function Writeform($name,$type,$num,$abroad,$datein,$cmnd,$address,$room_name){
global $conn;
$query= "INSERT INTO form
(Guest_name, Guest_type, Guest_num, Guest_abroad, datein, Guest_CMND, Guest_Address, room_name) VALUES
('$name', '$type','$num','$abroad','$datein','$cmnd','$address','$room_name')";
if(mysqli_query($conn,$query)){
echo " added !";
$sql="UPDATE room r INNER JOIN form f ON r.room_name=f.$room_name SET r.room_status='not avaliable'";
if (mysqli_query($conn, $sql)) {
echo " Record updated successfully ";
}
else{
echo "Couldn't Update !!";
}
}
else{
echo "Couldn't Add!!";
}
}
我试图插入数据库并同时更新,但我一直在添加,无法更新!
答案 0 :(得分:2)
您无需加入UPDATE
查询中的表格。
$sql = "UPDATE room SET room_status = 'not available'
WHERE room_name = '$room_name'";
但最好使用准备好的陈述。
$sql = "UPDATE room SET room_status = 'not available'
WHERE room_name = ?";
$stmt = mysqli_prepare($conn, $sql);
mysqli_stmt_bind_param($stmt, "s", $room_name);
if (mysqli_stmt_execute($stmt)) {
echo "Record updated successfully!";
} else {
echo "Couldn't update!!";
}