Question

我想通过填写最终用户在评论表单中输入的密码来显示评论内容，但是ajax提交按钮不起作用。他们有任何冲突吗？请帮我改进这段代码。

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js" type="text/javascript"></script>
<script type="text/javascript">
    $(document).ready(function(){
        $("#button").click(function() {
            var password = $("#search").val();
            if (password != "") {
                $("#result").html("<img alt="ajax search" src='ajax-loader.gif'/>");
                $.ajax({
                    type: "post",
                    url: "search.php",
                    data: "password=" + password,
                    success: function(data){
                        $("#result").html(data);
                        $("#search").val("");
                    }
                });
            }
        }

        $("#button").click(function(){
            search();
        });

        $('#search').keyup(function(e) {
            if (e.keyCode == 13) {
                search();
            } 
        });
    });
</script>
<body>
    <?php if(!$userdata->data->ID):?>
        <form method="post" action="">
            <input type="password" id="search" placeholder="This is a secret question." />
            <input type="button" id="button" value="Submit" />
            <ul id="result"></ul>
        </form>
    <?php endif?>
</body>

ajax提交按钮不起作用

0 个答案: