使用内部联接时的SQL设置变量

Question

I have 2 tables:

1. Coupons with columns {id, business_id, name}
2. Businesses with columns {business_id, name, latitude, longitude}

In general, it's a coupons managment site, every business can offer coupons ..

Coupons.business_id is the Business.business_id that the coupons linked to.

I want to get all the coupons that close enough (20KM) to the user location.

I have the following SQL code that give's me the businesses in this range:

NOTE: lat and lng is the user's longitude and latitude.

    String query = "SELECT *,(((acos(sin((" + lat + "*pi()/180)) * " 
      +" sin((`Latitude`*pi()/180))+cos((" + lat + "*pi()/180)) * " 
     + " cos((`Latitude`*pi()/180)) * cos(((" + lng + "- `Longitude`)* "
     + "  pi()/180))))*180/pi())*60*1.1515 " +
    ") as distance  " +
    "FROM `Businesses` HAVING distance < 20 ORDER BY distance";

I tried to use this code for getting the close coupons: (call it second query )

    String query = "SELECT * FROM COUPONS inner join Businesses on Coupons.business_id = Businesses.business_id"
            + " (((acos(sin((" + lat + "*pi()/180)) * " 
      +" sin((`Businesses.latitude`*pi()/180))+cos((" + lat + "*pi()/180)) * " 
     + " cos((`Businesses.latitude`*pi()/180)) * cos(((" + lng + "- `Businesses.longitude`)* "
     + "  pi()/180))))*180/pi())*60*1.1515 " +
    ") as distance HAVING distance < 20 ORDER BY distance";

I'm getting the following error:

Hibernate: SELECT * FROM COUPONS inner join Businesses on Coupons.business_id = Businesses.business_id (((acos(sin((33.207933*pi()/180)) *  sin((`Businesses.latitude`*pi()/180))+cos((33.207933*pi()/180)) *  cos((`Businesses.latitude`*pi()/180)) * cos(((35.570246- `Businesses.longitude`)*   pi()/180))))*180/pi())*60*1.1515 ) as distance HAVING distance < 20 ORDER BY distance
Dec 24, 2013 12:36:31 PM org.apache.catalina.core.StandardWrapperValve invoke
SEVERE: Servlet.service() for servlet [jsp] in context with path [/InternetProject] threw exception [org.hibernate.exception.SQLGrammarException: could not execute query] with root cause
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'as distance HAVING distance < 20 ORDER BY distance' at line 1

What is the problem in my second query and how can I fix it? thanks in advance.. Using mySQL.

Answer 1

You mixed up the inner join and select statements. Try this:

String query = "SELECT *, (((acos(sin((" + lat + "*pi()/180)) * " 
      +" sin((`Businesses.latitude`*pi()/180))+cos((" + lat + "*pi()/180)) * " 
     + " cos((`Businesses.latitude`*pi()/180)) * cos(((" + lng + "- `Businesses.longitude`)* "
     + "  pi()/180))))*180/pi())*60*1.1515 " +
    ") as distance FROM COUPONS inner join Businesses on Coupons.business_id = Businesses.business_id HAVING distance < 20 ORDER BY distance";

Answer 2

Your SQL Statement is broken

it needs to look like this

SELECT *, 
       [domaths] AS distance 
FROM   table 
       INNER JOIN other_table 
               ON id = id 

Answer 3

Have you try this ,

SELECT c.business_id, 
       c.agent_name, 
       bus.mybusinessid, 
       bus.latitude, 
       bus.longitude 
FROM   coupons c 
       INNER JOIN (SELECT business_id, 
                          Count(*) AS mybusinessid, 
                   FROM   businesses 
                   GROUP  BY business_id) bus 
               ON bus.business_id = c.business_id;