我正在尝试提交表单而不刷新页面,我希望在点击按钮id = fav时出现提醒消息。这是代码,但我不知道我做错了什么。它应该是在按钮点击或表单提交?无刷新提交表单(ajax,php)
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#f1").submit(function(){
// AJAX Code To Submit Form.
$.ajax({
type: "POST",
url: "bookpage.php",
data: {'fav':'fav'} ,
cache: false,
success: function(response){
if(response.message){
alert(response.message);
}
}
});
}
return false;
});
});
</script>
<form action="#read" method="post" id="f1">
<div class="r1">
<button class="down" name="download" title="Add to favorites">Download</button>
<li><a class="full" href="full.php?id=<?php echo $id; ?>">Full page</a>
</li>
<button class="d-later" name="dlater">Download later</button>
<button class="fav-button" type="submit" id="fav"></button>
</div>
</form>
PHP
if(isset($_POST['fav']) && $_POST['fav'] == 'fav' && fav_exists($u , $ii)== true){
$query = "DELETE FROM favorites WHERE bid='$ii' AND email='$u'";
$result = mysql_query($query);
if(! $result) {
die('Could not delete data: ' . mysql_error());
} $response['message'] = 'My message';
echo json_encode($response);
}else if(isset($_POST['fav']) && $_POST['fav'] == 'fav' && fav_exists($u , $ii)== false){
$query = "INSERT INTO favorites (email,book, bid) VALUES('$u','$bname','$ii')";
$result = mysql_query($query);
if(! $result) {
die('Could not enter data: ' . mysql_error());
}
$response['message'] = 'My message';
echo json_encode($response);
}
function fav_exists($u , $ii){
$query = "SELECT id FROM favorites WHERE email='$u' AND bid='$ii'";
$result = mysql_query($query);
$count = mysql_num_rows($result);
if($count >= 1) {
return true;
} else {
return false;
}
}
添加'按钮类型='submit'' – C2486
数据建议立即进行删除d是数据:{'fav':fav}, – JYoThI
没有,没有工作@ Rishi –