您好我想知道这是否可行?使用jquery帖子在同一页面上有两个不同的表单发送它来进行一些检查。第一个工作完美无缺,然而当我转到第二个表单时,我得到一个错误,说它是一个未定义的变量,但我使用的是与第一个表单完全相同的方法。它将加载在PHP页面中为源表单回显的任何内容,但不会回显我正在输入的内容。是否有更好,更正确的方法来执行此操作?
这不适用于真实的网站,只是测试我正在进行的项目。
HTML:
<form action="php/signup.php" method="post" class="form-inline" name="signupForm">
<input type="text" maxlength="20" name="username" id="user_in">
<input type="password" maxtlength="20" name="password" id="pass_in">
<input type="submit" name="submit" Value="Sign Up">
</form>
<div id="feedback"></div> <!-- Feedback for Sign Up Form -->
<br /><br />
<form name="feedForm">
<input type="text" id="feed_in" name="feed_me_in" placeholder="feed">
<div id="feedme"></div> <!-- FEEDback for feed form -->
</form>
<script src="js/jquery-1.9.1.js"></script>
JavaScript的:
<script>
$(document).ready(function() {
$('#feedback').load('php/signup.php').show();
//SIGN IN FORM
$('#user_in, #pass_in').keyup(function() {
$.post('php/signup.php', { username: document.signupForm.username.value,
password: document.signupForm.password.value },
function(result) {
$('#feedback').html(result).show
});
});
$('#feedme').load('php/feed.php').show();
//FEED FORM
$('#feed_in').keyup(function() {
$.post('php/feed.php', { feed: document.feedForm.feed_me_in.value },
function(result) {
$('$feedme').html(result).show
});
});
});
</script>
PHP for Feed Form:
<?php
$feed = mysql_real_escape_string($_POST['feed']);
if(isset($feed)) {
echo $feed;
} else {}
?>
注册表单的PHP:
<?php
if(isset($_POST['username'])) {
include_once('connect.php'); //Connect
$username = mysql_real_escape_string($_POST['username']);
$sql1 = "SELECT username FROM users WHERE username='$username'";
$check = mysql_query($sql1);
$numrows = mysql_num_rows($check);
if(strlen($username)<=4) {
echo "Username is too short";
} elseif($numrows == 0) {
echo "Username is available";
} elseif($numrows > 0) {
echo "Username is already taken";
}
} else {
echo "Please type a username";
}
?>
答案 0 :(得分:1)
$('$feedme').html(result).show
应该是
$('#feedme').html(result).show();