在提交表單按鈕後停留在同一頁面

Question

我想在提交表單按鈕後保持在同一頁面上。 取決於用戶點擊的按鈕，我想在單擊該按鈕的相同頁面上顯示某些信息。

出於某種原因，在點擊按鈕後，它會將我重定向到我的「ask_login.php」頁面。 我讀過，有人推薦使用AJAX或JQuery，但我不太瞭解它。如果我能得到一些幫助，我將不勝感激。謝謝。

loggedin.php

<?php 
    $username = $_POST['username']; 
    $password = $_POST['password']; 

    if($username && $password) { 
     //info is provided 
     $queryget = mysql_query("SELECT * FROM users WHERE id='$username' AND password='$password'"); 
      $numrows = mysql_num_rows($queryget); 

      if($numrows != 0) { 

       //something has been found 
       $_SESSION['id'] = $username; 



      } else { 
       echo "Wrong username/password"; 
       echo "<script>alert('Username/Password are wrong');</script>"; 
       echo "<script language='javascript'>window.location = 'ask_login.php';</script>"; 
      } 
    } 
    else { 
     echo "Wrong username/password"; 
     echo "<script>alert('Username/Password are wrong');</script>"; 
     echo "<script language='javascript'>window.location = 'ask_login.php';</script>"; 
    } 

    ?> 


       <form method="POST" action=""> 
       <input type="submit" name="details" value="Details"> 
       </form> 


       <form method="POST" action=""> 
      <input type="submit" name="optionB" value="option B"> 
       </form> 

      <form method="POST" action=""> 
      <input type="submit" name="optionC" value="option C"> 
       </form> 

Answer 1

我覺得這下面的代碼是不正確的...... 如果（$用戶名& & $密碼）

試試這個

if(!empty($username) && !empty($password)) 

Answer 2

<form action="#">。沒有行動意味着'提交到當前網址'，英鎊符號意味着'當前文件'，不應該導航。儘管我不確定表單在語義上是否允許這樣做。 我會建議張貼表單數據，但阿賈克斯。它不難做到，Jquery提供了最常用的ajax函數，並且基本上是通過表單中的每個輸入循環，並將值推送到uri編碼的字符串中，並通過ajax調用發佈。

Answer 3

我認爲它是喊出

<?php 
if($_POST) 
{ 
$username = $_POST['username']; 
$password = $_POST['password']; 

if($username && $password) { 
    //info is provided 
    $queryget = mysql_query("SELECT * FROM users WHERE id='$username' AND password='$password'"); 
     $numrows = mysql_num_rows($queryget); 

     if($numrows != 0) { 

      //something has been found 
      $_SESSION['id'] = $username; 



     } else { 
      echo "Wrong username/password"; 
      echo "<script>alert('Username/Password are wrong');</script>"; 
      echo "<script language='javascript'>window.location = 'ask_login.php';</script>"; 
     } 
} 
else { 
    echo "Wrong username/password"; 
    echo "<script>alert('Username/Password are wrong');</script>"; 
    echo "<script language='javascript'>window.location = 'ask_login.php';</script>"; 
} 
} 
?> 


      <form method="POST" action=""> 
      <input type="submit" name="details" value="Details"> 
      </form> 


      <form method="POST" action=""> 
     <input type="submit" name="optionB" value="option B"> 
      </form> 

     <form method="POST" action=""> 
     <input type="submit" name="optionC" value="option C"> 
      </form> 

Answer 4

你可以用jquery來處理表格

$.post("test.php", $("#testform").serialize(),function(e){alert(e);}); 

test.php的

<?php 
    $username = $_POST['username']; 
    $password = $_POST['password']; 

    if($username && $password) { 
     //info is provided 
     $queryget = mysql_query("SELECT * FROM users WHERE id='$username' AND password='$password'"); 
     $numrows = mysql_num_rows($queryget); 

     if($numrows != 0) { 

      //something has been found 
      $_SESSION['id'] = $username; 



     } else { 
      echo "Wrong username/password"; 
      echo "<script>alert('Username/Password are wrong');</script>"; 
      echo "<script language='javascript'>window.location = 'ask_login.php';</script>"; 
     } 
} 
else { 
    echo "Wrong username/password"; 
    echo "<script>alert('Username/Password are wrong');</script>"; 
    echo "<script language='javascript'>window.location = 'ask_login.php';</script>"; 
} 

?> 

http://api.jquery.com/jquery.post/

提防，沒有輸入驗證！

閱讀此頁

http://www.php.net/manual/en/function.mysql-real-escape-string.php