在一個while循環執行查詢.. MySQli

Question

我試圖更新我的當前數據庫從純文本密碼哈希密碼使用crypt() ..我試圖做到這一點，用戶不必更改他們的密碼（這是一個不穩定的方法）我的代碼是像這樣：

$Query = $Database->prepare("SELECT ID,Username,Password FROM userlist"); 
$Query->execute(); 
$Query->bind_result($ID,$Username,$Password); 
    while ($Query->fetch()){ 
     $Hashed = $FrameWork->Hash_Password($Password); 
     $Secondary_Query = $Database->prepare("UPDATE userlist SET Password=?, Salt=? WHERE ID=?"); 
     $Secondary_Query->bind_param('ssi', $Hashed['Password'],$Hashed['Salt'],$ID); 
     $Secondary_Query->execute(); 
     $Secondary_Query->close(); 
    } 
$Query->close(); 

，我發現了錯誤：

Fatal error: Call to a member function bind_param() on a non-object in C:\inetpub\www\AdminChangeTextPass.php on line 24

現在。我知道我的專欄名稱與我的數據庫名稱完全一致。我也知道我的變量是正確設置的。

調試

調試：

$Query = $Database->prepare("SELECT ID,Username,Password FROM userlist"); 
$Query->execute(); 
$Query->bind_result($ID,$Username,$Password); 
    while ($Query->fetch()){ 
     echo $Password."<br>"; 
    } 
$Query->close(); 
// Returns: 
//test 
//test 

Then: 
$Query = $Database->prepare("SELECT ID,Username,Password FROM userlist"); 
$Query->execute(); 
$Query->bind_result($ID,$Username,$Password); 
    while ($Query->fetch()){ 
     print_r($FrameWork->Hash_Password($Password)); 
    } 
$Query->close(); 

/* 
Returns: 
Array ([Salt] => ÛûÂÒs8Q-h¸Ý>c"ÿò [Password] => Ûûj1QnM/Ui/16) 

Array ([Salt] => ÛûÂÒs8Q-h¸Ý>c"ÿò [Password] => Ûûj1QnM/Ui/16) 

*/ 

數據庫模式

CREATE TABLE IF NOT EXISTS `userlist` (
    `ID` int(255) NOT NULL AUTO_INCREMENT, 
    `Username` varchar(255) NOT NULL, 
    `Password` varchar(255) NOT NULL, 
    `Salt` text NOT NULL, 
    PRIMARY KEY (`ID`) 
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=9 ; 

-- 
-- Dumping data for table `userlist` 
-- 

INSERT INTO `userlist` (`ID`, `Username`, `Password`, `Salt`) VALUES 
(1, 'test', 'test', ''), 
INSERT INTO `userlist` (`ID`, `Username`, `Password`, `Salt`) VALUES 
(2, 'test', 'test', ''); 

---

有我的代碼如下所示：

個

$Secondary_Query = $Database->prepare("UPDATE userlist SET Password=? WHERE ID=?"); 
$Query = $Database->prepare("SELECT ID,Username,Password FROM userlist LIMIT 1"); 
    var_dump($Secondary_Query); 
#$Query->execute(); 
#$Query->bind_result($ID,$Username,$Password); 
# while ($Query->fetch()){ 
#  $Hashed = $FrameWork->Hash_Password($Password); 
#  $Secondary_Query = $Database->prepare("UPDATE userlist SET Password=? WHERE ID=?"); 
#    $Secondary_Query->bind_param('ssi', $Hashed['Password'],$Hashed['Salt'],$ID); 
#  $Secondary_Query->execute(); 
     # $Secondary_Query->close(); 
# } 
#$Query->close(); 

的var_dump($Secondary_Query);回報：

object(mysqli_stmt)#3 (10) { ["affected_rows"]=> int(-1) ["insert_id"]=> int(0) ["num_rows"]=> int(0) ["param_count"]=> int(2) 

["field_count"]=> int(0) ["errno"]=> int(0) ["error"]=> string(0) "" ["error_list"]=> array(0) { } ["sqlstate"]=> string(5) "00000" ["id"]=> int(1) }

而且var_dump($Query);回報：

object(mysqli_stmt)#4 (10) { ["affected_rows"]=> int(-1) ["insert_id"]=> int(0) ["num_rows"]=> int(0) ["param_count"]=> int(0) ["field_count"]=> int(3) ["errno"]=> int(0) ["error"]=> string(0) "" ["error_list"]=> array(0) { } ["sqlstate"]=> string(5) "00000" ["id"]=> int(2) }

---

正如我不能提交答案的還..我的工作代碼如下：

$Query = $Database->prepare("SELECT ID,Username,Password FROM userlist"); 
$Query->execute(); 
$Query->bind_result($ID,$Username,$Password); 
$Query->store_result(); 
    while ($Query->fetch()){ 
     $Hashed = $FrameWork->Hash_Password($Password); 
     $Secondary_Query = $Database->prepare("UPDATE userlist SET Password=?, Salt=? WHERE ID=?"); 
     $Secondary_Query->bind_param('ssi', $Hashed['Password'],$Hashed['Salt'],$ID); 
     $Secondary_Query->execute(); 
     $Secondary_Query->close(); 
    } 
$Query->close(); 

Answer 1

編輯：啓用更詳細的錯誤報告是幫助調試問題的關鍵：mysqli_report(MYSQLI_REPORT_ALL)。

以下答案來自於另一個SO問題here。

你不能有兩個同時查詢，因爲默認情況下mysqli使用非緩衝查詢（對於預先準備的語句;它與vanilla mysql_query相反）。您可以將第一個數據讀取到數組中並循環訪問，或者告訴mysqli緩衝查詢（使用$Query->store_result()）。

有關詳細信息，請參見here。