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它不是报告数据库中的数据并在 excel 中显示它，而是在 excel 中显示屏幕上的内容.. 我希望它显示 $excel 变量中包含的内容.. 下面是我的代码

if(isset($_POST['excel_report'])){

    $select_query = "SELECT * FROM question_data";
    $result = mysql_query($select_query);
    $excel ='';
    $excel .= "<table border ='0'>";
    $excel .= "<tr> <td> Question Title </td> <td> Question Answer </td> </tr>";
    while($row = mysql_fetch_array($result, MYSQL_ASSOC)) {

        $excel .="<tr> <td>".$row['question_title']."</td> <td>".$row['question_response']."</tr>";

    }

    $excel .="</table>";
    $filename ="questiondata.xls";
    header('Content-type: application/ms-excel');
    header('Content-Disposition: attachment; filename='.$filename);
    echo $excel;

}

我的 HTML 表单，对于表单，元素是用 jquery 动态添加的。虽然我没有使用任何表单数据，我只是使用 excel 的提交按钮从数据库中获取数据并将其显示在电子表格..我哪里出错了？

<form action="/cms/form_test.php" method="post">

        <h1> Test Form </h1>

        <div class="test">
            <input type ="hidden" name ="form_id" value ="4" />
            <div id ="select_data">
                <select id="select"> </select> <br /><br />

            </div>
            <div id ="options"></div>
            <div id ="select_state">
                <div id ="state_title"> </div>
                <select id="select_state_dropdown"> </select> <br /><br />

            </div>
        </div>
        <input type="submit" name ="excel_report" value="Export Question Data to Excel" />
        <input type="submit" name ="submit" value="submit form" />
    </form>
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