添加幾個條件，找到在mysql中附近的地方 - 谷歌地圖

Question

我試圖用一個解決方案，我對谷歌的文檔here

的問題是我想在搜索

添加一些更多的過濾器找到致富附近的地方

原液

$sql2 = "SELECT *, (3959 * acos(cos(radians(" . $row1['tubeLan'] . ")) * cos(radians(mapLan)) * cos(radians(mapLon) - radians(" . $row1['tubeLon'] . ")) + sin(radians(" . $row1['tubeLan'] . ")) * sin(radians(mapLan)))) AS distance FROM " . $table . " HAVING distance < 2500 ORDER BY distance LIMIT 0 , 20; 

我希望它在上面的語句

SIZE between $firstrange and $secondrange" 

添加此過濾器是什麼0

我試過，但沒有提供所要求的結果

$sql2 = "SELECT *, (3959 * acos(cos(radians(" . $row1['tubeLan'] . ")) * cos(radians(mapLan)) * cos(radians(mapLon) - radians(" . $row1['tubeLon'] . ")) + sin(radians(" . $row1['tubeLan'] . ")) * sin(radians(mapLan)))) AS distance FROM " . $table . " HAVING distance < 2500 ORDER BY distance LIMIT 0 , 20 and SIZE between $firstrange and $secondrange"; 

我在做什麼錯在這裏？我不能通過將AND放在它們之間來添加這些條件嗎？

Answer 1

相信你可以使用添加條件和，但你必須將它們添加到您的WHERE或HAVING子句中，而不是在聲明的末尾：

SELECT *, 
(3959 * acos(cos(radians(" . $row1['tubeLan'] . ")) * cos(radians(mapLan )) * cos(radians(mapLon) - radians(" . $row1['tubeLon'] . ")) + sin(radians(" . $row1['tubeLan'] . ")) * sin(radians(mapLan)))) AS distance 
FROM " . $table . " 
WHERE SIZE between $firstrange and $secondrange 
HAVING distance < 2500 ORDER BY distance LIMIT 0 , 20 ; 

（我想SIZE是在你的表中的列）