我有不止一页向它传递两个值,这些值是 Sname 和 SID ..
我用过
$sql = "SELECT s.Sname, e.PID , s.SID
from student AS s
INNER JOIN evaluator AS e
WHERE (e.EID1 = '$id' AND s.PID = e.PID) OR (e.EID2 = '$id' AND s.PID = e.PID)
GROUP BY s.Sname ";
$result = mysql_query ($sql, $connection);
echo "<tr><th>Student Name </th>";
echo "<td><select id='Sname' name='Sname' >";
echo "<option value='' selected='selected'>--</option> ";
while( $row = mysql_fetch_array($result))
{
echo "<option value='$row[SID]|$row[Sname]' >$row[Sname]</option> ";
}
在我输入的接收页面中
list($SID, $Sname) = explode("|", $_POST['Sname']);
它可以工作,但是对于其他页面我想要相同的值,我试图放置相同的explode()但它不起作用..给我一个错误说未定义的索引:Sname +未定义的偏移量:1 ..我的问题是如何将相同的值 Sname 和 SID 也传递给其他页面?