我正在尝试为我使用PHP生成的表分配一个ID,但它会一直返回错误。这是到目前为止工作正常的完整代码。我想要的只是在表格中添加一个'id',这样我就可以在相关的表格中为它应用css样式
<?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);
有任何帮助吗?也许我会以错误的方式去做?
答案 0 :(得分:0)
请尝试下面的代码块。我添加了表格ID,并在循环
中关闭</tr>
<?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);