我正在尝试在我的页面上实现一个喜欢/不喜欢按钮。我设法让按钮工作(点击时它会变成喜欢，反之亦然)，它还会在数据库表上创建或删除类似的东西。现在的问题是喜欢的计数器。它仅在我第一次单击按钮时有效，即如果最初有 2 个赞 并且我不喜欢帖子显示 1 个赞，但如果我再次尝试单击在它上面，它一直显示 1 like，我必须重新加载页面才能看到它更改为 2 likes.

这是我目前所拥有的:

JQUERY

$(document).on('click', ".miPiace", function() {
    trova = '';
    commentoORisposta = '';
    valCOR = '';
    var comOrisp;       

    thử{
        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('
' + data + ' likes
');
        });


    }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('
' + data + ' likes
');
        });


    };
});

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);
    }
}

其他有 $commenti 类的 PHP 文件

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'];

}

câu trả lời hay nhất

您可能想尝试正常的 $.ajax 调用。并关闭缓存。有时这会导致您需要刷新才能看到更改的问题。

$.ajax({
    gõ: 'NHẬN',
    bộ đệm: sai,
    url: "lib/ottieniCose.php",
    data: { like: "", id: valCOR, comOrisp: comOrisp },
    dataType: "html",
    success: function(html){
        trova.find('.numDiLikes').html('
' + data + ' likes
');
    }
});

关于PHP 喜欢/不喜欢按钮只在第一次工作，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/34051940/