I am using this particular example
https://www.w3schools.com/sql/sql_orderby.asp
I want to get the list of customers ordered by country pattern.
For instance, Country starting with As should be first in number, followed by countries starting with Ge, followed by country starting with Fr.
If I need to order by using 1 pattern search in the country I can achieve using the following code :
SELECT * FROM Customers
ORDER BY (CASE WHEN Country LIKE 'Fr%' THEN 0
ELSE 1
END)ASC;
How can I add 2 more conditions to the ORDER BY clause ? Thank you.
You can add multiple CASE for your additional "conditions" by incrementing the number each time:
SELECT * FROM Customers
ORDER BY (CASE
WHEN Country LIKE 'As%' THEN 0
WHEN Country LIKE 'Ge%' THEN 1
WHEN Country LIKE 'Fr%' THEN 2
ELSE 3
END) ASC;
Related
Here is a sample Database
First_name country
Andy US
Manu India
Paul Pakistan
Ramesh Pakistan
Rich India
So, what i want is to select all records from the above table and display according to name.
Like :-
I want to select person name to be display first whose country name is India and after US, Pakistan.
How can i perform this task in single SQL query ?
Update
I don't know how many Country are there.
Country the need to be display first will be input by the user.
Use a CASE statement to give each record a sortkey. 0 for a country match, 1 for a mismatch, so the desired country comes first.
select *
from mytable
order by case when country = #country then 0 else 1 end, first_name
May be something like this
Select * From Table1
Order By CASE WHEN country = 'INDIA' THEN 0
WHEN country = 'US' THEN 1
Esle 2
END;
Or You can use FIELD
Select * From Table1 Order By FIELD(country, 'India', 'US', 'Pakistan') ;
Use FIELD Function
Try this:
SELECT fitst_name, country
FROM tableA
ORDER BY FIELD(country, 'India', 'US', 'Pakistan'), fitst_name
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 am attempting to query a table for a limited resultset in order to populate an autocomplete field in javascript. I am, therefore, using a LIKE operator with the partial string entered.
If I have, for example, a table such as:
tblPlaces
id country
1 Balanca
2 Cameroon
3 Canada
4 Cape Verde
5 Denmark
For the sake of this example, let's say I want two rows returning - and yeah, for this example, I made up a country there ;) I want to prioritize any instance where a partial string is matched at the beginning of country. The query I began using, therefore is:
SELECT id, country FROM tblPlaces WHERE country LIKE 'ca%' LIMIT 2
This returned 'Cameroon' and 'Canada' as expected. However, in instances where there are no two names in which the string is matched at the beginning of a word (such as 'de'), I want it to look elsewhere in the word. So I revised the query to become
SELECT id, country FROM tblPlaces WHERE country LIKE '%ca%' LIMIT 2
This then returned 'Cape Verde' and 'Denmark', but in doing so broke my original search for 'ca', which now returns 'Balanca' and 'Cameroon'.
So, my question is, how to go about this using a single query that will prioritize a match at the start of a word (perhaps I need to use REGEXP?) I am assuming also that if the 'country' column is indexed, these matches will at least be returned with subsequent alphabetical priority (i.e. Cameroon before Canada etc).
If you mean to prioritize matches that are Exactly at the start...
SELECT id, country
FROM tblPlaces
WHERE country LIKE '%ca%'
ORDER BY CASE WHEN country LIKE 'ca%' THEN 0 ELSE 1 END, country
LIMIT 2
EDIT
More generic and possibly faster (Assuming "closer to the start the 'better' the match")...
SELECT id, country
FROM tblPlaces
WHERE country LIKE '%ca%'
ORDER BY INSTR(country, 'ca'), country
LIMIT 2
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
Lets say I have a database of widgets. I am showing a list of the top ten groupings of each widget, separated by category.
So lets say I want to show a list of all widgets in category A, but I want to sort them based on the total number of widgets in that category and only show the top 10 groupings.
So, my list might look something like this.
Top groupings in Category A
100 Widgets made by company 1 in 1990.
90 Widgets made by company 1 in 1993.
70 Widgets made by company 3 in 1993.
etc...(for 10 groupings)
This part is easy, but now lets say I want a certain grouping to ALWAYS show up in the listings even if it doesnt actually make the top ten.
Lets say I ALWAYS want to show the number of Widgets made by company 1 in 2009, but I want this grouping to be shown somewhere in my list randomly (not first or last)
So the end list should look something like
Top groupings in Category A
100 Widgets made by company 1 in 1990.
90 Widgets made by company 1 in 1993.
30 Widgets made by company 1 in 2009.
70 Widgets made by company 3 in 1993.
How would i accomplish this in MySQL?
thanks
Edit:
Currently, my query looks like this
SELECT
year,
manufacturer,
MAX(price) AS price,
image_url,
COUNT(id) AS total
FROM
widgets
WHERE
category_id = A
AND
year <> ''
AND
manufacturer <> ''
GROUP BY
category_id,
manufacturer,
year
ORDER BY
total DESC,
price ASC
LIMIT
10
);
Thats without the mandatory grouping in there.
The placement doesnt necessarily have to be random, just shouldnt be on any extreme end. And the list should be 10 groupings including the mandatory listing. So 9 + 1
I would use an UNION query: your current query union the query for 2009, then handle the sorting in the presentation layer.
You can write 2 separate query (one for all companies and another just for company 1) and then use UNION to join them together. Finally, add ORDER BY RAND().
It will look like
SELECT * FROM
(
SELECT company_id, company_name, year, count(*) as num_widgets
....
LIMIT 10
UNION DISTINCT
SELECT company_id, company_name, year, count(*) as num_widgets
...
WHERE company_id =1
...
LIMIT 10
)x
ORDER BY RAND();
You could add a field that you make true for company 1 in 2009 and include it in the where clause. Something like
select * from companies where group = 'some group' or included = true order by included, widgets_made limit 10
For the random part you would have that as subquery then include a column that has a random number from 1 to 10 if the field that you made is true, and rownum otherwise, then sort by that column