无法在PHP中使用CSS样式表

最后发布: 2017-02-01 06:39:56


问题

即使我已为表格插入样式,也无法在CSS中设置表格样式。

<style>
table, th, td 
{
border: 1px solid black;
}
</style>

上面是代码的简单样式

<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的信息。 我可以知道我能做什么,以便它显示带有CSS的表。 谢谢。

php css
回答

这是因为您使用php动态插入表格。

尝试这个

<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,您的表代码就会随页面一起加载,并且数据库中的所有信息都会动态加载


回答

您可以轻松地设置样式。 html的回显是不好的做法。 因此请小孩子检查以下方式。

<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,您可以像以前一样使用。

table, th, td { border: 1px solid black; }

希望这个能对您有所帮助!!!