Assuming a table with a column where integers are stored.
-----------------------------
id | some_int | some_other_value
-----------------------------
1 5 hello
2 9 how
3 987 are
4 5 you
5 9 thanks
6 1 for
7 5 answering. :-)
Is SELECT * FROM mytable ORDER BY some_int; distinct? Meaning will it always return the rows in the same order, after each query?
To the best of my knowledge if there are duplicates in the order by clause, there is no guarantee for the order in which the duplicates are presented.
If this is a concern, you could modify the order by to include the primary key (id I am assuming).
ORDER BY some_int, id
Since id is a primary key, it should also be indexed. Thus the performance difference will be minimal.
Doing this order by does not always have to give you the same result order. It is not frequent that the order will be different but it's not a 100% safe to assume that the order will always be the same as it is not an unique value that you're ordering by. To achieve this you should also include the primary key after the initial order.
SELECT * FROM mytable ORDER BY some_int, id
Use this:
SELECT * FROM mytable ORDER BY some_int,id;
So it will sort for 'some_int' and then using 'id' Duplicates for 'some_id' will be set to fixed position using 'id' column
Related
Does the following query (specifically the ORDER BY price IS NULL DESC) use the index on the price column?
SELECT * FROM products ORDER BY price IS NULL DESC, price DESC
product_id | price
-----------|------
1 5
2 6
3 NULL
4 8
I would like the query to return
product_id | price
-----------|------
3 NULL
4 8
2 6
1 5
You are not purely "sorting by IS NULL"; you have a composite (multi-column) index.
That ORDER BY is too complex to avoid a "sort". (Note: 'filesort' does not necessarily involve a 'file', this example is likely to do all the work in RAM.)
The processing goes something like this:
Collect 3 columns: product_id, price, (price IS NULL)
Note that price IS NULL had to be evaluated, getting 0 (not null) or 1 (is null)
Logically combine the 0/1 with the price to make a 2-column pair of columns to drive the sorting
Sort
Deliver the results.
If you had said simply ORDER BY price DESC, INDEX(price) could be used, but it would sort the NULLs last. (I say "could" because the "filesort" might be faster than bouncing between the index's BTree and the data.)
Caveat: You have probably "simplified" the query for the Question. In response, I have addressed your simplified query. And I provided some hints that may Answer may or may not apply in the "general case".
I have a posts table, where the ids are not necessarily in sequence. The posts should be sorted by their created timestamp value, but it is possible, that two posts have the same created timestamp, in which case I would like to sort them by their id.
Given this ordering, I was trying to find the entry that immediately precedes a specified entry.
Example: (descending order)
id | created
------|------------
4 | 2017-01-05
15 | 2017-01-04
12 | 2017-01-04
2 | 2017-01-04
8 | 2017-01-02
11 | 2017-01-01
(I simplified the timestamp to only the date, but you get the idea.)
In this example, given the id 2, I would like to return the id 8. After some experimentation I have come up with the following:
SELECT id
FROM posts
WHERE created < ? OR id < ?
AND created = ?
ORDER BY created DESC, id DESC
LIMIT 1
This works, given the timestamp and id in the correct places in the query, but is very tedious.
My question is now:
Is there, given an ordering, an easier way to find the preceding element?
Clarification:
The above code does return the correct results, but I was wondering whether there was some more general way to achieve the same, one that will be more easily adapted if the sorting specification changes..
You can use tuples:
SELECT id
FROM posts
WHERE (created, id) < (?,?)
ORDER BY created DESC, id DESC
LIMIT 1
what you have is fine ... otherwise the only way i can think of is create an order out of that using concat and lpad but that's even more confusing to read than what you have.
the more confusing method using concat and lpad
SELECT id,created FROM POSTS
WHERE CONCAT(created,LPAD(id, 8, '0')) < CONCAT('2017-01-04',LPAD(2, 8, '0'))
ORDER BY created desc, id desc
LIMIT 1
For a dating application, I have a few tables that I need to query for a single output with a LIMIT 10 of both queries combined. It seems difficult to do at the moment, even though it's not an issue to query them separately, but the LIMIT 10 won't work as the numbers are not exact (ex. not LIMIT 5 and LIMIT 5, one query may return 0 rows, while the other 10, depending on the scenario).
