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如何从数据库中传递下拉值的下拉列表的值?如何从数据库中传递下拉值的下拉列表的值?
这里是我的代码..
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("thesis", $con);
@$setup=$_GET['pqsetup'];
?>
<form id="form1" name="form1" method="post" action="">
<table width="414" border="1" align="center">
<tr>
<td width="84" class="style1">
Set-Up: </td>
<td width="9">
<td width="49">
<?php
$pqsetupquer = mysql_query("SELECT * FROM pqsetup");
echo "<select name='pqsetup' onchange=\"reload(this.form)\">";
echo "<option selected>---------------------</option>";
while($noticia2 = mysql_fetch_array($pqsetupquer)) {
if($noticia2['pqs_no'][email protected]$setup){
echo "<option value='$noticia2[pqs_no]' selected>$noticia2[pqs_name]</option>"."<BR>";
}else
echo "<option value='$noticia2[pqs_no]'>$noticia2[pqs_name]</option>"."<BR>";
}
echo "</select>";
?> </td>
<td width="142">Menu Packages</td>
<td width="32"> <?php
$menuquer=mysql_query("SELECT pqs_no, menu_name FROM menu_packages");
//echo "<select name='menu_name' onchange=\"reload(this.form)\">";
echo "<select name='menu_name'>";
echo "<option selected>---------------------</option>";
while($noticia = mysql_fetch_array($menuquer))
{
if($noticia['pqs_no'][email protected]$setup)
{
echo "<option value='$noticia[pqs_no]' selected>$noticia[menu_name]</option>"."<BR>";
}
else
//echo "<option value='$noticia2[pqs_no]'>$noticia[menu_name]</option>"."<BR>";
echo " ";
}
echo"</select>";
?></td>
</tr>
</table>
</form>
我如何能转移的价值,只是在另一个页面显示出来?
转移如何?哪里? – 2011-04-27 20:48:01
转移到另一个页面? – gaildelposo 2011-04-27 21:02:31
'@ $ setup = $ _ GET ['pqsetup'];'不错! – Rudie 2011-04-27 22:02:15