Question

我需要使用数据库信息填充表。但是，回显的HTML没有被插入到我希望它去的DOM中（在表内）。相反，它落在桌子上方。奇怪的是，它低于<h1>标签，所以我知道在将内容加载到页面之前没有发生回声。

检查动态内容会发现我的所有标签都没有包含在内，并且所有数据都作为单个字符串转储。

我做错了什么，或者是否有不同的方法来确保表数据实际上在表中？

HTML

    <h1>Contact Us Form Data</h1>

    <table>

        <thead>
            <tr>
                 <th>First Name</th>
                 <th>Last Name</th>
                 <th>Email</th>
                 <th>Interest</th>
                 <th>Employment</th>
                 <th>City</th>
                 <th>State</th>
                 <th>Date</th>
            </tr>
        </thead>
        <tbody>
        <?php echo display_records() ?>
        </tbody>

    </table>

PHP

function fetch_records() {
    $dbhost = "localhost";
    $dbuname = "user";
    $dbpass = "pass";
    $dbname = "db_dev";
    $conn = mysqli_connect($dbhost, $dbuname, $dbpass, $dbname);

    $query = "
    SELECT * FROM `contact_us_form`
    ";

    return mysqli_query($conn, $query);
}

function record_to_html( $record ) {
    $html = "<td>" .
        "<tr>" . $record->first_name . "</tr>" .
        "<tr>" . $record->last_name . "</tr>" .
        "<tr>" . $record->email . "</tr>" .
        "<tr>" . $record->interest . "</tr>" .
        "<tr>" . $record->employment . "</tr>" .
        "<tr>" . $record->city . "</tr>" .
        "<tr>" . $record->state . "</tr>" .
        "<tr>" . $record->date . "</tr>" .
    "</td>";

    return $html;
}

function display_records() {
    $result = fetch_records();

    $html = "";

    while ( $row = $result->fetch_object() ) {
        $html .= record_to_html( $row );
    }

    return $html;
}

Answer 1

问题是这段代码：

$html = "<td>" .
    "<tr>" . $record->first_name . "</tr>" .
    "<tr>" . $record->last_name . "</tr>" .
    "<tr>" . $record->email . "</tr>" .
    "<tr>" . $record->interest . "</tr>" .
    "<tr>" . $record->employment . "</tr>" .
    "<tr>" . $record->city . "</tr>" .
    "<tr>" . $record->state . "</tr>" .
    "<tr>" . $record->date . "</tr>" .
"</td>";

<tr>和</tr>应为<td>和</td>，反之亦然。

表数据在表格而不是内部回应

1 个答案: