-1
我有以下代碼:PHP數據庫形式CSS
<?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 = $count + 1;
}
?>
它顯示了從數據庫表格的記錄。我有以下按鈕:
<FORM METHOD="LINK" ACTION="register.php">
<INPUT TYPE="submit" VALUE="Go back to register"></INPUT></form>
無論我把按鈕放在哪個DIV中,或者我給了什麼CSS。該按鈕始終保持高於數據庫的表格,並且我無法將其置於桌子下方。
任何想法?
如果您發佈的完整代碼這將有助於becsause沒有哪裏的表單元素是相對於數據庫輸出指示。一個快速而骯髒的修復方法是將提交按鈕放在表格頁腳中。 – 2015-04-01 19:27:38
看看這個:http://stackoverflow.com/questions/11582286/form-method-link-or-a-whats-the-difference不要認爲有一種方法稱爲「LINK」 – Maximus2012 2015-04-01 19:28:25
該鏈接不是這個問題與問題無關。表單在$ count = $ count + 1下; }和在同一個div中。如果我把它放在另一個div中,結果仍然是一樣的。是的,可以工作,但想知道是否有更好的方法 – David 2015-04-01 19:31:49