2
我试图在我的页面上实现一个喜欢/不喜欢的按钮。我设法让这个按钮起作用(当它被点击时,它会变成不喜欢的,反之亦然),它也会在数据库表上创建或删除。现在的问题是喜欢的反面。它仅在第一次点击按钮时起作用,即如果最初有2 likes
,我不喜欢帖子显示1 likes
,但是如果我尝试再次点击它,它会一直显示1 like
,我必须重新加载页面才能看到它更改为2 likes
。PHP喜欢/不喜欢按钮只能第一次工作
这是我到目前为止有:
JQUERY
$(document).on('click', ".miPiace", function() {
trova = '';
commentoORisposta = '';
valCOR = '';
var comOrisp;
try{
trova = $(this).parentsUntil("#fermamiQui");
commentoORisposta = trova.find(".idCommento");
comOrisp = 'commento';
valCOR = commentoORisposta.val();
} catch(err){
trova = $(this).parentsUntil(".infoCommento");
commentoORisposta = trova.find(".idRisposta");
comOrisp = 'risposta';
valCOR = commentoORisposta.val();
}
valCOR = commentoORisposta.val();
if ($(this).hasClass('fa-thumbs-o-up')) {
$(this).removeClass('fa-thumbs-o-up');
$(this).addClass('fa-thumbs-up');
$.get("lib/ottieniCose.php", { like: "", id: valCOR, comOrisp: comOrisp })
.done(function(data) {
trova.find('.numDiLikes').replaceWith('<p>' + data + ' likes</p>');
});
}else if($(this).hasClass('fa-thumbs-up')){
$(this).removeClass('fa-thumbs-up');
$(this).addClass('fa-thumbs-o-up');
$.get("lib/ottieniCose.php", { remLike: "", id: valCOR, comOrisp: comOrisp })
.done(function(data) {
trova.find('.numDiLikes').replaceWith('<p>' + data + ' likes</p>');
});
};
});
PHP
if (isset($_GET['like'])) {
if ($_GET['comOrisp'] == 'commento') {
$commento->set_likes($_GET['id'], true);
return print $commento->get_likes($_GET['id'], true);
} elseif ($_GET['comOrisp'] == 'risposta') {
$commento->set_likes($_GET['id'], false);
return print $commento->get_likes($_GET['id'], false);
}
} elseif (isset($_GET['remLike'])) {
if ($_GET['comOrisp'] == 'commento') {
$commento->remove_likes($_GET['id'], true);
return print $commento->get_likes($_GET['id'], true);
} elseif ($_GET['comOrisp'] == 'risposta') {
$commento->remove_likes($_GET['id'], false);
return print $commento->get_likes($_GET['id'], false);
}
}
其他PHP文件那里是$建议与类
public function get_likes($id, $commento){
$idComm = 0;
$idRisp = 0;
$retVal = ($commento) ? $idComm = $id : $idRisp = $id;
if ($idComm != 0){
$query = "SELECT commento,
(SELECT COUNT(*) FROM likes
WHERE commento = {$idComm})
AS like_count FROM likes";
} elseif($idRisp != 0){
$query = "SELECT risposta,
(SELECT COUNT(*) FROM likes
WHERE risposta = {$idRisp})
AS like_count FROM likes";
}
$trovaQuanti = mysqli_query($_SESSION['connessione'], $query);
$trovaDavveroQuanti = mysqli_fetch_assoc($trovaQuanti);
if ($trovaDavveroQuanti == null) {
return '0';
}
return $trovaDavveroQuanti['like_count'];
}
** WARNING **:当使用'mysqli'你应该使用参数化查询,而['bind_param'(http://php.net/manual/en/mysqli-stmt.bind-param。 PHP)将用户数据添加到您的查询。 **不要**使用字符串插值或连接来完成此操作,因为您将创建严重的[SQL注入漏洞](http://bobby-tables.com/)。 **绝不**将'$ _POST'数据直接放入查询中。 – tadman