我有一个单选框和2个下拉菜单,它们在提交时保存到mysql中。收音机框是yes或no以及在html中创建的2个下拉框。它目前正在全面运行并保存所有数据。从mysql中检索数据并以表格形式显示
我现在想要做的是当用户重新登录时,它会显示他们以前选择的内容(如果他们有)。
php脚本:
<?php
session_start();
require_once("config.php");
if(!isset($_SESSION['username'])){
header('Location: login.php');
exit;
}else{
$sql = "SELECT attendance1 FROM user WHERE username = '".mysql_real_escape_string($_SESSION['username'])."'";
$res = mysql_query($sql);
$row = mysql_fetch_array($res);
if(($row[0] == "Yes") || ($row[0] == "No")){
header("Location: errorsubmit.html");
exit;
}
}
if(isset($_POST['submit'])){
$sql = "UPDATE user SET attendance1 = '" . mysql_real_escape_string($_POST['attendance1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());
$sql = "UPDATE user SET colour1= '" . mysql_real_escape_string($_POST['colour1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());
$sql = "UPDATE user SET shade1= '" . mysql_real_escape_string($_POST['shade1']) . "' WHERE username = '" . mysql_real_escape_string($_SESSION['username']) . "'";
mysql_query($sql) or die("Error in SQL: " . mysql_error());
header("Location: thanks.html", true, 303);
}
?>
FORM:
<form>
<input name="attendance1" type="radio" id="Yes" value="Yes" checked="checked"/>Yes
<br />
<input name="attendance1" type="radio" id="No" value="No" />No
</h3></td>
<td>
<select name="colour1" id="colour1" >
<option selected="selected">Please Select</option>
<option>Red</option>
<option>White</option>
<option>Green</option>
</select>
</td>
<td><h3>
<select name="shade1" id="shade1" >
<option selected="selected">Please Select</option>
<option>light</option>
<option>heavy</option>
</select>
<td> </td>
<td><label>
<input type="submit" name="submit" id="button" value="Submit" />
</label></td>
</tr>
</table>
似乎没有工作,虽然我知道它应该只是出来作为请当我在 – Jacob1 2012-02-26 01:36:31