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。如果你能看到并告诉我这个问题，我会很高兴。原因是什么？

<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

你要听的 “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:)