Question

所以我的表格包含“活动名称”“活动描述”“活动日期”和复选框“很重要”。当我检查复选框值“是”时，它重要，它将sql value =“1”发送到表“is_important”。一切都很好，但是我为“is_important”= 1表提供了引导风格“bg-danger”，它并没有显示出来。有什么问题？你可以在代码中看到：

 <?php
    if (isset($_POST['important'])) {
        $error = array();
        $success = array();
        $eventTime = time();
        $important = $_POST['important'];
        $eventName = trim(mysql_real_escape_string($_POST['EventName']));
        $eventDesc = htmlentities(trim(mysql_real_escape_string($_POST['EventDesc'])), ENT_QUOTES);

        if (!isset($eventName) || empty($eventName)) {
            $error['eventName'] = "Prasome ivesti ivykio varda";
        } else if (strlen($eventName) > 32 || strlen($eventName) < 3) {
            $error['eventName'] = "Ivykio pavadinimas turi buti tarp 3 ir 32 simboliu";
        }
        if (!isset($eventDesc) || empty($eventDesc)) {
            $error['eventDesc'] = "Prasome ivesti ivykio aprasyma";
        }

        if (empty($error)) {
            $sql = "INSERT INTO notes_list (title, description, timestamp,is_important) VALUES ('$eventName', '$eventDesc','$eventTime','$important')";
            $result = mysqli_query($con, $sql);
            $success[] = "SEKME !";
        } else {

        }
    }
    ?>
 <table class="table table-striped">
            <thead>
                <tr>
                    <th>Event name</th>
                    <th>Event description</th>
                    <th>Event date</th>
                </tr>
            </thead>
            <tbody>
                <?php
                $query = "SELECT * FROM notes_list ORDER BY id DESC LIMIT 10";
                $result2 = mysqli_query($con, $query);
                print_r($_POST);
                if ($result2) {
                    while ($note = mysqli_fetch_assoc($result2)) {
                        ?>

                        <tr<?php echo (($note['is_important'] == 1) ? "class='bg-danger'" : ""); ?>>

                            <td><?php echo $note['title']; ?></td>
                            <td><?php echo $note['description'] ?></td>
                            <td><?php echo date('l M jS', $note['timestamp']); ?></td>
                        </tr>
                        <?php
                    }

                    mysqli_free_result($result2);
                }

                /* close connection */
                mysqli_close($con);
                ?>
            </tbody>
        </table>

此图片中的完整示例： https://www.dropbox.com/s/h650h2spy2487dm/chechbox.jpg?dl=0

Answer 1

这：

<tr<?php echo (($note['is_important'] == 1) ? "class='bg-danger'" : ""); ?>>

会渲染：

<trclass='bg-danger'>

如果is_important为1.您需要在class之前有空格。

选中复选框后如何为SQL表赋予样式？

1 个答案: