PHP数据库表单CSS
我有以下代码:PHP数据库表单CSS,php,html,css,forms,Php,Html,Css,Forms,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>U
<?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 = $count + 1;
}
?>
首先,您没有使用
标记关闭表;你这样做很重要
还有METHOD=“LINK”
,它不存在,也不是有效的方法
您可以使用method=“post”
或method=“get”
关于它,请参见堆栈上的以下答案:
如果您发布完整的代码会有所帮助,因为没有显示表单元素相对于db输出的位置。一个快速而肮脏的修复方法是将submit按钮放在表尾。请看:不要认为有一个名为“LINK”的方法链接不是问题所在,与问题无关..表单位于$count=$count+1;}在同一个div中,如果我把它放在另一个div中,结果仍然是一样的。是的,这可能行得通,但我想知道是否有更好的办法,拉尔夫@弗雷德:我一定会去的,我会接受的
<FORM METHOD="LINK" ACTION="register.php">
<INPUT TYPE="submit" VALUE="Go back to register"></INPUT></form>