Ordering by different column depending on value of another column - php

Question

There are two type of questions there 1.Passage and 2.Normal questions.
usally in test i want to pick random questions which consist type_id=0 in that if type=1 question come the the next passage should be relates to that question(Comprehension question should come in sequential). By using the below query i am able to get the questions
SELECT *
FROM tbl_testquestion
ORDER BY
CASE
WHEN type_id=0 THEN RAND()
WHEN type_id=1 THEN qu_id
END ASC
all the passage questions are coming last
and i have limit of 40 questions for test and in the table i have 50 passage questions and 70 Normal questions.
How can i write a query to call passage questions in between normal
questions.
EXAMPLE
1.who is the president of America.?(type_id=0)
2.A,B,C are 3 students Aname is "Arun" B name is "Mike" C name is "Jhon"(type_id=1)
who is C from the above passage
3.A,B,C are 3 students Aname is "Arun" B name is "Mike" C name is "Jhon"(type_id=1)
who is A from the above passage
4.Who is CEO of Facebook.?(type_id=0)
Form the Above 4 question we will pick random if Question 1 comes in that rand() no problem when the question 2 comes in the rand() the next question should be sequential. it means next question should be 3 after that passage questions completed it should switch back to rand() functionality

Answer 1

I think that the design of your database should be improved, but I’m going to answer your question as it stands.
I think I have a rather simple solution, which I can express in portable SQL without CTE’s.
It works this way: let’s assign two numbers to each row, call them major (an integer, just to be safe let’s make it a multiple of ten) and minor (a float between 0 and 1). For type 0 questions, minor is always 0. Each type 1 question relating to the same passage gets the same major (we do this with a join with a grouped subselect). We then order the table by the sum of the two values.
It will be slow, because it joins using a text field. It would be better if each distinct passage_description had an integer id to be used for the join.
I assume that all type 0 questions have empty or null passage_description, while type 1 questions have them non-empty (it would make no sense otherwise.)
I assume you have a RAND() function which yields floating values between 0 and 1.
Here we go:
SELECT u.qu_id, u.type_id,
u.passage_description, u.passage_image,
u.cat_id, u.subcat_id,
u.question, u.q_instruction, u.qu_status
FROM (
SELECT grouped.major, RAND()+0.001 AS minor, t1.*
FROM tbl_testquestion t1
JOIN (SELECT 10*FLOOR(1000*RAND()) major, passage_description
FROM tbl_testquestion WHERE type_id = 1
GROUP BY passage_description) grouped
USING (passage_description)
-- LIMIT 39
UNION
SELECT 10*FLOOR(1000*RAND()) major, 0 minor, t0.*
FROM tbl_testquestion t0 WHERE type_id = 0
) u ORDER BY u.major+u.minor ASC LIMIT 40;
With the above query without modifications, there is still a small probability that you get questions of only one type. If you want to be sure that you have at least one type 0 question, you can uncomment the LIMIT 39 on the first part of the UNION. If you want at least two, then say LIMIT 38, and so on. All type 1 questions related to the same passage will be grouped together in one test; it is not guaranteed that all questions in the database related to that passage will be in the test, but in a comment above you mention that this can be “broke”.
Edited:
I added a small amount to minor, just to bypass the rare but possible case in which RAND() returns exactly zero. Since major goes by tens, the fact that minor might now be greater than one is immaterial.

Answer 2

Use the following, I haven't tested this so, if there are any errors please report back, I will correct them. $r is a random value produced by PHP for this query. You could do $r = rand(); before calling the query
SELECT * FROM (
UNION((
SELECT *, RAND()*(SELECT COUNT(*) FROM tbl_testquestions) as orderid
FROM tbl_testquestion
WHERE type_id=0
ORDER BY orderid
LIMIT 20
),(
SELECT *, MD5(CONCAT('$r', passage_description)) as orderid
FROM tbl_testquestion
WHERE type_id=1
ORDER BY orderid
LIMIT 20
))
) AS t1
ORDER BY orderid
Explanation: orderid will keep type_id=1 entries together as it would produce the same random sequence for the same passage questions.
Warning: Unless you add passage_id to the table, this question will work quite slowly.
Edit: Fixed the ordering (I hope), forgot that MYSQL generates random numbers between 0 and 1.

Answer 3

This is the solution for mysql,
sorry it is not so readable because mysql does not supports CTE like sql-server.
Maybe you can compare with sql-server CTE syntax to the bottom to better understand how it works.
select
d.*
, o.q_ix, rnd_ord -- this is only for your reference
from (
select *, floor(rand()*1000) as rnd_ord -- this is main order for questions and groups
from (
select * from (
select
(#r1 := #r1 - 1) as q_ix, -- this is row_number() (negative so we can keep group separated)
passage_description, 0 qu_id, type_id
from (
select distinct passage_description, type_id
from tbl_testquestion,
(SELECT #r1 := 0) v, -- this is the trick for row_number()
(SELECT #rnd_limit := -floor(rand()*3)) r -- this is the trick for dynamic random limit
where type_id=1
) p
order by passage_description -- order by for row_number()
) op
where q_ix < #rnd_limit
union all
select * from (
select
(#r2 := #r2 + 1) as q_ix, -- again row_number()
'' as passage_description, qu_id, type_id
from tbl_testquestion,
(SELECT #r2 := 0) v -- var for row_number
where type_id=0
order by qu_id -- order by for row_number()
) oq
) q
) o
-- look at double join for questions and groups
join tbl_testquestion d on
((d.passage_description = o.passage_description) and (d.type_id=1))
or
((d.qu_id=o.qu_id) and (d.type_id=0))
order by rnd_ord
limit 40
and this is the more readable sql-server syntax:
;with
p as (
-- select a random number of groups (0-2) and label groups (-1,-2)
select top (abs(checksum(NEWID())) % 3) -ROW_NUMBER() over (order by passage_description) p_id, passage_description
from (
select distinct passage_description
from d
where type_id=1
) x
),
q as (
-- label questions (1..n)
select ROW_NUMBER() over (order by qu_id) q_ix, qu_id
from d
where type_id=0
),
o as (
-- calculate final order
select *, ROW_NUMBER() over (order by newid()) rnd_ord
from (
select p.q_ix, passage_description, 0 qu_id from p
union all
select q.q_ix, '', qu_id from q
) x
)
select top 40
d.*
, o.rnd_ord, o.q_ix
from o
join d on
((d.passage_description = o.passage_description) and (d.type_id=1))
or
((d.qu_id = o.qu_id) and (d.type_id=0))
order by
rnd_ord
that's all