Question

我有这个代码。我的问题是当我检查复选框仍然是令人耳目一新，卷轴上升，并没有返回他的提交页面后的位置。谁能帮我？对不起，我是ajax的新手。

    <style>
        .container {
            border:2px solid #ccc;
            width:600px;
            height: 1000px;
            overflow-y: scroll;
        }
    </style>
  <script src="https://code.jquery.com/jquery-1.10.2.js"></script>
</head>
<body>

<form action="ajaxtry.php" id="searchForm">
 <div id="Html2" style="position:absolute;left:0x;top:60px;width:250px;height:100px;z-index:12" class="container">
    <input onclick="javascript: submit()" type="checkbox">Checkbox1
    <br>
    <input onclick="javascript: submit()" type="checkbox">Checkbox1
    <br>
    <input onclick="javascript: submit()" type="checkbox">Checkbox1
    <br>
    <input onclick="javascript: submit()" type="checkbox">Checkbox1
    <br>
    <input onclick="javascript: submit()" type="checkbox">Checkbox1
    <br>
    <input onclick="javascript: submit()" type="checkbox">Checkbox1
    <br>
    <input onclick="javascript: submit()" type="checkbox">Checkbox1
</form>
<!-- the result of the search will be rendered inside this div -->
<div id="result"></div>

<script>
// Attach a submit handler to the form
$( "#searchForm" ).submit(function( event ) {

  // Stop form from submitting normally
  event.preventDefault();

  // Get some values from elements on the page:
  var $form = $( this ),
    term = $form.find( "input[name='s']" ).val(),
    url = $form.attr( "action" );

  // Send the data using post
  var posting = $.post( url, { s: term } );

  // Put the results in a div
  posting.done(function( data ) {
    var content = $( data ).find( "#content" );
    $( "#result" ).empty().append( content );
  });
});
</script>

ajax仍在加载onclick提交

0 个答案: