成員激活函數給出可捕獲的致命錯誤

Question

我已經將我的激活腳本從mysql轉換爲mysqli，但是我的激活腳本在電子郵件中單擊鏈接後沒有更新數據庫。這是我的舊代碼

function activate($email, $email_code) { 
global $myConnection; 
$email = mysqli_real_escape_string($email); 
$email_code = mysqli_real_escape_string($email_code); 

if (mysqli_result(mysqli_query($myConnection, "SELECT COUNT(`mem_id`) FROM `members` WHERE `mem_email` = '$email' AND `email_code` = '$email_code' AND `mem_emailactivated` = 0"), 0) == 1) { 
    mysqli_query($myConnection, "UPDATE `members` SET mem_emailactivated = 1 WHERE `mem_email` = '$email'"); 
    return true; 
} else { 
    return false; 
} 
} 

這是我的新代碼

function activate($email, $email_code) { 
$myNewConnection = myConnection(); 
$email = mysqli_real_escape_string($myNewConnection, $email); 
$email_code = mysqli_real_escape_string($myNewConnection, $email_code); 
$query = mysqli_query($myNewConnection, "SELECT COUNT(`mem_id`) FROM `members` WHERE `mem_email` = '$email' AND `email_code` = '$email_code' AND `mem_emailactivated` = '0'"); 
$result = mysqli_num_rows($myNewConnection, $query); 
echo $query; 
exit(); 
if ($result == 1) { 
    $sql = "UPDATE `members` SET mem_emailactivated = 1 WHERE `mem_email` = '$email'"; 
    $query = mysqli_query($myConnection, $sql) or die(mysql_error($myConnection)); 
    return true; 
} else { 
    return false; 
} 
} 

如果我附和了查詢，你可以在我的代碼中看到我的錯誤

Catchable fatal error: Object of class mysqli_result could not be converted to string in /homepages/24/d156351352/htdocs/test/storescripts/users.php on line 211 

然而我很困惑，因爲我沒有在函數或查詢中使用mysqli_result，所以我不確定錯誤在哪裏

Answer 1

我越來越困惑與mysqli的，在哪裏建立連接。當mysqli_num_rows中有$ myNewConnection時，它不應該在那裏。它應該看起來像這樣

function activate($email, $email_code) { 
$myNewConnection = myConnection(); 
$email = mysqli_real_escape_string($myNewConnection, $email); 
$email_code = mysqli_real_escape_string($myNewConnection, $email_code); 
$query = mysqli_query($myNewConnection, "SELECT COUNT(`mem_id`) FROM `members` WHERE `mem_email` = '$email' AND `email_code` = '$email_code' AND `mem_emailactivated` = '0'"); 
$result = mysqli_num_rows($query); 
print_r ($result); 
exit(); 
if ($result == 1) { 
$sql = "UPDATE `members` SET mem_emailactivated = 1 WHERE `mem_email` = '$email'"; 
$query = mysqli_query($myConnection, $sql) or die(mysql_error($myConnection)); 
return true; 
} else { 
return false; 
} 
} 

Answer 2

因爲SELECT查詢給result在array使用print_r查看結果

print_r($query);