即使我有表格的插入样式,我也无法在css中设置表格样式。
<style>
table, th, td
{
border: 1px solid black;
}
</style>
以上是代码的简单样式
<div class="container">
<table class="u-full-width">
<?php
$result=mysqli_query($connection,"SELECT * FROM register");
}
else
{
$result=mysqli_query($connection,"SELECT * FROM Persons WHERE username LIKE '".$Search."%'");
}
echo "<table border='1'>
<tr>
<th>username</th>
<th>password</th>
<th>Name</th>
<th>gender</th>
<th>age</th>
<th>Contact</th>
<th>address</th>
<th>email</th>
<th>occupation</th>
</tr>";
while($row=mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['Username'] . "</td>";
echo "<td>" . $row['password'] . "</td>";
echo "<td>" . $row['Name'] . "</td> ";
echo "<td>" . $row['gender'] . "</td>";
echo "<td>" . $row['age'] . "</td> ";
echo "<td>" . $row['contact'] . "</td>";
echo "<td>" . $row['address'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "<td>" . $row['occupation'] . "</td>";
echo "</tr>";
}
echo"</table>";
mysqli_close($connection);
?>
</div>
</table>
以上是用于显示表格。我可以知道我能做什么,以便用css显示表格。感谢。
答案 0 :(得分:0)
因为你用php动态插入表。
试试这个
<div class="container">
<table class="u-full-width" border='1'>
<?php
if () {//note code is incomplete, might be copy pasting issue
$result=mysqli_query($connection,"SELECT * FROM register");
}
else
{
$result=mysqli_query($connection,"SELECT * FROM Persons WHERE username LIKE '".$Search."%'");
}
?>
<tr>
<th>username</th>
<th>password</th>
<th>Name</th>
<th>gender</th>
<th>age</th>
<th>Contact</th>
<th>address</th>
<th>email</th>
<th>occupation</th>
</tr>
<?php
while($row=mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['Username'] . "</td>";
echo "<td>" . $row['password'] . "</td>";
echo "<td>" . $row['Name'] . "</td> ";
echo "<td>" . $row['gender'] . "</td>";
echo "<td>" . $row['age'] . "</td> ";
echo "<td>" . $row['contact'] . "</td>";
echo "<td>" . $row['address'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "<td>" . $row['occupation'] . "</td>";
echo "</tr>";
}
mysqli_close($connection);
?>
</table>
</div>
</table>
这样,您的表代码就会加载页面,并且数据库中的任何信息都是动态加载的
答案 1 :(得分:0)
你可以轻松设计风格。回声的HTML是不好的做法。小心翼翼地检查以下方式。
<table border='1'>
<tr>
<th>username</th>
<th>password</th>
<th>Name</th>
<th>gender</th>
<th>age</th>
<th>Contact</th>
<th>address</th>
<th>email</th>
<th>occupation</th>
</tr>
while($row=mysqli_fetch_array($result)){ ?>
<tr>
<td><?php echo $row['Username']; ?></td>
<td><?php echo $row['password']; ?></td>
<td><?php echo $row['Name']; ?></td>
<td><?php echo $row['gender']; ?></td>
<td><?php echo $row['age']; ?></td>
<td><?php echo $row['contact']; ?></td>
<td><?php echo $row['address']; ?></td>
<td><?php echo $row['email']; ?></td>
<td><?php echo $row['occupation']; ?></td>
</tr>
<?php } ?>
</table>
<?php mysqli_close($connection); ?>
?>
在CSS中你可以像以前一样使用。
table, th, td
{
border: 1px solid black;
}
希望这会对你有帮助!!!