>我有一个房地产项目,其中表格如下:
-
apartment
表具有建筑物 ID -
building
表具有邻域 ID -
neighborhood
表具有城市 ID
现在,如果我想获得某个城市的所有公寓,最好的方法是什么?我正在使用PHP
和MySQL
这是我尝试和失败的
public static function getAllApartmentsInACity ($city_id) {
$results = array();
$db = DB::getInstance(); //My DB class nothing wrong here
$n = self::getAllNeighborhoodsInACity($city_id); // returns an array of objects of all neighborhoods
$b = array();
$a = array();
foreach ($n as $singleN) {
$b[] = self::getAllBuildingsInANeighborhood($singleN->id); //returns an array of objects of all buildings
}
foreach ($b as $singleB) {
echo $singleB->id . '-';
}
return $a;
}
毕竟,当我var_dump
这个函数的回报时,我得到了Null
,前提是我在表中有数据。任何帮助将不胜感激
使用以下
查询:
$query = "SELECT apartment.*
FROM apartment
JOIN building ON
building.id = apartment.building_id
JOIN neighborhood ON
neighborhood.id = building.neighborhood_id
JOIN city ON
city.id = neighborhood.city_id AND
city.id = 'SOME CITY ID'
";