PHP/MySQL代碼問題

-2

編輯問題解決了，使用了錯誤報告，發現這是雙引號與單引號的問題。PHP/MySQL代碼問題

我在獲取這個PHP/mySQL代碼顯示任何問題。因爲它只是顯示一個空白的白頁。我插入一個簡單的打印命令到代碼的不同部分，只要它的外面將打印聲明：

} else { 
    print "<p>Error: Couldn't find Confirmation code</p>"; 
    } 

    if($updater){ 
    print "<p>Thank you" . $username . ", you have been confirmed and can now access our members page.</p>"; 

    }

部分。這讓我相信這是這個領域的一個問題，或者我的if/else語句。

<html> 
    <head> 
     <title>CS148 "Where Are You From?" Confirmation Page</title> 

     <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> 
     <meta name="author" content="Stephen B. Wakita" /> 

     <meta name='description' content='Confirm Page for CS148 PHP Form Assignment.'/> 

     <link rel="stylesheet" href="mystyle.css" type="text/css" media="screen" /> 
    </head> 
    <body> 
    <?php 

    $dbh = mysql_connect('webdb.uvm.edu','swakita','password'); 

    if(!$dbh) 
    { 
     die('Could not connect: ' . mysql_error()); 
    } 

    mysql_select_db('SWAKITA', $dbh); 

    $confnum = mysql_real_escape_string($_GET['confnum']); 

    $tblname = mysql_real_escape_string('tblWhere'); 

    $sql = "SELECT * FROM $tblname WHERE fldConfirm='$confnum'"; 
    $check = mysql_query($sql); 

    if($check) { 
     $count = mysql_num_rows($check); 

     if($count == 1) { 

      $rows = mysql_fetch_array($check); 
      $username = $rows['pk_Username']; 


      $sql2 = "UPDATE $tblname(fldAdminLevel)VALUES('4')"; 
      $updater = mysql_query($sql2); 
     } else { 
      print "<p>Error: Couldn't find Confirmation code</p>"; 
     } 

     if($updater) { 
      print "<p>Thank you" . $username . ", you have been confirmed and can now access our members page.</p>"; 
     } 
    } 
    ?> 
    </body> 
</html>

來源

2012-03-07 Wakeeta

在php.ini中打開錯誤報告。嘗試向else語句前面的if語句添加一些調試輸出，以便測試是否存在邏輯錯誤。 – 2012-03-07 22:25:05

請正確縮進您的代碼。 – Ryan 2012-03-07 22:25:22

你確定dB連接有效嗎？ – 2012-03-07 22:26:17

我傾向於

$sql2 = "UPDATE $tblname(fldAdminLevel)VALUES('4')"; 
$updater = mysql_query($sql2);

是問題。

請試試這個，看看如果您有任何有用的反饋：

if($check) { 
    $count = mysql_num_rows($check); 

    if($count == 1) { 

     $rows = mysql_fetch_array($check); 
     $username = $rows['pk_Username']; 


     $sql2 = "UPDATE $tblname(fldAdminLevel)VALUES('4')"; 
     $updater = mysql_query($sql2); 
     if(mysql_errno()){ 
      echo mysql_error(); 
     } 
    } else { 
     print "<p>Error: Couldn't find Confirmation code</p>"; 
    } 

    if($updater) { 
     print "<p>Thank you" . $username . ", you have been confirmed and can now access our members page.</p>"; 
    } 
} else { 
    echo "Variable $check fail."; 
}

來源

2012-03-07 22:53:34 Blake

此外，嘗試將您的$ sql2複製/粘貼到phpmyadmin中以測試查詢，或者與您的mysql服務器進行交互。逐個排除任何問題。根據你在php中的錯誤處理，它可能會殺死那個函數並阻止任何響應。 – Blake 2012-03-07 22:57:22

PHP/MySQL代碼問題

回答

相關問題