Question

我有一个使用jquery和ajax提交给php脚本的表单，但在php脚本中显示为空白。

形式：

<table id="coachapplication">

<form id="coachapplicationform" action="" method="post">

<tr><td>Briefly explain your teaching methods:<br /> <textarea maxlength="100" name="coachmethods" id="coachmethods"></textarea></td></tr>

<tr><td><input type="checkbox" value="yes" name="agree" id="agree" /> I am fully prepared to take on the responsibilities of being a coach which include logging into the website daily and hosting sessions.

<tr><td><button id="submitcoachapplication" type="button">Submit</button></td></tr>

</form>

</table>

jquery脚本：

$('#submitcoachapplication').click(function() {

$.ajax({

type: "POST",

url: "includes/sendcoachapplication.php",

data: $("form#coachapplicationform").serialize(),

success: function(msg){

    $("#coachapplication").html(msg);

}

});

});

php脚本：

<?php

session_start();

$user = $_SESSION['username'];

include("dbcon.php");

include("user.php");

$result = mysql_query("SELECT * FROM coachapplications ORDER BY username");

while($row = mysql_fetch_array($result)) {

$username = $row['username'];

if($username == $user) die("You already have a pending application.");

}

$coachmethods = $_POST['coachmethods'];

$coachmethods = mysql_real_escape_string($coachmethods);

$agree = $_POST['agree'];

if($coachmethods == "") die("Please enter some info about how you intend to teach.");

if($agree != "yes") die(" Please agree to the terms.");

$sql="INSERT INTO coachapplications (username, methods) VALUES ('$user', '$coachmethods')";

if (!mysql_query($sql,$con)) die('Error: ' . mysql_error());

echo 'Your application has been submitted, and is awaiting admin approval. If you are found suitable for the program, you will be taught how to use the system.';

mysql_close($con);

?>

Answer 1

不允许<form>元素作为<table>的子元素。您的浏览器可能会通过将其移到表外而纠错，这意味着<input>元素不再位于其中。

将<table>放在<form>内，而不是相反。

更好的是，don't use tables layout。

表格没有被序列化？ jquery和ajax

1 个答案: