2
主键我想输入数据到表id_ss
ss_kat
并在表中输入ss_kat
id_kat
,这样的数据将输入kat
ss_kat
复选框菜单显示,但我已经多次尝试2不能,约有erornya在哪里?如何插入从复选框
我想从形象在这里
<?php
include "../../inc/conn.php";
?>
<div class="row">
<div class="col-md-6">
<form action="proses.php" method="POST" enctype="multipart/form-data" style="margin-top:20px;">
<input type="hidden" name="id_ss_kat" />
<?php
$ss = mysql_query("SELECT * from ss");
?>
<select class="form-control" name="id_ss">
<?php
while ($rows=mysql_fetch_array($ss))
{ ?>
<option class="form-control" value="<?php echo $rows['id_ss'];?>" name="id_ss"><?php echo $rows['name'];?></option>
<?php
}
?>
</select>
<br>
<?php
$kat = mysql_query("SELECT * from kat WHERE kat_status='1'");
?>
<?php
while ($rows=mysql_fetch_array($kat))
{ ?>
<input type="checkbox" name="id_kat[<?php echo $rows['id_kat'];?>]" value="<?php echo $rows['id_kat'];?>"><?php echo $rows['nama'];?>
<?php
}
?>
<br>
<input class="btn btn-primary" type="submit" name="ssf" value="SAVE">
<br>
</form>
<br>
</div>
</div>
,并在这里提交的复选框表kat
和样本指数proses.php
<?php
if (isset($_POST['ssf'])) {
$id_ss_kat = $_POST['id_ss_kat'];
$id_kat = $_POST['id_kat'];
$id_ss = $_POST['id_ss'];
foreach($id_kat as $kat) {
$source.=$kat.", ";
}
$one=substr($source,0,-1);
$q = "INSERT into ss_kat VALUES('$id_ss_kat','$one','$id_sekolah')";
$dq = mysql_query($q);
if ($dq) {
echo "berhasil";
}else{
mysql_error();
}
}
?>
指数样子的图片如下: