PHP中的表ID

Question

我試圖給我用PHP生成的表分配一個ID，但它一直返回一個錯誤。這是迄今爲止工作正常的完整代碼。我只需要在表格中添加一個'id'，這樣我就可以在相關表格中應用css樣式

 <?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); 

任何幫助？也許我會以錯誤的方式去談論它？

Answer 1

請嘗試下面的代碼塊。我已經加入表ID和也接近</tr>內while循環

<?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);