thinkphp ajax問題

Question

<p class="main-wrap">
    <p class="crumb-wrap">
        <p class="crumb-list"><i class="icon-font"></i><a href="/jscss/admin">首页</a><span class="crumb-step">></span><span class="crumb-name">链接管理</span></p>
    </p>
    <p class="search-wrap">
        <p class="search-content">
            <form action="/jscss/admin/design/index" method="post">
                <table class="search-tab">
                    <tr>
                        <th width="120">选择分类:</th>
                        <td>
                            <select name="search-sort" id="">
                                <option value="">全部</option>
                                <option value="19">精品界面</option><option value="20">推荐界面</option>
                            </select>
                        </td>
                        <th width="70">关键字:</th>
                        <td><input class="common-text" placeholder="关键字" name="keywords" value="" id="" type="text"></td>
                        <td><input class="btn btn-primary btn2" name="sub" value="查询" type="submit"></td>
                    </tr>
                </table>
            </form>
        </p>
    </p>
    <p class="result-wrap">
        <form name="myform" id="myform" method="post" action="__CONTROLLER__/sort">
            <p class="result-title">
                <p class="result-list">
                    <a href="__CONTROLLER__/add"><i class="icon-font"></i>新增链接</a>
                    <a id="updateOrd" href="javascript:void(0)"><i class="icon-font"></i><input type="submit" name="" value="更新排序"></a>
                </p>
            </p>
            <p class="result-content">
                <table class="table-bordered table">
                    <tr class="active">
                        <th>排序</th>
                        <th>ID</th>
                        <th>链接名称</th>
                        <th>链接地址</th>
                        <th>链接描述</th>
                        <th>操作</th>
                    </tr>
                    <volist name="links" id="arr">
                        <tr class="yjsTr">
                            <td>
                                <input class="common-input sort-input" name="{$arr.id}" value="{$arr.sort}" type="text">
                            </td>
                            <td>{$arr.id}</td>
                            <td><a target="_blank" href="#" title="{$arr.title}">{$arr.title}</a> …
                            </td>
                            <td><a target="_blank" href="#" title="{$arr.url}">{$arr.url}</a> …
                            </td>
                            <td><a target="_blank" href="#" title="{$arr.desc}">{$arr.desc}</a> …
                            </td>
                            <td>
                                <a class="link-update" href="__CONTROLLER__/edit">修改</a>
                                <a class="link-del" onClick="return confirm('确定删除该条数据吗？');" href="__CONTROLLER__/del/id/{$arr.id}">删除</a>
                            </td>
                        </tr>
                    </volist>
                </table>
                <p class="list-page"> 2 条 1/1 页</p>
            </p>
        </form>
    </p>
</p>

<script type="text/javascript" src="__PUBLIC__/js/libs/angularjs.js"></script>
<script type="text/javascript">
    $(function(){
        $('a.link-update').on('click', function(){
            var $obj = $(this).parents('tr.yjsTr').children();
            console.log(5);
            var data = {
                'title': $obj.eq(2).text(),
                'url': $obj.eq(3).text(),
                'desc': $obj.eq(4).text(),
            }
            $.ajax({
                type: 'post',
                data: data,
                url: '__CONTROLLER__/edit',
                success: function(){
                    window.location.href = '__CONTROLLER__/edit';
                },
                error: function(){
                }
            })
            return false;
        })
    })
</script>

上述是lst.html中的程式碼，要將表單資料傳送到__CONTROLLER__/edit中，下面是edit函數

public function edit(){
    $link = D('Link');
    if(IS_POST){
        var_dump($_POST);
        die;
    }
    // $this->display();
}

1、請問為什麼沒收到表單post提交的資料？
2、如果要實現應該怎麼實現？

Answer 1

你谷歌瀏覽器按下F12找到NetWork tab下再執行你的這個ajax 看看包內容和地址

Answer 2

if($_POST){
var_dump($_POST[]);
}