members table
member_id | member_name
------------------------
1 Herb
2 Karen
3 Megan
dating_requests
request_id | member1 | member2 | request_time
----------------------------------------------------
1 1 2 2012-12-21 12:51:45
dating_alerts
alert_id | alerter_id | alertee_id | type | alert_time
-------------------------------------------------------
5 3 2 platonic 2012-12-21 10:25:32
dating_alerts_status
status_id | alert_id | alertee_id | viewed | viewed_time
-----------------------------------------------------------
4 5 2 0 0000-00-00 00:00:00
Imagine you are Karen and just logged in, you should see these 2 items:
1. Herb requested a date with you.
2. Megan wants a platonic relationship with you.
In one query with a LIMIT of 10. Instead here are two queries that need to be combined:
1. Herb requested a date with you.
-> query = "SELECT dr.request_id, dr.member1, dr.member2, m.member_name
FROM dating_requests dr
JOIN members m ON dr.member1=m.member_id
WHERE dr.member2=:loggedin_id
ORDER BY dr.request_time LIMIT 5";
2. Megan wants a platonic relationship with you.
-> query = "SELECT da.alert_id, da.alerter_id, da.alertee_id, da.type,
da.alert_time, m.member_name
FROM dating_alerts da
JOIN dating_alerts_status das ON da.alert_id=das.alert_id
AND da.alertee_id=das.alertee_id
JOIN members m ON da.alerter_id=m.member_id
WHERE da.alertee_id=:loggedin_id AND da.type='platonic'
AND das.viewed='0' AND das.viewed_time<da.alert_time
ORDER BY da.alert_time LIMIT 5";
Again, sometimes both tables may be empty, or 1 table may be empty, or both full (where LIMIT 10 kicks in) and ordered by time. Any ideas on how to get a query to perform this task efficiently? Thoughts, advice, chimes, optimizations are welcome.
You can combine multiple queries with UNION, but only if the queries have the same number of columns. Ideally the columns are the same, not only in data type, but also in their semantic meaning; however, MySQL doesn't care about the semantics and will handle differing datatypes by casting up to something more generic - so if necessary you could overload the columns to have different meanings from each table, then determine what meaning is appropriate in your higher level code (although I don't recommend doing it this way).
When the number of columns differs, or when you want to achieve a better/less overloaded alignment of data from two queries, you can insert dummy literal columns into your SELECT statements. For example:
SELECT t.cola, t.colb, NULL, t.colc, NULL FROM t;
You could even have some columns reserved for the first table and others for the second table, such that they are NULL elsewhere (but remember that the column names come from the first query, so you may wish to ensure they're all named there):
SELECT a, b, c, d, NULL AS e, NULL AS f, NULL AS g FROM t1
UNION ALL -- specify ALL because default is DISTINCT, which is wasted here
SELECT NULL, NULL, NULL, NULL, a, b, c FROM t2;
You could try aligning your two queries in this fashion, then combining them with a UNION operator; by applying LIMIT to the UNION, you're close to achieving your goal:
(SELECT ...)
UNION
(SELECT ...)
LIMIT 10;
The only issue that remains is that, as presented above, 10 or more records from the first table will "push out" any records from the second. However, we can utilise an ORDER BY in the outer query to solve this.
Putting it all together:
(
SELECT
dr.request_time AS event_time, m.member_name, -- shared columns
dr.request_id, dr.member1, dr.member2, -- request-only columns
NULL AS alert_id, NULL AS alerter_id, -- alert-only columns
NULL AS alertee_id, NULL AS type
FROM dating_requests dr JOIN members m ON dr.member1=m.member_id
WHERE dr.member2=:loggedin_id
ORDER BY event_time LIMIT 10 -- save ourselves performing excessive UNION
) UNION ALL (
SELECT
da.alert_time AS event_time, m.member_name, -- shared columns
NULL, NULL, NULL, -- request-only columns
da.alert_id, da.alerter_id, da.alertee_id, da.type -- alert-only columns
FROM
dating_alerts da
JOIN dating_alerts_status das USING (alert_id, alertee_id)
JOIN members m ON da.alerter_id=m.member_id
WHERE
da.alertee_id=:loggedin_id
AND da.type='platonic'
AND das.viewed='0'
AND das.viewed_time<da.alert_time
ORDER BY event_time LIMIT 10 -- save ourselves performing excessive UNION
)
ORDER BY event_time
LIMIT 10;
Of course, now it's up to you to determine what type of row you're dealing with as you read each record in the resultset (suggest you test request_id and/or alert_id for NULL values; alternatively one could add an additional column to the results that explicitly states from which table each record originated, but it should be equivalent provided those id columns are NOT NULL).
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
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