将MySql数据放入多维数组中

Question

可以是PHP或Mysql解决方案... 我希望能够将已从mysql数据库中选择的一些数据存储在php数组中。到目前为止，我只能够放于一个“假”阵列

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

然后我得到的每一个大陆阵列和我大陆阵列获得国家命名..

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

警告这个扩展被废弃在PHP 5.5.0，它是PHP 7.0.0中删除。应该使用MySQLi或PDO_MySQL扩展名。有关更多信息，另请参阅MySQL: choosing an API guide和related FAQ。替代该功能包括：

• mysqli_fetch_array（）
• PDOStatement对象::获取（）

但一般的概念（取只是associtive阵列）：

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

在此外，如果您只想要大洲和国家名称：

SELECT continents_name, country_name FROM continents . . .