插入sql数据库后表类不起作用答案

【问题标题】：table class not working after inserting sql database inside插入sql数据库后表类不起作用
【发布时间】：2018-03-02 13:18:19
【问题描述】：

我首先做了表格布局并确保一切正常，在将表格连接到数据库并尝试将记录放入其中后，一切正常，但是课程没有，我现在有没有布局的无聊表格制作。

    <body>
     <h1>Employees</h1>
     <table class="responstable">
    <?php
    require 'connection.php';
    $conn    = Connect();



    $result = mysqli_query($conn,"SELECT * FROM employee");


    echo "<table border='1'>


      <tr>
        <th>Id</th>
        <th>First name</th>
        <th>Last name</th>
        <th>Salary</th>
        <th>Start Date</th>
        <th>Department</th>
      </tr>";


    while($row = mysqli_fetch_array($result))
    {
    echo "<tr>";
    echo "<td>" . $row['id'] . "</td>";
    echo "<td>" . $row['firstname'] . "</td>";
    echo "<td>" . $row['lastname'] . "</td>";
    echo "<td>" . $row['salary'] . "</td>";
    echo "<td>" . $row['startdate'] . "</td>";
    echo "<td>" . $row['department'] . "</td>";
    echo "</tr>";
    }
    echo "</table>";

    mysqli_close($conn);
    ?>



      <script src='http://cdnjs.cloudflare.com/ajax/libs/respond.js/1.4.2/respond.js'></script>


    </body>
    </html>

表类是“负责任的”

【问题讨论】：

标签： php html css sql class

【解决方案1】：

table 标记在 2 个地方定义。请尝试删除它。

echo "<table border='1'>

或者删除h1标签后顶部的table标签，并将class添加到echo中定义的table标签中，完整代码

<body>
     <h1>Employees</h1>
    <?php
    require 'connection.php';
    $conn    = Connect();



    $result = mysqli_query($conn,"SELECT * FROM employee");


    echo "<table border='1' class='responstable'>


      <tr>
        <th>Id</th>
        <th>First name</th>
        <th>Last name</th>
        <th>Salary</th>
        <th>Start Date</th>
        <th>Department</th>
      </tr>";


    while($row = mysqli_fetch_array($result))
    {
    echo "<tr>";
    echo "<td>" . $row['id'] . "</td>";
    echo "<td>" . $row['firstname'] . "</td>";
    echo "<td>" . $row['lastname'] . "</td>";
    echo "<td>" . $row['salary'] . "</td>";
    echo "<td>" . $row['startdate'] . "</td>";
    echo "<td>" . $row['department'] . "</td>";
    echo "</tr>";
    }
    echo "</table>";

    mysqli_close($conn);
    ?>



      <script src='http://cdnjs.cloudflare.com/ajax/libs/respond.js/1.4.2/respond.js'></script>


    </body>
    </html>

【讨论】：

我真笨！谢谢！当我可以时将其标记为答案