MySQL的 - 1：N的關係

Question

A table : id, name 
B table : id, a_id, name 

1A：NB

我要選擇像下面

array( 
    0 => aID_1, 
    aName_1, 
    array (
     bName_1, 
     bName2 , .v.v.v 
    ), 
    ... 
    ... 
) 

我可以只在做B中與導致記錄和相關信息查詢？

Answer 1

是這樣的：

SELECT * FROM tableA as a LEFT JOIN tableB as b ON a.id = b.a_id; 

Answer 2

$qry = mysql_query ("SELECT a.id, a.name, b.name as b_name FROM A LEFT JOIN B on A.id = B.a_id"); 
$data = array(); 
while ($r = mysql_fetch_rows($qry)) 
{ 
    if (!array_key_exists($r["id"], $data) 
    { 
     $data[$r["id"]] = array (
      "id" => $r["id"], 
      "name" => $r["name"], 
      "b" => array() 
     ); 
    } 

    $data[$r["id"]]["b"][] = $r["b_name"]; 
} 

，現在你可以通過迭代的foreach：

foreach ($data as $d) 
{ 
    print_r($d); 
}