php - 获取MySql数据并放入多维数组中

Question

可以是 PHP 或 Mysql 解决方案... 我希望能够将从 mysql 数据库中选择的一些数据存储在 php 数组中。到目前为止，我只能放入“假”数组

$SQL = "SELECT *  FROM continents RIGTH JOIN Country ON Country_Continents = continents_ID";
       while ($CONT = mysql_fetch_array($DataSet)){
            $array_cont[] = $CONT["continents_name"];
            $country_ID_rry [] = $CONT;
         }

然后我得到每个大陆的数组，大陆数组得到国家名称..

Array
(
    [0] => Array
        (
            [0] => America
            [continents_name] => America
            [1] => 3
            [country_id] => 3
            [2] => México
            [country_name] => México
        )

    [1] => Array
        (
            [0] => SouthAmérica
            [continents_name] => SouthAmerica
            [1] => 2
            [country_id] => 2
            [2] => Argentina
            [country_name] => Argentina
        )

    [2] => Array
        (
            [0] => SotuhAmerica
            [continents_name] => SouthAmerica
            [1] => 5
            [country_id] => 5
            [2] => Venezuela
            [country_name] => Venezuela
        )

    [3] => Array
        (
            [0] => SouthAmerica
            [continents_name] => SouthAmerica
            [1] => 6
            [country_id] => 6
            [2] => Colombia
            [country_name] => Colombia
        )

    [4] => Array
        (
            [0] => Caribe
            [continents_name] => Caribe
            [1] => 1
            [country_id] => 1
            [2] => Cuba
            [country_name] => Cuba
        )

)

但我想要类似的东西......

Array
(
    [SouthAmerica] => Array
        (
            [0] => Argentina
            [1] => Brazil
            [2] => Colombia
        )

    [NorthAmerica] => Array
        (
            [0] => Usa
            [1] => Mexico
            [2] => Canada
        )

    [Europa] => Array
        (
            [0] => Ukraine
            [1] => Germany
            [2] => England
        )

)

Answer 1

警告此扩展在 PHP 5.5.0 中已弃用，并在 PHP 7.0.0 中被删除。相反，MySQLi或 PDO_MySQL应使用扩展名。另请参阅MySQL: choosing an API guide和 related FAQ了解更多信息。此功能的替代方案包括:

• mysqli_fetch_array()
• PDOStatement::fetch()

但是总体思路(仅获取关联数组):

while ($CONT = mysql_fetch_assoc($DataSet)){
    $Pais_ID_rry[$CONT["continents_name"]][] = $CONT['country_name'];
}

此外，如果您只想要大陆和国家名称:

SELECT continents_name, country_name FROM continents . . .