我正在尝试在页面上制作一个“赞”按钮,但似乎无法使其正常工作。基本上有三个函数使用ajax将数据发送到更新数据库的php页面。我已经检查了数据库并且所有三个更新都正确。如果用户最初不喜欢并单击,它会正确显示“不喜欢”按钮,但如果您单击“不喜欢”,它不会切换回来(尽管它会更新数据库)。
这是设置的正确方法吗?我对 ajax 还很陌生,不确定这是否是正确的方法。提前致谢 史蒂夫
public function likesScript($p){?>
<script>
//display list of people who like this
function getLikes(){
$.ajax({
type: "POST",
url: "likelist.php",
data: { p: "<?php echo $_GET['p']?>"}
}).success(function(res) {
//check to see if current user likes this
if($('li#<?PHP echo $_SESSION['userId']; ?>').length){
$(".Like").addClass('hidden');
$(".UnLike").removeClass('hidden');
}
else{
$(".UnLike").addClass('hidden');
$(".Like").removeClass('hidden');
}
$("#likedBy").append(res);
console.log(res);
});
}
function removeLike() {
$.ajax({
type: "POST",
url: "likedata.php",
data: { arg1: "<?php echo $_SESSION['userId']?>", arg2: "<?php echo $p;?>", arg3: "0" }
})
getLikes();
return false;
}
function addLike() {
$.ajax({
type: "POST",
url: "likedata.php",
data: { arg1: "<?php echo $_SESSION['userId']?>", arg2: "<?php echo $p;?>", arg3: "1" }
})
getLikes();
return false;
}
$(document).ready(function() { getLikes();
$(".UnLike").live('click',removeLike);
$(".Like").live('click',addLike);
});
</script>
likelist.php:
<?php
require $_SERVER['DOCUMENT_ROOT'].'/view.class.php';
$view = new view();
include $_SERVER['DOCUMENT_ROOT'].'/profile.class.php';
include $_SERVER['DOCUMENT_ROOT'].'/init.php';
$profile = new profile($dbh);
if(isset($_POST)){
$p = $_POST['p'];
$view->printLikes($profile->getLikes($p));
}
likedata.php:
<?php
include $_SERVER['DOCUMENT_ROOT'].'/profile.class.php';
include $_SERVER['DOCUMENT_ROOT'].'/init.php';
$profile = new profile($dbh);
if(isset($_POST)){
$liker = $_POST['arg1'];
$likee = $_POST['arg2'];
$likeYesNo = $_POST['arg3'];
$profile->insertLikes($liker, $likee, $likeYesNo);
}
?>
最佳答案
AJAX 是 ayshcronous,因此 getLikes
函数将在 addLike
和 removeLike
中的 AJAX 完成之前触发。您绝对需要将 getLikes
放入 $.ajax
的成功回调中,这样它就不会检索可能尚未更新的数据
function addLike() {
$.ajax({
type: "POST",
url: "likedata.php",
data: { arg1: "<?php echo $_SESSION['userId']?>", arg2: "<?php echo $p;?>", arg3: "1" },
success: getLikes
})
}
关于php - ajax 喜欢/不喜欢按钮不切换回来,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13187615/