另一个MYSQL加入问题

Question

嗨，我真的很挣扎让下面的SQL查询工作，我提前道歉，如果ive做了这个捣烂，但即时通讯仍然学习SQL的全部高级领域。

这里是我的代码...

"SELECT *,(((acos(sin((".$latitude."*pi()/180)) * 
sin((`latitude`*pi()/180))+cos((".$latitude."*pi()/180)) * 
cos((`latitude`*pi()/180)) * cos(((".$longitude."- `longitude`)*pi()/180)))) 
*180/pi())*60*1.1515) as distance 
FROM `locations` l HAVING distance <= '".$distance."' JOIN 
(SELECT * users) u 
ON (l.id = u.basic_location) 
      WHERE u.id != A $AND2 
      ORDER BY distance ASC" 

我不断收到以下错误消息......

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 
'JOIN (SELECT * users) u ON (l.id = u.basic_location) WHERE u.id' at line 1 

我用尽许多这样的组合，但香港专业教育学院成为难倒，并寻找一些帮帮我？

Answer 1

1. JOIN应该FROM tablename之后跟随或其他JOIN
2. 你应该用WHERE，不HAVING在这种情况下
3. 你应该JOIN users u，而不是子查询（在这种情况下，它是没有意义的）

Answer 2

试试这个

"SELECT *, 

     ((((SELECT 
     acos(
      sin((".$latitude."*pi()/180)) * 
      sin((`latitude`*pi()/180))+ 
      cos((".$latitude."*pi()/180)) * 
      cos((`latitude`*pi()/180)) * 
      cos(((".$longitude."- `longitude`)*pi()/180)) 
      ))*180/pi())*60*1.1515) as distance 
    FROM `locations` l)X 
    JOIN users u 
ON (X.id = u.basic_location) 
WHERE u.id != A $AND2 
AND distance <= '".$distance."' 
ORDER BY distance ASC" 

个

变化

1）HAVING distance <=需要改变，以WHERE distance <=。请注意，使用HAVING子句过滤只有当你有一个像SUM任何聚合函数，COUNT等。对于任何正常的过滤操作使用WHERE子句

2）加入语法是：

SELECT t1.* 
FROM Table1 t1 
JOIN Table2 t2 
ON t1.Id = t2.Id 
Where t1.<Some condition> 
AND t2.<some Condition> 

而且不

SELECT t1.* 
FROM Table1 t1 Where t1.<Some condition> 
JOIN Table2 t2 
ON t1.Id = t2.Id 
Where t2.<some Condition> 

如果需要加入，然后不喜欢

SELECT X.*,Y.* 
FROM (SELECT * FROM Table1 t1 Where t1.<Some condition>) X 
JOIN (SELECT * FROM Table2 t2) Y 
ON X.Id = Y.Id 
Where Y.<some Condition> 

希望你能理解之前，从第一个表进行筛选。

Answer 3

select * from 
(SELECT *, 
(((acos(sin((".$latitude."*pi()/180)) * 
sin((`latitude`*pi()/180))+ 
cos((".$latitude."*pi()/180)) * 
cos((`latitude`*pi()/180)) * 
cos(((".$longitude."- `longitude`)*pi()/180))))* 
180/pi())*60*1.1515) as distance 
FROM `locations` l INNER JOIN 
`users` u 
ON l.id = u.basic_location 
WHERE u.id != A $AND2) derived_table dt 
where dt.distance <= '".$distance."' 
ORDER BY dt.distance ASC