jQuery 适用于 Chrome、Firefox、IE 但不适用于 iPhone

Question

此代码适用于计算机浏览器，但不适用于 iPhone。如果您能看到并告诉我问题，我将非常高兴。可能是什么原因？

<ol style="margin: 0px; padding: 0px; display:inline-block" id="update" class="timeline">
<form id="begenform" action="#" method="post">
<input type="hidden" id="feedid" value="<?php echo $feed[$i]['i']; ?>"/>
<input type="hidden" id="userid" value="<?php echo $_SESSION['u']; ?>" />
<input type="hidden" id="i" value="<?php echo $i; ?>" />
<?php if(!empty($liked['u']))echo"<span style='color:#cd2122'>Beğendin</span>"; else echo '<input id="begenbutton" type="submit" class="likesubmit" value="Beğen" />'?>
</form></ol>


<script type="text/javascript" >
$(function() {
$(".likesubmit").click(function() 
{
var gideni = $("#i").val();
var userid = $("#userid").val();
var feedid = $("#feedid").val();
var dataString = '&userid='+ userid + '&feedid=' + feedid + '&gideni=' + gideni;

$.ajax({
type: "POST",
url: "profilelike.php",
data: dataString,
cache: false,
success: function(html){
$("ol#update").append(html);
$("span.like").text("Ok");
}
});
return false;
}); });
</script>

Answer 1

你需要监听“touchstart”和“touchend”事件：

$('.myElement').bind("touchstart", function(e){alert('Clicked!')} );

我的实现：

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script type="text/javascript">

$(document).ready(function() {

    $('#begenform').bind('touchstart click', function(e) {

        e.stopPropagation(); 
        e.preventDefault();

        var gideni = $("#i").val();
        var userid = $("#userid").val();
        var feedid = $("#feedid").val();
        var dataString = '&userid='+ userid + '&feedid=' + feedid + '&gideni=' + gideni;

        $.ajax({
            type: "POST",
            url: "profilelike.php",
            data: dataString,
            cache: false,
            success: function(html) {

                $("ol#update").append(html);
                $("span.like").text("Ok");
            }
        });

        return false;
    });
});

</script>

<form id="begenform" method="post" action="#">
    <input id="begenbutton" type="submit" class="likesubmit" value="Beğen" />
</form>

对我来说工作正常。 试试看:)