Sql query combining - mysql

i have two sql queries those are following
1) SELECT a.* FROM modzzz_listing_main as a LEFT JOIN modzzz_listing_rating as b ON a.id=b.gal_id WHERE LTRIM(a.city) = 'Houston' AND a.state = 'TX' AND a.tags LIKE '%Barber Shop%' ORDER BY b.gal_rating_sum DESC LIMIT 0 ,10
2) SELECT zip_code ,( 3959 * acos( cos( radians('41.97734070') ) * cos( radians( latitude ) ) * cos( radians( longitude ) - radians('-70.97234344') ) + sin( radians('41.97734070') ) * sin( radians( latitude ) ) ) ) AS distance FROM city_finder WHERE latitude IS NOT NULL AND longitude IS NOT NULL HAVING distance < 20 ORDER BY distance ASC
how can i combine this two queries by the condition `
modzzz_listing_main.zip=city_finder.zip_code
` .i am totally confused..please any one help me..

to see the join easier:
select * from
(
SELECT a.* FROM modzzz_listing_main as a LEFT JOIN modzzz_listing_rating as b ON a.id=b.gal_id WHERE LTRIM(a.city) = 'Houston' AND a.state = 'TX' AND a.tags LIKE '%Barber Shop%' ORDER BY b.gal_rating_sum DESC LIMIT 0 ,10
) queryA
left join
(
SELECT zip_code ,( 3959 * acos( cos( radians('41.97734070') ) * cos( radians( latitude ) ) * cos( radians( longitude ) - radians('-70.97234344') ) + sin( radians('41.97734070') ) * sin( radians( latitude ) ) ) ) AS distance FROM city_finder WHERE latitude IS NOT NULL AND longitude IS NOT NULL HAVING distance < 20 ORDER BY distance ASC
) queryB
on queryA.zip=queryB.zip_code
proper formatting
SELECT *
FROM
( SELECT a.*
FROM modzzz_listing_main AS a
LEFT JOIN modzzz_listing_rating AS b ON a.id=b.gal_id
WHERE LTRIM(a.city) = 'Houston'
AND a.state = 'TX'
AND a.tags LIKE '%Barber Shop%'
ORDER BY b.gal_rating_sum DESC LIMIT 0 ,
10 ) queryA
LEFT JOIN
( SELECT zip_code ,
(3959 * acos(cos(radians('41.97734070')) * cos(radians(latitude)) * cos(radians(longitude) - radians('-70.97234344')) + sin(radians('41.97734070')) * sin(radians(latitude)))) AS distance
FROM city_finder
WHERE latitude IS NOT NULL
AND longitude IS NOT NULL HAVING distance < 20
ORDER BY distance ASC ) queryB ON queryA.zip=queryB.zip_code

Related

Mysql query with multiple having conditions creating error

I am trying multiple having but problem is with multiple having conditions any one guide me how to adjust two having in same query
Here is SQL Query :
SELECT schools.*,
schools.required_gpa AS gpa,
(college_act_scores.min_act + college_act_scores.max_act)/2 AS act_avrg ,
(college_sat_scores.min_sat + college_sat_scores.max_sat)/2 AS sat_avrg,
(6371 * Acos( Cos( Radians(31.4699398) ) * Cos( Radians( latitude ) ) * Cos( Radians( longtitude ) - Radians(74.3096108) ) + Sin( Radians(31.4699398) ) * Sin( Radians( latitude ) ) ) ) AS distance
FROM schools
LEFT JOIN college_paying1
ON college_paying1.college_id = schools.id
LEFT JOIN college_act_scores
ON (
college_act_scores.college_id = schools.id
AND college_act_scores.college_child_sub_cat_id = 144)
LEFT JOIN college_sat_scores
ON (
college_sat_scores.college_id = schools.id
AND college_sat_scores.college_child_sub_cat_id = 136)
WHERE college_paying1.on_campus >= 0
AND college_paying1.on_campus <=80348
AND college_paying1.college_child_sub_cat_id =120
HAVING (
act_avrg BETWEEN 0 AND 36)
having (
sat_avrg BETWEEN 0 AND 1600)
GROUP BY schools.id
ORDER BY distance ASC limit 0, 10
You need to move the HAVING clause after GROUP BY. Document here
This should work:
select schools.*,schools.required_gpa as gpa,
(college_act_scores.min_act + college_act_scores.max_act)/2 as act_avrg ,
(college_sat_scores.min_sat + college_sat_scores.max_sat)/2 as sat_avrg,
(6371 * acos( cos( radians(31.4699398) ) * cos( radians( latitude ) ) * cos( radians( longtitude ) - radians(74.3096108) ) + sin( radians(31.4699398) ) * sin( radians( latitude ) ) ) ) AS distance
from schools
left join college_paying1 on college_paying1.college_id = schools.id
left join college_act_scores on (college_act_scores.college_id = schools.id
AND college_act_scores.college_child_sub_cat_id = 144)
left join college_sat_scores on (college_sat_scores.college_id = schools.id
AND college_sat_scores.college_child_sub_cat_id = 136)
where college_paying1.on_campus >= 0
and college_paying1.on_campus <=80348
and college_paying1.college_child_sub_Cat_id =120
GROUP BY schools.id
HAVING ( act_avrg BETWEEN 0 AND 36) AND ( sat_avrg BETWEEN 0 AND 1600)
order by distance asc
LIMIT 0 , 10
Hope this helps! Cheers!
Instead of
HAVING ( act_avrg BETWEEN 0 AND 36) HAVING ( sat_avrg BETWEEN 0 AND 1600)
Put
HAVING (act_avrg BETWEEN 0 AND 36) AND (sat_avrg BETWEEN 0 AND 1600)

