PHP数据库表单CSS


PHP Database form CSS

我有以下代码:

<?php
 require "dbconn.php";
 $register="SELECT * from register WHERE username = '". $_SESSION['username']."'";
$re = $connect->query($register);
$numrow = $re->num_rows; 
echo "<table>";
echo "<tr>";
echo"<th>Username</th>";
echo"<th>Forename</th>";
echo"<th>Surname</th>";
echo"<th>Course</th>";
echo"<th>Subject</th>";
echo"<th>Level</th>";
echo"<th>Date</th>";
echo"<th>Time</th>";
echo"</tr>";
$count = 0;
while ($count < $numrow) 
{
$row = $re->fetch_assoc();
extract($row);    
echo "<tr>";
echo "<td>";
echo $username;
echo "</td>";
echo "<td>";
echo $firstName;
echo "</td>";
echo "<td>";
echo $surname;
echo "</td>";
echo "<td>";
echo $course;
echo "</td>";
echo "<td>";
echo $subject;
echo "</td>";
echo "<td>";
echo $level;
echo "</td>";
echo "<td>";
echo $date;
echo "</td>";
echo "<td>";
echo $time;
echo "</td>";
echo "</tr>";
$count = $count + 1;
}
?>

它以表格的形式显示数据库中的记录

<FORM METHOD="LINK" ACTION="register.php">
<INPUT TYPE="submit" VALUE="Go back to register"></INPUT></form>

无论我把按钮放在什么DIV中,或者我给它什么CSS。按钮总是位于数据库中表单的上方,我无法将其放在表下方。

有什么想法吗?

首先,您没有使用</table>标记关闭表;你这样做很重要。

然后是METHOD="LINK",它不存在,也不是一个有效的方法。

您要使用method="post"method="get"

在Stack上看到这个答案:

  • https://stackoverflow.com/a/11582427/