Array對象的PHP

Question

我是新來的PHP，並試圖產生以下JSON輸出：

{ 
    "contacts": [ 
    { 
     "id": "1", 
     "name": "John Bob", 
     "email": "john@bob.com" 
    }, 
    { 
     "id": "2", 
     "name": "Johnny Mac", 
     "email": "johnny@mac.com" 
    }, 
     ..... 
    ] 
} 

，我有下面的PHP代碼：

$final = array(); 
foreach ($contacts as $contact) 
{ 
    ..... 
    $final[] = array(
     'id'   => (string)$id, 
     'name'   => $name, 
     'email'  => $email 
    ); 
} 
header('Content-Type: application/json; charset=UTF-8'); 
echo json_encode($final, JSON_PRETTY_PRINT); 

然而

，此輸出：

[ 
    { 
    "id": "1", 
    "name": "John Bob", 
    "email": "john@bob.com" 
    }, 
    { 
    "id": "2", 
    "name": "Johnny Mac", 
    "email": "johnny@mac.com" 
    }, 
] 

我讀了this tutorial，改變$final[]到$final['contacts']但ST生病無法產生所需的json文件。

Answer 1

或者：

$final['contacts'][] = array(..); 

或：

$contacts_output = array(); 
foreach ($contacts as $contact) { 
    $contacts_output[] = array(..); 
} 
$final = array('contacts' => $contacts_output); 
echo json_encode($final, JSON_PRETTY_PRINT); 

Answer 2

foreach ($contacts as $contact) 
{ 
    ..... 
    $final['contacts'][] = array(
      array(//insert all your array in contacts 
       'id'  => (string)$id, 
       'name'  => $name, 
       'email' => $email 
    ); 
} 

Answer 3

試試這個：

$final = array(); 
$final['contacts'] = array(); 
foreach ($contacts as $contact) 
{ 
    ......... 
    $final['contacts'][] = array( 
    'id'  => (string)$id, 
    'name'  => $name, 
    'email' => $email 
); 
} 

Answer 4

$final = array('contacts'=>[]); 

foreach ($contacts as $contact) 
{ 
    $final['contacts'] []= 
      array(//insert all your array in contacts 
       'id'  => (string)$id, 
       'name'  => $name, 
       'email' => $email 
      ); 
}