Php 在关联数组中联接，而不是在分离的记录中联接

表

存储

id  name  date
1   foo   2011-06-15 15:10:34
2   bar   2011-07-02 16:45:18

表

位置

storeid  zipcode  latitude  longitude
1        90001    xxxxx     xxxxx
1        45802    xxxxx     xxxxx
2        32843    xxxxx     xxxxx

如何生成包含名为

locations

的键的关联数组，该键是存储区所有位置的数组

我当前的SQL（将记录中的每个位置分隔开）：

我想要的示例：

array(
 [0] => array(
           "id" => 1,
           "name" => "foo",
           "locations" => array(
                           [0] => array(
                                    "zipcode" => 90001,
                                    "latitude" => -45.48513,
                                    "longitude" => 82.12432
                                   )
                           [1] => array(
                                    "zipcode" => 42802,
                                    "latitude" => -31.48513,
                                    "longitude" => 77.12432
                                   )
                          ) 
         )
)

其他商店也是如此

---

谢谢

所以您不能在一个查询中提取数据，因为SQL通常每行工作，并且没有像PHP数组那样的数据结构。无法使用JOIN嵌套记录。这就是为什么必须在PHP循环中使用单独的查询。像这样：

$query = "SELECT s.id,s.name FROM stores AS s";

$result = mysql_query($query);
$data = array();
while($row = mysql_fetch_assoc( $result )) {
    $data[] = $row['id'];
    $data[] = $row['name'];

    $query2 = "SELECT l.zipcode, l.latitude, l.longitude FROM locations AS l WHERE storeid=".$row['id'];

    $result2 = mysql_query($query2);
    while($row2 = mysql_fetch_assoc( $result )) {
        $data['locations']['zipcode'] = $row2['zipcode'];
        $data['locations']['latitude'] = $row2['latitude'];
        $data['locations']['longitude'] = $row2['longitude'];
    }
}

否则，您可以使用JOIN获取所有结果，并执行以下操作：

$query = "SELECT * FROM stores AS s 
LEFT JOIN locations AS l
ON s.id = l.storesid";

$result = mysql_query($query);
$data = array();
while($row = mysql_fetch_assoc( $result )) {
    $data[$row[id]]['id'] = $row['id'];
    $data[$row[id]]['name'] = $row['name'];
    $data[$row[id]]['locations'][] = array($row['zipcode'], $row['latitude'], $row['longitude']);
}

---

但这将使主数组的索引设置为从0开始不连续，但每个索引将等于“存储”项的ID，该项将在所有位置的单独记录中重复存储，与我当前的代码完全相同。我花了一些时间编辑了答案。看一看，将在其中添加更多详细信息。谢谢，伙计，你认为哪一个最快？用第二个。您只有一个SQL查询，其余的由PHP计算，占用的资源更少。干杯

$query = "SELECT * FROM stores AS s 
LEFT JOIN locations AS l
ON s.id = l.storesid";

$result = mysql_query($query);
$data = array();
while($row = mysql_fetch_assoc( $result )) {
    $data[$row[id]]['id'] = $row['id'];
    $data[$row[id]]['name'] = $row['name'];
    $data[$row[id]]['locations'][] = array($row['zipcode'], $row['latitude'], $row['longitude']);
}