I want to search for records where a particular field either STARTS WITH some string (let's say "ar") OR that field CONTAINS the string, "ar".
However, I consider the two conditions different, because I'm limiting the number of results returned to 10 and I want the STARTS WITH condition to be weighted more heavily than the CONTAINS condition.
Example:
SELECT *
FROM Employees
WHERE Name LIKE 'ar%' OR Name LIKE '%ar%'
LIMIT 10
The catch is that is that if there are names that START with "ar" they should be favored. The only way I should get back a name that merely CONTAINS "ar" is if there are LESS than 10 names that START with "ar"
How can I do this against a MySQL database?
You need to select them in 2 parts, and add a Preference tag to the results. 10 from each segment, then merge them and take again the best 10. If segment 1 produces 8 entries, then segment 2 of UNION ALL will product the remaining 2
SELECT *
FROM
(
SELECT *, 1 as Preferred
FROM Employees
WHERE Name LIKE 'ar%'
LIMIT 10
UNION ALL
SELECT *
FROM
(
SELECT *, 2
FROM Employees
WHERE Name NOT LIKE 'ar%' AND Name LIKE '%ar%'
LIMIT 10
) X
) Y
ORDER BY Preferred
LIMIT 10
Assign a code value to results, and sort by the code value:
select
*,
(case when name like 'ar%' then 1 else 2 end) as priority
from
employees
where
name like 'ar%' or name like '%ar%'
order by
priority
limit 10
Edit:
See Richard aka cyberkiwi's answer for a more efficient solution if there are potentially lots of matches.
My solution is:
SELECT *
FROM Employees
WHERE Name LIKE '%ar%'
ORDER BY instr(name, 'ar'), name
LIMIT 10
The instr() looks for the first occurrence of the pattern in question. AR% will come before xxAR.
This prevents:
Should only do table scan 1 time. Unions and derived tables do 3. The first two on the columns to filter out the patterns and then the 3rd on the subset to find where they equal - since union filters out dupes.
Gives a true sort based on the location of the pattern. Wx > xW > xxW > etc...
Try this (don't have a MySQL instance immediately available to test with):
SELECT * FROM
(SELECT * FROM Employees WHERE Name LIKE 'ar%'
UNION
SELECT * FROM Employees WHERE Name LIKE '%ar%'
)
LIMIT 10
There are probably better ways to do it, but that immediately sprang to mind.
SELECT *
FROM Employees
WHERE Name LIKE 'ar%' OR Name LIKE '%ar%'
ORDER BY LIKE 'ar%' DESC
LIMIT 10
Should work orders by the binary true / false for like and if index'ed should benefit from the index
Related
I am trying to do SQL code in mysqli query to select rows with higher priority more often. I have a DB where all posts are sorted by priority, but I want it select like this (10 - the highest priority):
**Priority**
10
3
10
9
7
10
9
1
10
How can I do this? I have tried that to solve by more ways but no result. Thank you.
If you want to sample your data with preference to higher priorities, you could do something like this:
SELECT *
FROM (
SELECT OrderDetailID
,mod(OrderDetailID, 10) + 1 AS priority
,rand() * 10 AS rand_priority
FROM OrderDetails
) A
WHERE rand_priority < priority
ORDER BY OrderDetailID
This query runs in MySQL Tryit from W3Schools.
mod(OrderDetailID, 10) + 1 simulates a 1-10 priority - your table just has this value in it already
rand() * 10 gives you a random number between 0 and 10
Then by filtering to only ones where the random number is less than the priority, you get a result set where the higher priorities are more likely.
You may use rank function if your MySQL version supports it. It will order your data by priority in descending order and ranks each row. If the two rows have same priority then both rows will have same ranking. Then you can filter out the first rank data which will give you highest priority rows always.
Select * FROM
(
SELECT
col1,
col2,
priority,
RANK() OVER w AS 'rank'
FROM MyTable
WINDOW w AS (ORDER BY priority)
) MyQuery
Where rank = 1
Note : Syntax might be incorrect, please feel to edit the query.
This post might help you for ranking if your MySql version doesn't support Rank.
I'm getting grey hair by now...
I have a table like this.
ID - Place - Person
1 - London - Anna
2 - Stockholm - Johan
3 - Gothenburg - Anna
4 - London - Nils
And I want to get the result where all the different persons are included, but I want to choose which Place to order by.
For example. I want to get a list where they are ordered by LONDON and the rest will follow, but distinct on PERSON.
Output like this:
ID - Place - Person
1 - London - Anna
4 - London - Nils
2 - Stockholm - Johan
Tried this:
SELECT ID, Person
FROM users
ORDER BY FIELD(Place,'London'), Person ASC "
But it gives me:
ID - Place - Person
1 - London - Anna
4 - London - Nils
3 - Gothenburg - Anna
2 - Stockholm - Johan
And I really dont want Anna, or any person, to be in the result more then once.
