JSON陣列到陣列PHP多維

Question

我有這樣的JSON數據：

{ 
     "response": { 
     "count": 212, 
     "list": [ 
      { 
      "code": "02007", 
      "name": "swept the room", 
      "rate": 750000, 
      "withValue": false 
      }, 
      { 
      "code": "02005", 
      "name": "mop room", 
      "rate": 600000, 
      "withValue": false 
      }, 
      { 
      "code": "02003", 
      "name": "buying food", 
      "rate": 175000, 
      "withValue": false 
      } 
     ] 
     }, 
     "metaData": { 
     "message": "OK", 
     "code": 200 
     } 
} 

，我有表模式是這樣的：

 
mysql> desc master_service; 

    +----------------+-------------+------+-----+---------+----------------+ 
    | Field   | Type  | Null | Key | Default | Extra   | 
    +----------------+-------------+------+-----+---------+----------------+ 
    | id    | int(25)  | NO | PRI | NULL | auto_increment | 
    | code   | varchar(10) | YES |  | NULL |    | 
    | name   | varchar(20) | YES |  | NULL |    | 
    | rate   | double  | YES |  | NULL |    | 
    | withvalue  | tinyint(1) | YES |  | NULL |    | 
    +----------------+-------------+------+-----+---------+----------------+ 

和我的編碼是這樣。

//使用PHP PDO

include_once 'db_connect.php'; 
$data = json_decode($response, true); 

$tempservice = array(); 
if(isset($data['response']) && isset($data['response']['list'])) 
{ 
    //use foreach on ['response']['list'] index - here are teachers data stored 
    foreach($data['response']['list'] as $service)  
     $tempservice[$kesadaran['code']] = $service['name']; 
    } 

    foreach($tempservice as $key =>$value) { 
     $sql = "insert into master_service(code,name) Values ('$key','$value')"; 
     $st = $dbh->prepare($sql); 
     $st->execute ($data); 
    } 

它只能保存在代碼和名稱形式的數據庫。我想率和withValue可以在數據庫中保存

Answer 1

foreach($data['response']['list'] as $value) { 
    $sql = "insert into master_service(code,name,rate,withvalue) Values ('{$value['code']}','{$value['name']}', '{$value['rate']}', '{$value['withValue']}')"; 
    $st = $dbh->prepare($sql); 
    $st->execute ($data); 
} 

對不起，我忘了上json_decode真實參數..編輯

Answer 2

你可以試試這個： -

include_once 'db_connect.php'; 
$data = json_decode($response, true); 

$tempservice = []; 
if(isset($data['response']) && isset($data['response']['list'])){ 
    //use foreach on ['response']['list'] index - here are teachers data stored 
    foreach($data['response']['list'] as $service){ 
     $withValue = ($service['withValue']) ? 1 : 0; 
     $tempservice[$service['code']] = ['name'=>$service['name'], 'rate'=>$service['rate'], 'withvalue'=>$withValue]; 
    } 
} 

foreach($tempservice as $key =>$value) { 
    $sql = "insert into master_service(code, name, rate, withvalue) Values ('$key', '".$value['name']."', '".$value['rate']."', '".$value['withvalue']."')"; 
    $st = $dbh->prepare($sql); 
    $st->execute(); 
} 

注：請修復任何語法錯誤。你也可以跳過創建tempservice陣列，可以執行INSERT SQL有

Answer 3

只需使用一個foreach循環，並添加缺少的列，無需使用重組$tempservice陣列：

include_once 'db_connect.php'; 
$data = json_decode($response, true); 

// Simplified if, checking for isset($data['response') is redundant 
if (isset($data['response']['list'])) 
{ 
    // Prepare the statement using placeholders 
    $sql = "INSERT INTO master_service (code,name,rate,withValue) VALUES (:code,:name,:rate,:withValue)"; 
    $stmt = $dbh->prepare($sql); 

    foreach($data['response']['list'] as $row) 
    { 
     // Execute query with parameters 
     $stmt->execute($row); 
    } 
} 

重要：我更換了使用佔位符查詢變量。這樣可以防止SQL Injection的安全。我還將prepare放在循環之外，以便它在每次迭代時不會「重複」。