在提交表单按钮后停留在同一页面

Question

我想在提交表单按钮后保持在同一页面上。 取决于用户点击的按钮，我想在单击该按钮的相同页面上显示某些信息。

出于某种原因，在点击按钮后，它会将我重定向到我的“ask_login.php”页面。 我读过，有人推荐使用AJAX或JQuery，但我不太了解它。如果我能得到一些帮助，我将不胜感激。谢谢。

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

我觉得这下面的代码是不正确的...... 如果（$用户名& & $密码）

试试这个

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

Answer 2

<form action="#">。没有行动意味着'提交到当前网址'，英镑符号意味着'当前文件'，不应该导航。尽管我不确定表单在语义上是否允许这样做。 我会建议张贴表单数据，但阿贾克斯。它不难做到，Jquery提供了最常用的ajax函数，并且基本上是通过表单中的每个输入循环，并将值推送到uri编码的字符串中，并通过ajax调用发布。

Answer 3

我认为它是喊出

<?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来处理表格

$.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，没有输入验证！

阅读此页

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