在mysql中的連接

-2

我正在嘗試做一個註冊表單。如果你可以給我一些提示，如果我的代碼錯誤或不正確在mysql中的連接

它仍然顯示我這個準備（$ sqlinsert）; $ user = $ _POST ['user']; $ pass = $ _POST ['pass']; $ email = $ _POST ['email']; $ lname = $ _POST ['lname']; $ fname = $ _POST ['fname']; $ stmt-> bind_param（$ user，$ pass，$ email，$ lname，$ fname）; $ stmt->執行（）; $ rowcount = $ stmt-> affected_rows;如果（$ rowcount> 0）{echo「Success！」; } else {echo「Error！」; } //關閉執行$ stmt-> close（）; ？>

這裏是我的代碼：

<?php 
    $mysqli = new mysqli('local','root','','dbtform'); 
    if (mysqli_connect_errno()) { 
    printf("Connect failed: %s\n", mysqli_connect_error()); 
    exit(); 
    $sqlinsert = "INSERT INTO tblform (user, pass, email, lname, fname) 
    VALUES ('sssss',?,?,?,?,?)"; 
    $stmt = $mysqli->prepare($sqlinsert); 

    $user = $_POST['user']; 
    $pass = $_POST['pass']; 
    $email = $_POST['email']; 
    $lname = $_POST['lname']; 
    $fname = $_POST['fname']; 



    $stmt->bind_param($user, $pass, $email, $lname, $fname); 
    $stmt->execute(); 

    $rowcount = $stmt->affected_rows; 


    if ($rowcount > 0) 
{ 
echo "Success!"; 
} 

else 
{ 
echo "Error!"; 
} 

//CLOSE EXECUTE 
$stmt->close(); 

?> 


    <head> 
     <title>Animated Form Switching with jQuery</title> 
     <meta http-equiv="Content-Type" content="text/html; charset=UTF-8"/> 
     <meta name="description" content="Expand, contract, animate forms with jQuery wihtout leaving the page" /> 
     <meta name="keywords" content="expand, form, css3, jquery, animate, width, height, adapt, unobtrusive javascript"/> 
     <link rel="shortcut icon" href="../favicon.ico" type="image/x-icon"/> 
     <link rel="stylesheet" type="text/css" href="css/style.css" /> 
     <script src="js/cufon-yui.js" type="text/javascript"></script> 
     <script src="js/ChunkFive_400.font.js" type="text/javascript"></script> 
     <script type="text/javascript"> 
      Cufon.replace('h1',{ textShadow: '1px 1px #fff'}); 
      Cufon.replace('h2',{ textShadow: '1px 1px #fff'}); 
      Cufon.replace('h3',{ textShadow: '1px 1px #000'}); 
      Cufon.replace('.back'); 
     </script> 
    </head> 
    <body> 
     <div class="wrapper"> 
      <h1>Rythreion</h1> 
      <h2>Demo: click the <span>orange links</span> to see the form animating and switching</h2> 
      <div class="content"> 
       <div id="form_wrapper" class="form_wrapper"> 
        <form class="register"> 
         <h3>Register</h3> 
         <div class="column"> 
          <div> 
           <label>First Name:</label> 
           <input type="text" name="fname" /> 
           <span class="error">This is an error</span> 
          </div> 
          <div> 
           <label>Last Name:</label> 
           <input type="text" name="lname"/> 
           <span class="error">This is an error</span> 
          </div> 

         </div> 
         <div class="column"> 
          <div> 
           <label>Username:</label> 
           <input type="text" name="user"/> 
           <span class="error">This is an error</span> 
          </div> 
          <div> 
           <label>Email:</label> 
           <input type="text" name="email"/> 
           <span class="error">This is an error</span> 
          </div> 
          <div> 
           <label>Password:</label> 
           <input type="password" name="pass" /> 
           <span class="error">This is an error</span> 
          </div> 
         </div> 
         <div class="bottom"> 
          <div class="remember"> 
           <input type="checkbox" /> 
           <span>Send me updates</span> 
          </div> 
          <input type="submit" value="Register" /> 
          <a href="index.html" rel="login" class="linkform">You have an account already? Log in here</a> 
          <div class="clear"></div> 
         </div> 
        </form> 
        <form class="login active"> 
         <h3>Login</h3> 
         <div> 
          <label>Username:</label> 
          <input type="text" name="user" /> 
          <span class="error">This is an error</span> 
         </div> 
         <div> 
          <label>Password: <a href="forgot_password.html" rel="forgot_password" class="forgot linkform">Forgot your password?</a></label> 
          <input type="password" name="pass" /> 
          <span class="error">This is an error</span> 
         </div> 
         <div class="bottom"> 
          <div class="remember"><input type="checkbox" /><span>Keep me logged in</span></div> 
          <input type="submit" value="Login"></input> 
          <a href="register.html" rel="register" class="linkform">You don't have an account yet? Register here</a> 
          <div class="clear"></div> 
         </div> 
        </form> 
        <form class="forgot_password"> 
         <h3>Forgot Password</h3> 
         <div> 
          <label>Email:</label> 
          <input type="text" name="email" /> 
          <span class="error">This is an error</span> 
         </div> 
         <div class="bottom"> 
          <input type="submit" value="Send reminder"></input> 
          <a href="index.html" rel="login" class="linkform">Suddenly remebered? Log in here</a> 
          <a href="register.html" rel="register" class="linkform">You don't have an account? Register here</a> 
          <div class="clear"></div> 
         </div> 
        </form> 
       </div> 
       <div class="clear"></div> 
      </div> 
      <a class="back" href="http://google.com">back to the Codrops tutorial</a> 
     </div> 


