我正在尝试实现一个利用jquery的post功能来动态更新数据库的表单。我意识到的是，在用户点击“更新”按钮后，成功函数被回复就好了“更新成功”消息。
我对stackoverflow世界的问题是为什么在后续点击（没有刷新页面）我没有得到同样的成功消息。而且，具有讽刺意味的是我的数据库正在更新，所以我知道AJAX调用正在进行中。
我在下面发布了我的代码：
JS
var TEAM = {
update: function() {
var form_data = $('form').serialize();
$.ajax({
type: "POST",
url: "../manager/edit_team.php",
data: form_data,
error: function() {
$('#status').text('Update failed. Try again.').slideDown('slow');
},
success: function() {
$('#status').text('Update successful!');
},
complete: function() {
setTimeout(function() {
$('#status').slideUp('slow');
}, 3000);
},
cache: false
});
}
}
// jQuery Code for when page is loaded
$(document).ready(function()
{
$("#update").on("click", function() {
TEAM.update();
});
});
PHP（我也欢迎任何其他评论）
require '../includes/config.php';
include '../includes/header.html';
// autoloading of classes
function __autoload($class) {
require_once('../classes/' . $class . '.php');
}
// Site access level -> Manager
$lvl = 'M';
// Assign user object from session variable
if (isset($_SESSION['userObj']))
{
$manager = $_SESSION['userObj'];
}
else
{
session_unset();
session_destroy();
$url = BASE_URL . 'index.php';
ob_end_clean();
header("Location: $url");
exit();
}
// Establish database connection
require_once MYSQL2;
// Assign Database Resource to object
$manager->setDB($db);
// Authorized Login Check
if (!$manager->valid($lvl))
{
session_unset();
session_destroy();
$url = BASE_URL . 'index.php';
ob_end_clean();
header("Location: $url");
exit();
}
// Check for a valid game sch ID, through GET or POST:
if ( (isset($_GET['z'])) && (is_numeric($_GET['z'])) )
{
// Point A in Code Flow
// Assign variable from myteams-m.php using GET method
$id = $_GET['z'];
}
elseif ( (isset($_POST['z'])) && (is_numeric($_POST['z'])) )
{
// Point C in Code Flow
// Assign variable from edit_team.php FORM submission (hidden id field)
$id = $_POST['z'];
}
else
{
// No valid ID, kill the script.
echo '<p class="error">This page has been accessed in error.</p>';
include '../includes/footer.html';
exit();
}
$team = new ManagerTeam();
$team->setDB($db);
$team->setTeamID($id);
$team->pullTeamData();
$flag = 0;
echo $flag . "<br />";
// Confirmation that form has been submitted:
if ($_SERVER['REQUEST_METHOD'] == 'POST')
{ // Point D in Code Flow
// Assume invalid values:
$tname = FALSE;
// Validate team name
if ($_POST['tname'])
{
$tname = $_POST['tname'];
}
else
{
echo '<p class="error"> Please enter a team name.</p>';
}
// Validate about team information
if ($_POST['abouttm'])
{
$abtm = trim($_POST['abouttm']);
}
else
{
$abtm = '';
}
// Check if user entered information is valid before continuing to edit game
if ($tname)
{
if($team->editTeam($tname, $abtm) == True)
{
echo '<p>Team was successfully updated</p>';
$flag = 1;
}
else
{
echo '<p>No changes were made</p>';
$flag = 2;
}
}
else
{ // Errors in the user entered information
echo '<p class="error">Please try again.</p>';
}
} // End of submit conditional.
echo $flag . "<br />";
// Point B in Code Flow
// Always show the form...
// Get team name attribute
$team->pullTeamData();
$teamname = $team->getTeamAttribute('tmname');
$about = $team->getTeamAttribute('about');
if ($teamname != '') // Valid user ID, show the form.
{
// Headliner
echo '<h2>Edit Team</h2>';
// Create the form:
echo '
<div id="EditTeam"></div>
<div id="Team">
<fieldset id="TeamDetails">
<legend>Edit Team</legend>
<form method="post" id="information">
<p id="status"></p>
<input type="hidden" name="z" value="' . $id . '" />
<p>
<label for="tname">New Team Name:</label><br/>
<input type="text" name="tname" id="tname" size="10" maxlength="45" value="' . $teamname . '" />
</p>
<p>
<label for="abouttm">Team Information:</label><br/>
<textarea id="abouttm" name="abouttm" cols="30" rows="2">"' . $about . '"</textarea><br />
<small>Enter something cool about your team.</small>
</p>
<p>
<input type="hidden" name="id" id="id">
<input type="button" value="update" id="update" />
</p>
</form>
</fieldset>
</div>';
}
else
{ //Not a valid user ID, kill the script
echo '<p class="error">This page has been accessed in error.</p>';
include '../includes/footer.html';
exit();
}
// Close the connection:
$db->close();
unset($db);
include '../includes/footer.html';
?>
您会注意到我还定义了一个$ flag来帮助调试，但具有讽刺意味的是，无论“更新”按钮的点击次数如何，它都会输出0。所以没有迹象表明数据库正在更新，但是当我检查表时肯定是。
我感谢任何帮助或指示。谢谢，
答案 0 :(得分：2)
#status消息未显示，因为您已将其隐藏 slideUp()，要再次显示，您需要slideDown()他们。
success: function() {
$('#status').text('Update successful!');
-ADD-> $('#status').slideDown('slow');
},
complete: function() {
setTimeout(function() {
$('#status').slideUp('slow');
}, 3000);
以与错误处理程序相同的方式执行操作：
success: function(){
$('#status').text('Update successful!').slideDown('slow');
...
似乎你已经知道了，只是忘记了......
可能有用的其他方法是stop()，以确保在新动画启动时停止上一个动画。，在使用 long 超时时尤其重要/动画。
（有用= 可以防止其他可见性问题，并确保消息不会开始跳入和跳出）
（long = 大约0.5-1.5秒或更长时间，如果在这段时间内可能发生其他事情那么它很长...... ）
例如，这将clear fx queue，finish running animation immediately和slideUp()：
$('#status').stop(true, true).slideUp('slow');
如果您使用相同的代码至少两次，或者如果它是可以重复使用的一般方法，则使其可重用：
function redirect_to( $page ) {
session_unset();
session_destroy();
$url = BASE_URL . $page;
ob_end_clean();
header("Location: $url");
exit();
}
if ($condition == true) {
redirect_to( 'index.php' );`
}