mysqlの結果をPHPでピボットテーブルに変換

Question

products(prod_id, prod_name,...) 
    attribute_key(attr_key_id, attr_name) 
    attributes_values(attr_val_id, attr_key_id, value) 
    product_attr_values(id,product_id,attr_val_id) 

MYSQLの結果を配列定義として定義しました。問題は私が に同じ商品の異なる値を同じ属性に連結して、 のカンマ（;）で接続することです。しかし、私はそれを連結せずに結果を得ることができます。ヘルプは高く評価しました。

$shop = Array 
(
"0" => Array 
    (
     "id" => 9, 
     "prodname" => 'Ramkinkar Baij', 
     "attribute_name" => 'Author', 
     "value" => 'sabsyasachi das' 
    ), 

"1" => Array 
    (
     "id" => 10, 
     "prodname" => 'Ramkinkar Baij', 
     "attribute_name" => 'Author', 
     "value" => 'Ela Datta' 
    ), 

"2" => Array 
    (
     "id" => 11, 
     "prodname" => 'Ramkinkar Baij', 
     "attribute_name" => 'Editor', 
     "value" => 'Sathi Basu' 
    ), 

"3" => Array 
    (
     "id" => 12, 
     "prodname" => 'Ramkinkar Baij', 
     "attribute_name" => 'Editor', 
     "value" => 'Radha Prasad Gupta' 
    ), 

"4" => Array 
    (
     "id" => 13, 
     "prodname" => 'Ramkinkar Baij', 
     "attribute_name" => 'Place_of_Publication', 
     "value" => 'Panagarh, Burdwan, West Bengal' 
    ), 

"5" => Array 
    (
     "id" => 14, 
     "prodname" => 'Rendava Prapancha Telugu Mahasabhalu, 1981', 
     "attribute_name" => 'Author', 
     "value" => 'sabsyasachi das' 
    ), 

"6" => Array 
    (
     "id" => 15, 
     "prodname" => 'Rendava Prapancha Telugu Mahasabhalu, 1981', 
     "attribute_name" => 'Author', 
     "value" => 'Ela Datta' 
    ), 

"7" => Array 
    (
     "id" => 16, 
     "prodname" => 'Rendava Prapancha Telugu Mahasabhalu, 1981', 
     "attribute_name" => 'Translated_by', 
     "value" => 'partha dasgupta' 
    ), 

"8" => Array 
    (
     "id" => 17, 
     "prodname" => 'Rendava Prapancha Telugu Mahasabhalu, 1981', 
     "attribute_name" => 'Editor', 
     "value" => 'Bandana Mukhopadhay' 
    ), 

"9" => Array 
    (
     "id" => 18, 
     "prodname" => 'Rendava Prapancha Telugu Mahasabhalu, 1981', 
     "attribute_name" => 'Editor', 
     "value" => 'Sathi Basu' 
    ), 

"10" => Array 
    (
     "id" => 19, 
     "prodname" => 'Rendava Prapancha Telugu Mahasabhalu, 1981', 
     "attribute_name" => 'Translated_Title', 
     "value" => '(Second World Telugu Conference, 1981)' 
    ), 

"11" => Array 
    (
     "id" => 20, 
     "prodname" => 'Rendava Prapancha Telugu Mahasabhalu, 1981', 
     "attribute_name" => 'Place_of_Publication', 
     "value" => 'Hyderabad' 
    ), 

"12" => Array 
    (
     "id" => 21, 
     "prodname" => 'Rendava Prapancha Telugu Mahasabhalu, 1981', 
     "attribute_name" => 'Foreword', 
     "value" => 'Information on Government organisations of Art and Culture in Andhra Pradesh' 
    ), 

"13" => Array 
    (
     "id" => 22, 
     "prodname" => 'Rendava Prapancha Telugu Mahasabhalu, 1981', 
     "attribute_name" => 'Descriptive_Tags', 
     "value" => 'Conference, Language' 
    ) 

); 


$groups = array(); 

foreach($shop as $key => $array){ 
//$type is not necessary, it's just for clarity below 
$product = $array['prodname']; 



if(!isset($groups[$product])){ 
    $groups[$product] = array(); 
    $groups[$product]['prod'] = $array['prodname']; 
} 


$groups[$product][$array["attribute_name"]] = $array['value']; 

} 
//then combine the groups into a master array 
$out = array(); 
foreach($groups as $g){ 
$out[] = $g; 
} 
echo '<pre>'. print_r($out, true).'</pre>'; 
print "<pre>"; 
//print_r($groups); 
print "</pre>"; 

Answer 1

使用MySQLs GROUP BYとGROUP_CONCAT機能。

SELECT prod_name, GROUP_CONCAT(CONCAT(attr_name, ': ', value) SEPARATOR ', ') AS attributes FROM [your tables and joins] GROUP BY prod_id