2011-08-05 32 views
1

我有一个MySQL查询的问题。从MySQL中的子查询抓取查询

我写的SQL查询来获取从tablebusiness ID和时间戳与不同的查询

SELECT `ID`,`TimeStamp`,pow(-6-`Latitude`,2)+pow(106-`Longitude`,2)*cos(-6*0.017453292519943)*cos(`Latitude`*0.017453292519943) as DistanceSquare FROM `tablebusiness` WHERE Title LIKE '%sushi%' AND -6.8996746410157 < `Latitude` AND `Latitude` < -5.1003253589843 AND 105.10032535898 < `Longitude` AND `Longitude` < 106.89967464102 ORDER BY DistanceSquare LIMIT 2 ,20 

现在能正常工作在我的情况。

但是我不想在最终结果中提到DistanceSquare以节省带宽。所以我做

SELECT `ID`,`TimeStamp` FROM (SELECT `ID`,`TimeStamp`,pow(-6-`Latitude`,2)+pow(106-`Longitude`,2)*cos(-6*0.017453292519943)*cos(`Latitude`*0.017453292519943) as DistanceSquare FROM `tablebusiness` WHERE Title LIKE '%sushi%' AND -6.8996746410157 < `Latitude` AND `Latitude` < -5.1003253589843 AND 105.10032535898 < `Longitude` AND `Longitude` < 106.89967464102 ORDER BY DistanceSquare LIMIT 2 ,20) 

所以基本上我从内SELECT生成的表拿起ID和时间戳。它不起作用。我应该如何格式化它?

我只想从tablebusiness

获得ID和时间戳任何一个可以帮我解决这个问题呢?

不是一个大的,但我只是想学习。

这是PHP代码我用来生成这个查询

$phiper180=pi()/180; 
$formula="pow(".$_GET['lat']."-`Latitude`,2)+pow(".$_GET['long']."-`Longitude`,2)*cos(".$_GET['lat']."*".$phiper180.")*cos(`Latitude`*".$phiper180.")"; 

$query="SELECT `ID`,`TimeStamp`,$formula as DistanceSquare FROM `tablebusiness` WHERE Title LIKE '%".$_GET['keyword']."%' AND ". ($_GET['lat']-$distanceindegrees). " < `Latitude` AND `Latitude` < " . ($_GET['lat']+$distanceindegrees) . " AND " . ($_GET['long']-$distanceindegrees). " < `Longitude` AND `Longitude` < " . ($_GET['long']+$distanceindegrees)." ORDER BY DistanceSquare LIMIT ".($startFrom)." ,20"; 
$query="SELECT `ID`,`TimeStamp` FROM (".$query.")"; 

回答

2

为什么不你刚刚增加(POW(-6- Latitude,2)+ POW(106- Longitude,2)* COS (-6 * 0.017453292519943)* cos(Latitude * 0.017453292519943))按子句排序不需要使用子查询。请尝试以下操作:

SELECT `ID`,`TimeStamp` FROM `tablebusiness` WHERE Title LIKE '%sushi%' AND -6.8996746410157 < `Latitude` AND `Latitude` < -5.1003253589843 AND 105.10032535898 < `Longitude` AND `Longitude` < 106.89967464102 ORDER BY (pow(-6-`Latitude`,2)+pow(106-`Longitude`,2)*cos(-6*0.017453292519943)*cos(`Latitude`*0.017453292519943)) LIMIT 2 ,20 
+0

Ah效果不错。工作得很好。其实我一直在想用存储过程来做这件事,但似乎有几种方法可以做到这一点。 –