我有一个弹出形式从用户获得数据并将其添加到MySQL的phpMyAdmin的表,我希望能够以一个HTML表格一旦弹出关闭中显示这个数据,在我点击提交后,我被引导回主页,在那里我想让数据显示在桌面上。显示MySQL数据与HTML表格
M.html
<thead>
<tr>
<th scope="col" colspan="2">CRN</th>
<th scope="col" colspan="6">Title</th>
<th scope="col" rowspan="2">Co-Ordinator</th>
<th scope="col" colspan="6">Coursework Number</th>
<th scope="col" rowspan="2">Contribution</th>
<th scope="col" colspan="6">Edit</th>
<th scope="col" rowspan="2">Upload</th>
<th scope="col" colspan="6">Manage Grades</th>
</tr>
</table>
add.php
$display_query = "SELECT CRN, Title, Co-Ordinator, CourseworkNumber, Contribution FROM Modules";
$displayresult = mysqli_query($con, $display_query);
$num = mysql_numrows($displayresult);
mysqli_close($con);
header("Location: ../views/M.html");
我新的HTML和PHP不确定如何链接这到HTML
有啥问题只是选择另一页上的数据和显示或发送JSON http://stackoverflow.com/a/19027741/3841803 – silentprogrammer 2015-02-09 19:14:14