提交后成功/失败消息弹出框?

mun*_*nue 16 javascript php

基本上单击提交按钮后,我想弹出一个弹出框说成功或失败,然后单击确定以确认消息.目前我得到一个弹出框"未定义",然后是失败的消息弹出框. 请帮忙!

这是脚本

<?php
include ('config.php');

if (isset($_POST['name'])) {

$name = "name";

$query = "INSERT INTO pop ('id','name') VALUES ('','$name')";
    $result = mysql_query($query,$cn);
    if ($result) {
    echo "<script type='text/javascript'>alert('submitted successfully!')</script>";
}
else
{
    echo "<script type='text/javascript'>alert('failed!')</script>";
}
}       
?>

<html>
<head>
</head>
<body>

    <form action="" method="post">
    Name:<input type="text" id="name" name="name"/>
    <input type="submit" value="submit" name="submit" onclick="alert();"/>
    </form>
</body>
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Dan*_*nyB 20

您正在回显HTML的body标签之外.把你的回声放在那里,你应该没事.

另外,onclick="alert()"从提交中删除.这是您第一条undefined消息的原因.

<?php
  $posted = false;
  if( $_POST ) {
    $posted = true;

    // Database stuff here...
    // $result = mysql_query( ... )
    $result = $_POST['name'] == "danny"; // Dummy result
  }
?>

<html>
  <head></head>
  <body>

  <?php
    if( $posted ) {
      if( $result ) 
        echo "<script type='text/javascript'>alert('submitted successfully!')</script>";
      else
        echo "<script type='text/javascript'>alert('failed!')</script>";
    }
  ?>
    <form action="" method="post">
      Name:<input type="text" id="name" name="name"/>
      <input type="submit" value="submit" name="submit"/>
    </form>
  </body>
</html>
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