我想选择测试名称,在选择测试名称后,想要从“mst_result”表中获取所有值,其中测试名称等于我在drop sown列表中选择的值。目前我的上帝是这样的,但我没有得到如何获得所选测试名称的值。
<form name="form1" method="post" onSubmit="return check();">
<table>
<tr>
<td>Test Name </strong></div></td>
<select name="test" id="test">
<option value=""><------></option>
<?php
$rs=mysql_query("Select * from mst_test order by test_name",$cn);
while($row=mysql_fetch_array($rs))
{
if($row[0]== $rs['test_id'])
{
echo "<option value='$row[0]' selected>$row[2]</option>";
}
else
{
echo "<option value='$row[0]'>$row[2]</option>";
}
}
?>
</select>
<td align="left" ><input type="submit" name="res" value="Submit" /> </td>
</tr>
</table>
</form>
if(isset($_POST['res']))
{
$qr=mysql_query("Select * from mst_result order by login where test_id='$test' ",$cn);
$ii=1;
while($rr1=mysql_fetch_array($qr)){
$qr1=mysql_query("SELECT * FROM mst_result order by login");
$rr1 = mysql_fetch_array($qr1);
$qr2 = mysql_query("SELECT * FROM mst_result order by score");
$rr2 = mysql_fetch_array($qr2);
?>
<table>
<tr>
<td><?php echo $ii;?></td>
<td><?php echo $rr1['login'];?></td>
<td><?php echo $rr2['score'];?></td>
</tr>
<?php
$ii++;
}}
?>