我的形式有學生證,電子郵件,姓名...和複選框一個php/MySQL表格與一個公共字段的兩個表?
我設法保存studentid和複選框在表中並創建了一個新表來存儲所有相關學生證其他信息
表1 studentid ----- checkboxselections ....
表2 studentid -----電子郵件----名
我必須插入查詢,一個用於studentid和複選框和一個用於(表2)
$sql="INSERT INTO courses (studentid, ckb)
VALUES ('$studentid', '$cc')";
$sql2="INSERT INTO studentinfo (studentid, email, name)
VALUES ('$studentid', '$email', $fname)";
$ sql2無法存儲數據但是$ sql存儲數據就好,我該如何解決這個問題?
這裏是完整的代碼
$studentid = mysqli_real_escape_string($dbcon, $_GET['studentid']); //echo $studentid;
$email = $dbcon->real_escape_string($_GET['email']);
$fname = $dbcon->real_escape_string($_GET['fname']);
$name = $_GET['ckb'];
if(isset($_GET['ckb'])) //checkboxes
{
foreach ($name as $courcess){
$cc=$cc. $courcess.',';
}
}
$sql="INSERT INTO courses (studentid, ckb)
VALUES ('$studentid', '$cc')";
$sql2="INSERT INTO studentinfo (studentid, email, name)
VALUES ('$studentid', '$email', $fname)";
if (!mysqli_query($dbcon,$sql)) {
die('Error: ' . mysqli_error($dbcon));
}
echo " Thank you for using IME Virtual Registeration ";
mysqli_close($dbcon);
?>
我的表單方法是GET
你在哪裏運行$ sql2? mysqli_query($ dbcon,$ sql2) – Masoomian 2014-09-06 07:05:55