PHP中的表ID

我试图将anID分配给我用PHP生成的表，但它总是返回一个错误。以下是迄今为止运行良好的完整代码。我只想在表中添加一个“id”，这样我就可以在相关的工作表中将css样式应用到表中

        <?php
        $con=mysqli_connect("localhost","<un>","<pw>","monitor");
        // Check connection
        if (mysqli_connect_errno())
          {
          echo "Failed to connect to MySQL: " . mysqli_connect_error();
          }

        $result = mysqli_query($con,"SELECT * FROM presnationalresults ORDER BY Percentage DESC");

        echo "<table border='1'>
        <tr>
        <th>President</th>
        <th>Party</th>
        <th>Votes</th>
        <th>Percentage</th>
        </tr>";

        while($row = mysqli_fetch_array($result))
          {
          echo "<tr>";
          echo "<td>" . $row['PresidentName'] . "</td>";
          echo "<td>" . $row['PartyCode'] . "</td>";
          echo "<td>" . $row['Votes'] . "</td>";
          echo "<td>" . $row['Percentage'] . "</td>";
          }
        echo "</table>";

        mysqli_close($con);

---

有什么帮助吗？也许我做得不对？

请尝试下面的代码块。我添加了表id，并在while循环中关闭了内部

<?php

$con = mysqli_connect("localhost", "<un>", "<pw>", "monitor");
// Check connection
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con, "SELECT * FROM presnationalresults ORDER BY Percentage DESC");

echo "<table border='1' id='table-id'>
        <tr>
        <th>President</th>
        <th>Party</th>
        <th>Votes</th>
        <th>Percentage</th>
        </tr>";

while ($row = mysqli_fetch_array($result)) {
    echo "<tr>";
    echo "<td>" . $row['PresidentName'] . "</td>";
    echo "<td>" . $row['PartyCode'] . "</td>";
    echo "<td>" . $row['Votes'] . "</td>";
    echo "<td>" . $row['Percentage'] . "</td>";
    echo "</tr>"; // you forget close tr
}
echo "</table>";

mysqli_close($con);

---

我们应该猜到确切的错误吗？还有MySQL代码吗？为什么还要麻烦使用echo输出静态HTML？你所做的只是让你的报价很难正确。对不起。解析错误：语法错误，意外“，你认为这是一段有效的php代码吗？