首页 > 解决方案 > 如何使用 1 或 0 的 php mysql 查询填充状态表?

问题描述

我想用php显示一个active和inactive的状态,active的值为1,inactive的值为0;我怎么能$status = ($row['is_active'])?'<span>Active</span>':'<span>Deactivited</span>';在这里 pu 变量<td><?php echo $row->role; ?> <?php echo $status; ?></td> 是我的代码

<?php
    $status = ($row['is_active'])?'<span class="label label-success pull-right">Active</span>':'<span class="label label-danger pull-right">Deactivited</span>';
    $no = 1;
    $select = $pdo->prepare("SELECT * FROM user");
    $select->execute();
    while($row=$select->fetch(PDO::FETCH_OBJ)){
?>
<tr>
    <td><?php echo $no++; ?></td>
    <td><?php echo $row->username; ?></td>
    <td><?php echo $row->fullname; ?></td>
    <td><?php echo $row->role; ?> <?php echo $status; ?></td>
    <td>
        <a href="#"
            onclick="return confirm('Delete User?')"
            class="btn"</a>
    </td>
</tr>
<?php
    }
?>

标签: phpmysql

解决方案


只需将检查移到$row['is_active']for 循环中即可。

这是一个实现。

<?php
    $no = 1;
    $select = $pdo->prepare("SELECT * FROM user");
    $select->execute();
    while ($row = $select->fetch(PDO::FETCH_OBJ)) {
        $status = ($row->is_active) ? '<span class="label label-success pull-right">Active</span>' : '<span class="label label-danger pull-right">Deactivited</span>';
        ?>
        <tr>
            <td><?php echo $no++; ?></td>
            <td><?php echo $row->username; ?></td>
            <td><?php echo $row->fullname; ?></td>
            <td><?php echo $row->role; ?><?php echo $status; ?></td>
            <td>
                <a href="#"
                   onclick="return confirm('Delete User?')"
                   class="btn"</a>
            </td>
        </tr>
        <?php
    } 
?>

推荐阅读