Query to get nearest distance record

I have following mysql query:
SELECT a.id
, ( 3959 * acos( cos( radians(a.latitude) ) * cos( radians( w.latitude ) )
* cos( radians( w.longitude ) - radians(-a.longitude) )
+ sin( radians(a.latitude) ) * sin(radians(w.latitude)) ) ) distance
FROM global_restaurants a
JOIN webgeocities w
ON w.name = a.locality
AND w.country_code = a.country
AND a.latitude LIKE w.latitude
JOIN states s
ON s.state_code = w.state_code
AND w.country_code = s.country_code
WHERE a.city_id = 0
After execution it returns:
id distance
70 6665.478678743614
70 6496.46971480875
70 6725.900646648246
70 6733.5156930808
90 6969.449661399672
90 7252.889875588891
I want it to only return 2 rows with minimum distance like:
id distance
70 6496.46971480875
90 6969.449661399672
add order by at the end of your query:
order by distance ASC LIMIT 2;
now you calculating the distance but
at end of the query you must sort them and limit the returns row count..
.
NOTE:
i think you have a small mistake on your query
the 3959 is for miles but i think the values are in meter
your full query can be like this:
select a.id, ( 6371000 * acos( cos( radians(a.latitude) ) * cos( radians( w.latitude ) )
* cos( radians( w.longitude ) - radians(-a.longitude) ) + sin( radians(a.latitude) ) * sin(radians(w.latitude)) ) ) AS distance
from `global_restaurants` as a INNER JOIN webgeocities as w ON (w.name = a.locality AND w.country_code = a.country and a.latitude like w.latitude) INNER JOIN
states AS s ON (s.state_code = w.state_code and w.country_code = s.country_code) where a.city_id = '0' ORDER BY distance ASC LIMIT 2;
For simplicity, let's assumed that you have your query in view - stack, then your query should look like this
select stack.* from stack join (
SELECT id, GROUP_CONCAT(distance ORDER BY
distance asc) grouped_distance FROM stack GROUP BY id
) group_distance on group_distance.id=stack.id
where find_in_set(stack.distance,group_distance.grouped_distance)between 1 and 2;

Merging 2 sql queries into a single one

I have two sql queries which when run independent produces the correct results
Query 1
SELECT id,
(6371 * acos( cos( radians(9.977364864079215) ) * cos( radians( latitude ) ) * cos( radians( longitude ) - radians(76.58620953448485) ) + sin( radians(9.977364864079215) ) * sin( radians( latitude ) ) ) )
AS distance
FROM geodata HAVING distance < 20
ORDER BY distance
LIMIT 0 , 20;
Query 2
SELECT DISTINCT e.* FROM schools e
WHERE (
(e.type = 'preprimary')
)
AND(
e.title LIKE '%government%'
)
LIMIT 0, 10
I want to merge the first query with the second one, so that it should return all "preprimary" type schools with title like "government" located within 20KM radius and the result needs to be ordered by the distance.
How can I merge the two queries? I tried using JOINING the geodata table on the school table. But I dont know the remaining. Sorry, if this is a silly question. I am pretty new to SQL world.
SELECT DISTINCT school.* FROM
( SELECT geodata.id,
(6371 * acos( cos( radians(9.977364864079215) ) * cos( radians( latitude ) ) * cos( radians( longitude ) - radians(76.58620953448485) ) + sin( radians(9.977364864079215) ) * sin( radians( latitude ) ) ) )
AS distance ,school.*
FROM geodata LEFT JOIN school on geodata.id=school.id
WHERE
(school.type = 'preprimary')
AND(
school.title LIKE '%government%'
)
AND school.id IS NOT NULL
HAVING distance < 20 )x
ORDER BY x.distance
LIMIT 0 , 10;
Try this:
SELECT *
From (
SELECT DISTINCT e.* ,
(6371 * acos( cos( radians(9.977364864079215) ) * cos( radians( latitude ) ) * cos( radians( longitude ) - radians(76.58620953448485) ) + sin( radians(9.977364864079215) ) * sin( radians( latitude ) ) )
) as distance
FROM schools e
LEFT JOIN geodata g ON e.id=g.id
WHERE (e.type = 'preprimary')
AND ( e.title LIKE '%government%' )
) as s
Where s.distance < 20
Order by s.distance

