通过JQuery

Question

使用ajax我在下拉菜单中插入表行。它工作并显示它。我想在点击行时附加事件，但是，我尝试过的东西不起作用。可能是什么问题？

PHP

<?php

for($i=1; $i<=2; $i++){
    echo "<div class='medium-block dropdown'>
            <div class='dropdown-toggle stat-dropdown not-added' id='away" . $i . "' data-toggle='dropdown'>
                <p class='vertical-center add-player' id='add-player" . $i . "' style='margin:0;'>Add Player</p>
            </div>
            <ul class='dropdown-menu drop-scroll' style='margin:0; padding:0; border-radius:0;'>
                <table class='table table-hover' style='margin:0;'>
                    <tbody id='choose-player-away" . $i . "'>
                    </tbody>
                </table>
            </ul>
          </div>";
}
for($i=3; $i<=5; $i++){
    echo "<div class='medium-block dropup'>
            <div class='dropdown-toggle stat-dropdown not-added' id='away" . $i . "' data-toggle='dropdown'>
                <p class='vertical-center add-player' id='add-player" . $i . "' style='margin:0;'>Add Player</p>
            </div>
            <ul class='dropdown-menu drop-scroll' style='margin:0; padding:0; border-radius:0;'>
                <table class='table table-hover' style='margin:0;'>
                    <tbody id='choose-player-away" . $i . "'>
                    </tbody>
                </table>
            </ul>
          </div>";
}?>

JQuery的

<script type="text/javascript">
$(document).ready(function(){
        scalability();
        $(window).resize(scalability);
        $(".not-added").each(function(i){

            $(this).click(function(){
                var identifier = $(this).attr('id').charAt(4);
                var teamid = $(this).attr('id').substring(0,4);
                if($(this).hasClass("not-added")){
                    if (window.XMLHttpRequest) {
                        // code for IE7+, Firefox, Chrome, Opera, Safari
                        xmlhttp = new XMLHttpRequest();
                    } else {
                        // code for IE6, IE5
                        xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");                   
                        }

                    xmlhttp.onreadystatechange = function() {
                        if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
                            document.getElementById("choose-player-"+teamid+identifier).innerHTML = xmlhttp.responseText;
                        }
                    };
                    xmlhttp.open("GET","getPlayerlist.php?team="+teamid+"&id="+identifier,true);
                    xmlhttp.send(); 

                    $(".player").on('click',function(){
                        alert("working");
                    });
                }

            });

        });

    });
</script>

由AJAX提取的PHP文件

<?php

$team = $_GET["team"];
$id = $_GET["id"];

$con = mysqli_connect('localhost','root','','sportacus');

$query = "SELECT * FROM " . $team . "_team WHERE incourt = 0 ORDER BY number";
$result = mysqli_query($con,$query);

while($row = mysqli_fetch_assoc($result)){
    echo "<tr class='player' id='" . $team . "-player" . $row['id'] . "'>
                <td>" . $row['number'] . "</td>
                <td>" . $row['first_name'] . " " . $row['last_name'] . "</td>
          </tr>";
}

mysqli_close($con);

?>

我希望可以点击由AJAX获取的php文件创建的类.player的行。

注意：如果有帮助，我正在使用bootstrap库。其他一切都有效，除了部分：

$(".player").on('click',function(){
    alert("working");
});

Answer 1

与event binding on dynamically created elements中报告的一样，您需要委派活动。

在您的情况下，您需要更改：

$(".player").on('click',function(){

为：

$(document).on('click', ".player", function () {
     alert("working");
});