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.
select * from "Test"."EMP"
id
1
2
3
4
5
Select SUM(1) FROM "Test"."EMP";
Select SUM(2) FROM "Test"."EMP";
Select SUM(3) FROM "Test"."EMP";
why the output of these queries is?
5
10
15
And
I don't understand why they write table name like this "Test"."EMP"
your table has 5 records. the statement select 1 from test.emp returns 5 records with values as 1 for all 5 records.
id
1
1
1
1
1
This is because db engine simply returns 1 for each existing record without reading the contents of the cell. and same happens for select <any static value> from test.emp
same happens for 2 and 3
id
2
2
2
2
2
hence there are 5 records returned with the static values and sum of those values will be the product of static number passed in the select statement and total records in the table
additional fact: It is always recommended to perform count(1) than count(*) as it consumes less resource and hence less load on the server
I don't think it's "Test"."EMP" with double quotes.. it's probably `Test`.`EMP` with backticks instead. The definition means its database_name.table_name. This is the recommended format to get the correct table_name from database_name; in this case, you're specifically making the syntax to query from `Test`.`EMP`. Read more about identifier qualifiers.
As for SUM(x), the x get's repeated according to the rows present in the table. So SUM(1) on 5 rows is 1+1+1+1+1, SUM(2) on 5 rows is 2+2+2+2+2, and so on.
Here is a table
id date name
1 180101 josh
2 180101 peter
3 180101 julia
4 180102 robert
5 180103 patrick
6 180104 josh
7 180104 adam
I need to get all the names whom having the same days as 'josh'. how can i achieve it without groupping the whole table together. i need to keep it efficient (this is not my real table, i just simplified my problem here, and i have hundred thousands of records, and 99% of the rows have different dates, so groupable rows by date is kind of rare).
So basicaly what i want is: if 'josh' is the target, i need to get 'josh,peter,julia,adam' (actually the first 10 distinct names sharing the same date with josh).
SELECT
COUNT(date) as datecount,
GROUP_CONCAT(DISTINCT name) as names,
FROM
table
GROUP BY
date
HAVING
datecount>1
// && name IN ('josh') would work nice for me, but im getting error because 'name' is not in GROUPED BY
LIMIT 10
Any idea ? As i mentioned it needs to be fast, and most of the rows have unique dates
Join the table with itself on date:
select distinct t1.name
from tbl t1
join tbl t2 using (date)
where t2.name = 'josh'
Demo
For the best performance you would have indexes on (name) and (date, name).
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
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