIELD
FROM SOME_TABLE
WHERE ID_FIELD IN(11,10,14,12,13)
ORDER BY
CASE WHEN ID_FIELD = 11 THEN 0
WHEN ID_FIELD = 10 THEN 1
WHEN ID_FIELD = 14 THEN 2
WHEN ID_FIELD = 12 THEN 3
WHEN ID_FIELD = 13 THEN 4
END

You may need to create a temp table with an autonumber field and insert into it in the desired order. Then sort on the new autonumber field.

Erm, not really. Closest you can get is probably:
SELECT * FROM table WHERE id IN (3, 2, 1, 4) ORDER BY id=4, id=1, id=2, id=3
But you probably don't want that :)
It's hard to give you any more specific advice without more information about what's in the tables.

It's hacky (and probably slow), but you can get the effect with UNION ALL:
SELECT id FROM table WHERE id = 7
UNION ALL SELECT id FROM table WHERE id = 2
UNION ALL SELECT id FROM table WHERE id = 5
UNION ALL SELECT id FROM table WHERE id = 9
UNION ALL SELECT id FROM table WHERE id = 8;
Edit: Other people mentioned the find_in_set function which is documented here.

You get answers fast around here, don't you…
The reason I'm asking this is that it's the only way I can think of to avoid sorting a complex multidimensional array. I'm not saying it would be difficult to sort, but if there were a simpler way to do it with straight sql, then why not.

One Oracle solution is:
SELECT id FROM table WHERE id in (7,2,5,9,8)
ORDER BY DECODE(id,7,1,2,2,5,3,9,4,8,5,6);
This assigns an order number to each ID. Works OK for a small set of values.

Best I can think of is adding a second Column orderColumn:
7 1
2 2
5 3
9 4
8 5
And then just do a ORDER BY orderColumn