Search by alias without showing the alias

I have a table of categories and a table of items.
Each item has latitude and longitude to allow me to search by distance.
What I want to do is show each category and how many items are in that category, within a distance chosen.
E.g. Show all TVs in Electronics category within 1 mile of my own latitude and longitude.
Here's what I'm trying but I cannot have two columns within an alias, obviously, and am wondering if there is a better way to do this?
Here is a SQL fiddle
Here's the query:
SELECT *, ( SELECT count(*),( 3959 * acos( cos( radians(52.993252) )
* cos( radians( latitude ) )
* cos( radians( longitude ) - radians(-0.412470) )
+ sin( radians(52.993252) )
* sin( radians( latitude ) ) ) ) AS distance
FROM items
WHERE category = category_id group by item_id
HAVING distance < 1 ) AS howmanyCat,
( SELECT name FROM categories WHERE category_id = c.parent ) AS parname
FROM categories c ORDER BY category_id, parent
First, start with the distance calculation for each item, then join in the category information and aggregate and filter
select c.*, count(i.item_id) as numitems
from category c left outer join
(SELECT i.*, ( 3959 * acos( cos( radians(52.993252) ) * cos( radians( latitude ) )
* cos( radians( longitude ) - radians(-0.412470) ) + sin( radians(52.993252) )
* sin( radians( latitude ) ) )
) AS distance
FROM items i
) i
on c.category_id = i.category_id and distance < 1
group by category_id;
Is this what you're looking for:
SELECT categories.name, count(items.item_id) as cnt
FROM items
JOIN categories
ON categories.category_id=items.category
WHERE ( 3959 * acos( cos( radians(52.993252) )
* cos( radians( latitude ) )
* cos( radians( longitude ) - radians(-0.412470) )
+ sin( radians(52.993252) )
* sin( radians( latitude ) ) ) ) < 1
GROUP BY categories.category_id;
this gives:
Tvs | 1
You can put the expression for computing the distance inside a nested SELECT, and then join the results to the categories table, like this:
SELECT COUNT(*), cc.name FROM (
SELECT
i.item_id
, c.category_id
, ( 3959 * acos( cos( radians(52.993252) )
* cos( radians( latitude ) )
* cos( radians( longitude ) - radians(-0.412470) )
+ sin( radians(52.993252) )
* sin( radians( latitude ) ) ) ) AS distance
FROM items i
JOIN categories c ON c.category_id = i.category
) raw
JOIN categories cc ON raw.category_id = cc.category_id AND raw.distance < 1
GROUP BY cc.name
The nested query pairs up items and categories, and adds the calculated distance column. The outer query then filters the rows by distance, and groups them by category to produce the desired output:
COUNT(*) NAME
-------- ----
1 TVs
Demo on sqlfiddle.

SELECT statements have a different number of columns

I have tried using the code below and it says
The used SELECT statements have a different number of columns
because distance column in added is more in 2nd query .How do I resolve something like this or can I get the order right without distance column?I really appreciate any help.THanks in Advance.
SELECT * FROM
(
SELECT * FROM user u
INNER JOIN employee e ON (u.empid = e.empid)
INNER JOIN awards a ON (u.empid = a.empid)
WHERE u.empid = 123
UNION
SELECT * ( 3959 * acos( cos( radians(37) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(-122) )+ sin( radians(37) ) * sin( radians( lat ) ) ) ) AS distance
FROM user u
INNER JOIN employee e ON (u.empid = e.empid)
INNER JOIN awards a ON (u.empid = a.empid)
HAVING distance < 25
ORDER BY distance
) a
ORDER BY timestamp DESC
It's because your 1st SELECT and 2nd SELECT have varying number of columns, since you introduced Distance in 2nd SELECT.
So try making Distance in 1st query null or 0 as follows.
SELECT * FROM
(
SELECT *, 0 as Distance
FROM user u
INNER JOIN employee e ON (u.empid = e.empid)
INNER JOIN awards a ON (u.empid = a.empid)
WHERE u.empid = 123
UNION
SELECT *, ( 3959 * acos( cos( radians(37) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(-122) )+ sin( radians(37) ) * sin( radians( lat ) ) ) ) AS distance
FROM user u
INNER JOIN employee e ON (u.empid = e.empid)
INNER JOIN awards a ON (u.empid = a.empid)
HAVING distance < 25
ORDER BY distance
) a
ORDER BY timestamp DESC
You would fix this by adding distance to the first query:
SELECT * FROM
(
SELECT *, NULL as distance FROM user u
INNER JOIN employee e ON (u.empid = e.empid)
INNER JOIN awards a ON (u.empid = a.empid)
WHERE u.empid = 123
UNION
SELECT *, ( 3959 * acos( cos( radians(37) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(-122) )+ sin( radians(37) ) * sin( radians( lat ) ) ) ) AS distance
FROM user u
INNER JOIN employee e ON (u.empid = e.empid)
INNER JOIN awards a ON (u.empid = a.empid)
HAVING distance < 25
ORDER BY distance
) a
ORDER BY timestamp DESC;