Question

我提交表单按钮后试图保持同一页面。根据用户点击的按钮，我想在单击按钮的同一页面上显示某些信息。

出于某种原因，它一直将我重新引导回我的问题＆＃34; ask_login.php＆＃34;单击按钮后的页面。我已经阅读过，有些人建议使用AJAX或JQuery，但我并不是很了解它。如果能得到一些帮助，我将不胜感激。感谢。

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

<form action="#">。没有行动意味着提交到当前网址，英镑符号表示当前文件＆＃39;而且不应该导航。虽然我不确定表格在语义上是否允许这样做。我建议在ajax上发布表单数据。它不难做到，Jquery提供了最常用的ajax函数，你基本上循环遍历表单中的每个输入，并将值推送到uri编码的字符串并通过ajax调用发布。

Answer 2

我认为以下代码不正确... if（$ username＆amp;＆amp; $ password）

试试这个

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

Answer 3

我认为应该是

 <?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

您可以使用jquery处理表单

$.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/

要注意，没有输入验证！

阅读本页

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

提交表单按钮后保持在同一页面

4 个答案: