將兩個數組合併成單個內部json_encode

我必須將兩個數組合併成單個內部Json編碼。我的代碼，將兩個數組合併成單個內部json_encode

$email = $_GET["email"]; 
    $password = $_GET["password"]; 

    $query  = "SELECT * FROM tbl_user_login WHERE email='$email' AND password='$password' AND verification='1'"; 
    $result  = mysqli_query($c, $query) or die(mysqli_error($c)); 
    $length  = mysqli_num_rows($result); 

    if($length == 1) 
    { 
     $var[] = array('status'=>"success"); 

     while($obj = mysqli_fetch_object($result)) 
     { 
      $var[] = $obj; 
     } 
     echo '{"login":'.json_encode($var).'}'; 
    } 
    else 
    { 
     $arr = array('status'=>"notfound"); 
     echo '{"login":['.json_encode($arr).']}'; 
    }

現在的結果是，

{"login":[{"status":"success"},{"login_id":"1","name":"Jithin Varghese","password":"some","phone":"","email":"[email protected]","addr":"","city":"","state":"","pincode":"0","type":"STD","verification":"1"}]}

而且需要輸出，

{"login":[{"status":"success","login_id":"1","name":"Jithin Varghese","password":"some","phone":"","email":"[email protected]","addr":"","city":"","state":"","pincode":"0","type":"STD","verification":"1"}]}

如何組合排列。我嘗試了很多。

來源

2015-12-16 Jithin Varghese

您不能在沒有驗證的情況下在您的SQL查詢中使用用戶輸入，那麼您可以操作SQL注入攻擊。請看[準備好的語句]（http://php.net/manual/de/mysqli.quickstart.prepared-statements.php） – KhorneHoly

變化

$var[] = array('status'=>"success"); 
while($obj = mysqli_fetch_object($result)) 
{ 
    $var[] = $obj; 
}

到

$var['status'] = "success"; 
// use the assoc fetch here.. to avoid casting to array 
while($arr = mysqli_fetch_assoc($result)) 
{ 
    $var = array_merge($var, $arr); 
}

來源

2015-12-16 11:43:20

現在結果是'{「login」：{「status」：「success 「，」0「：{」login_id「：」1「，」name「：」Jithin Varghese「，」password「：」some「，」phone「：」「，」email「：」[email protected]「，「addr」：「」，「city」：「」，「state」：「」，「pincode」：「0」，「type」：「STD」，「verification」：「1」}}}' –

好。謝謝。現在它正在工作。 –

這是一個瘋狂的猜測，但嘗試。

來源

2015-12-16 11:47:09 memo

您可以使用array_merge得到你要求精確的輸出：

$email = $_GET["email"]; 
$password = $_GET["password"]; 

$query  = "SELECT * FROM tbl_user_login WHERE email='$email' AND password='$password' AND verification='1'"; 
$result  = mysqli_query($c, $query) or die(mysqli_error($c)); 
$length  = mysqli_num_rows($result); 

$response = []; 
if($length == 1) 
{ 
    $response['login'] = arry(array_merge(array('status'=>"success"), mysqli_fetch_assoc($result))); 
} 
else 
{ 
    $response['login'] = array(array('status'=>"notfound")); 
} 

header('Content-Type: application/json'); 
echo json_encode($response);

注意，似乎沒有必要有login屬性是一個數組時，只有一個結果，所以它纔有意義刪除外部陣列包裝：

if($length == 1) 
{ 
    $response['login'] = array_merge(array('status'=>"success"), mysqli_fetch_assoc($result)); 
} 
else 
{ 
    $response['login'] = array('status'=>"notfound"); 
}

來源

2015-12-16 11:47:22 Steve

將兩個數組合併成單個內部json_encode

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