$(document).ready(function(){
var fid = 'pList';
$("#btn_addpassenger").bind("click",function(){
create_newtable(fid);
});
$("#btn_delpassenger").bind("click",function(){
$("#"+fid).remove();
});
});
function create_newtable(nowid){
var newid = "pList_" + new Date().getTime();
var fuzhi = $("#"+nowid).clone(false).attr("id",newid);
fuzhi.find("#btn_addpassenger").bind("click",function(){
create_newtable(newid);
});
fuzhi.find("#btn_delpassenger").bind("click",function(){
$("#"+newid).remove();
});
$("#"+nowid).after(fuzhi);
}
}<div id="passenger_conetent">
<table id="pList"></table>
</div>有人帮忙给简化一下这段代码吗
我总觉得在新复制的table里不需要在绑一次事件
这样写很怪
可是我不知道改怎么改
<script language="JavaScript">
$(document).ready(function(){
var fid = 'pList';
$("#btn_addpassenger").bind("click",function(){
create_newtable(fid);
});
$("#btn_delpassenger").bind("click",function(){
$("#"+fid).remove();
});
});
function create_newtable(nowid){
var newid = "pList_" + new Date().getTime();
var fuzhi = $("#"+nowid).clone(false).attr("id",newid);
$("#"+nowid).after(fuzhi);
}</script>
<input type="button" value="add" id="btn_addpassenger">
<div id="passenger_conetent">
<table id="pList"><tr><td>xx</td></tr></table>
</div>
<script src="http://ajax.Microsoft.com/ajax/jQuery/jquery-1.3.2.min.js" type="text/javascript"> </script>
<script language="JavaScript">
$(document).ready(function(){
$("#btn_addpassenger").bind("click",function(){
create_newtable($(this).parent().parent().parent().parent());
});
$("#btn_delpassenger").bind("click",function(){
$(this).parent().parent().parent().parent().remove();
});
});
function create_newtable(obj){
var newid = "pList_" + new Date().getTime();
var fuzhi = obj.clone(true).attr("id",newid);
obj.after(fuzhi);
}</script><div id="passenger_conetent">
<table id="pList">
<tr>
<td>xxx</td>
<td>
<input type="button" value="add" id="btn_addpassenger">
<input type="button" value="del" id="btn_delpassenger">
</td>
</tr>
</table>
</div>