This is one way to get the specified output, but this uses MySQL specific behavior which is not guaranteed:
SELECT q.ID
, q.Place
, q.Person
FROM ( SELECT IF(p.Person<=>#prev_person,0,1) AS r
, #prev_person := p.Person AS person
, p.Place
, p.ID
FROM users p
CROSS
JOIN (SELECT #prev_person := NULL) i
ORDER BY p.Person, !(p.Place<=>'London'), p.ID
) q
WHERE q.r = 1
ORDER BY !(q.Place<=>'London'), q.Person
This query uses an inline view to return all the rows in a particular order, by Person, so that all of the 'Anna' rows are together, followed by all the 'Johan' rows, etc. The set of rows for each person is ordered by, Place='London' first, then by ID.
The "trick" is to use a MySQL user variable to compare the values from the current row with values from the previous row. In this example, we're checking if the 'Person' on the current row is the same as the 'Person' on the previous row. Based on that check, we return a 1 if this is the "first" row we're processing for a a person, otherwise we return a 0.
The outermost query processes the rows from the inline view, and excludes all but the "first" row for each Person (the 0 or 1 we returned from the inline view.)
(This isn't the only way to get the resultset. But this is one way of emulating analytic functions which are available in other RDBMS.)
For comparison, in databases other than MySQL, we could use SQL something like this:
SELECT ROW_NUMBER() OVER (PARTITION BY t.Person ORDER BY
CASE WHEN t.Place='London' THEN 0 ELSE 1 END, t.ID) AS rn
, t.ID
, t.Place
, t.Person
FROM users t
WHERE rn=1
ORDER BY CASE WHEN t.Place='London' THEN 0 ELSE 1 END, t.Person
Followup
At the beginning of the answer, I referred to MySQL behavior that was not guaranteed. I was referring to the usage of MySQL User-Defined variables within a SQL statement.
Excerpts from MySQL 5.5 Reference Manual http://dev.mysql.com/doc/refman/5.5/en/user-variables.html
"As a general rule, other than in SET statements, you should never assign a value to a user variable and read the value within the same statement."
"For other statements, such as SELECT, you might get the results you expect, but this is not guaranteed."
"the order of evaluation for expressions involving user variables is undefined."
Try this:
SELECT ID, Place, Person
FROM users
GROUP BY Person
ORDER BY FIELD(Place,'London') DESC, Person ASC;
You want to use group by instead of distinct:
SELECT ID, Person
FROM users
GROUP BY ID, Person
ORDER BY MAX(FIELD(Place, 'London')), Person ASC;
The GROUP BY does the same thing as SELECT DISTINCT. But, you are allowed to mention other fields in clauses such as HAVING and ORDER BY.
I'm looking for a mysql select that will allow me to select (LIMIT 8) records after some changing number of first few matches;
select id
from customers
where name LIKE "John%"
Limit 8
So if i have a table with 1000 of johns with various last names
I want to be able to select records 500-508
You can send the offset to the limit statement, like this:
SELECT id
FROM customers
WHERE name LIKE "John%"
LIMIT 8 OFFSET 500
Notice the OFFSET 500 on the limit. That sets the 'start point' past the first 500 entries (at entry #501).
Therefor, entries #501, #502, #503, #504, #505, #506, #507 and #508 will be selected.
This can also be written:
LIMIT 500, 8
Personally, I don't like that as much and don't understand the order.
Pedantic point: 500-508 is 9 entries, so I had to adjust.
As a solution please try executing the following sql query
select id from customers where name LIKE "John%" Limit 500,8
I am trying to make a SQL statement that selects and orders numbers in DESC but put all strings last
so mysql is:
"SELECT * FROM posts ORDER BY price DESC"
and that gives me something like
test
best offer
6000
100
10
what I want is:
6000
100
10
test
best offer
or (doesnt matter)
6000
100
10
best offer
test
SELECT * FROM posts
ORDER BY price * 1 DESC
SQLFiddle demo
I think want you actually want is:
SELECT * FROM t ORDER BY value ASC;
which returns:
10
100
6000
best offer
test
SQL Fiddle
If you generally do not care about performance, you can do something like this:
SELECT * FROM posts
ORDER BY
CASE CAST(price AS int) WHEN 0 THEN ' ' ELSE price END DESC,
price ASC;
You might need the casts to int depending on your database.
You can also fiddle with the first expression to get the results you want. The point is that you have to sort first by whether it's a number or not and then the actual content.
The problem is that this you can't use any indexes this way. If you are allowed to change the database structure, it would be better to split the data into two columns. An int column with the price and a varchar column with the price description (in which case the first column is NULL). Then you can do faster queries.
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