     <!-- The JavaScript --> 
     <script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script> 
     <script type="text/javascript"> 
      $(function() { 
        //the form wrapper (includes all forms) 
       var $form_wrapper = $('#form_wrapper'), 
        //the current form is the one with class active 
        $currentForm = $form_wrapper.children('form.active'), 
        //the change form links 
        $linkform  = $form_wrapper.find('.linkform'); 

       //get width and height of each form and store them for later       
       $form_wrapper.children('form').each(function(i){ 
        var $theForm = $(this); 
        //solve the inline display none problem when using fadeIn fadeOut 
        if(!$theForm.hasClass('active')) 
         $theForm.hide(); 
        $theForm.data({ 
         width : $theForm.width(), 
         height : $theForm.height() 
        }); 
       }); 

       //set width and height of wrapper (same of current form) 
       setWrapperWidth(); 

       /* 
       clicking a link (change form event) in the form 
       makes the current form hide. 
       The wrapper animates its width and height to the 
       width and height of the new current form. 
       After the animation, the new form is shown 
       */ 
       $linkform.bind('click',function(e){ 
        var $link = $(this); 
        var target = $link.attr('rel'); 
        $currentForm.fadeOut(400,function(){ 
         //remove class active from current form 
         $currentForm.removeClass('active'); 
         //new current form 
         $currentForm= $form_wrapper.children('form.'+target); 
         //animate the wrapper 
         $form_wrapper.stop() 
            .animate({ 
             width : $currentForm.data('width') + 'px', 
             height : $currentForm.data('height') + 'px' 
            },500,function(){ 
             //new form gets class active 
             $currentForm.addClass('active'); 
             //show the new form 
             $currentForm.fadeIn(400); 
            }); 
        }); 
        e.preventDefault(); 
       }); 

       function setWrapperWidth(){ 
        $form_wrapper.css({ 
         width : $currentForm.data('width') + 'px', 
         height : $currentForm.data('height') + 'px' 
        }); 
       } 

       /* 
       for the demo we disabled the submit buttons 
       if you submit the form, you need to check the 
       which form was submited, and give the class active 
       to the form you want to show 
       */ 
       $form_wrapper.find('input[type="submit"]') 
          .click(function(e){ 
           e.preventDefault(); 
          });  
      }); 
     </script> 
    </body> 
</html>

來源

2013-02-26 user2111371

什麼錯誤信息你好嗎？ – christopher 2013-02-26 13:33:06

你把你的報價搞亂了'$ sqlinsert'。 – favoretti 2013-02-26 13:33:59

@ChrisCooney它不斷顯示準備（$ sqlinsert）; $ user = $ _POST ['user']; $ pass = $ _POST ['pass']; $ email = $ _POST ['email']; $ lname = $ _POST ['lname']; $ fname = $ _POST ['fname']; $ stmt-> bind_param（$ user，$ pass，$ email，$ lname，$ fname）; $ stmt->執行（）; $ rowcount = $ stmt-> affected_rows;如果（$ rowcount> 0）{echo「Success！」; } else {echo「Error！」; } //關閉執行$ stmt-> close（）; ？> – user2111371 2013-02-26 13:34:39

-1

如果你能給我一些提示，如果我的代碼是錯誤的或不正確

肯定。
你的代碼是錯誤的和不正確的，因爲它使用mysqli。
您必須改用PDO。

只是一個草圖，給你一個想法：
第一，創建一個連接

$dsn = 'mysql:dbname=test;host=localhost;charset=utf8'; 
$opt = array(
    PDO::ATTR_ERRMODE   => PDO::ERRMODE_EXCEPTION, 
    PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC 
); 
$pdo = new PDO($dsn, $user, $pass, $opt);

然後，運行查詢

$sql = "INSERT INTO tblform (user, pass, email, lname, fname) VALUES (?,?,?,?,?)"; 
$stmt = $pdo->prepare($sql); 
$stmt->execute($_POST); // you may wish to unset non-used fields from POST first 
if ($stmt->rowCount()) 
{ 
    echo "Success!"; 
}

來源

2013-02-26 14:31:09

感謝您的幫助 – user2111371 2013-02-26 14:53:13

在mysql中的連接

回答

相關問題