使用单个php表单在多个Mysql表中插入数据

I have a form with data I will like to insert in 2 different tables (order and order_etails). Here is what I did. But it is inserting in only 1 form.

<?php
include '../db/connect.php';

$sql="INSERT INTO order_etails (part_id, quantity, price, status_id,order_id)
  VALUES
        ('$_POST[part_id]','$_POST[quantity]','$_POST[price]','$_POST[status_id]','$_POST[order_id]     '),
('$_POST[part_id2]','$_POST[quantity2]','$_POST[price2]','$_POST[status_id2]','$_POST[order    _id2]'),
('$_POST[part_id3]','$_POST[quantity3]','$_POST[price3]','$_POST[status_id3]','$_POST[order    _id3]')";

$sql1="INSERT INTO order (platform)
VALUE
('$_POST[platform]')";

if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
 }
echo "record(s) added";

mysqli_close($con);

?>

I have also tried this:

<?php
include '../db/connect.php';


$sql="INSERT INTO order_details (part_id, quantity, price, status_id,order_id)
VALUES
('$_POST[part_id]','$_POST[quantity]','$_POST[price]','$_POST[status_id]','$_POST[order_id]'),
('$_POST[part_id2]','$_POST[quantity2]','$_POST[price2]','$_POST[status_id2]','$_POST[order_id2]'),
   ('$_POST[part_id3]','$_POST[quantity3]','$_POST[price3]','$_POST[status_id3]','$_POST[order_id3]');

INSERT INTO order (platform)
VALUE
('$_POST[platform]')";


mysql_query($sql1);

if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "record(s) added";

mysqli_close($con);

?>

Try VALUES instead of VALUE in your second query.

Also, you don't seem to actually execute both queries in either of your examples. You should have something like:

if (!mysqli_query($con,$sql))
{
    die('Error: ' . mysqli_error($con));
}
if (!mysqli_query($con,$sql1))
{
    die('Error: ' . mysqli_error($con));
}
echo "record(s) added";

You should also consider wrapping the two executions in a transaction, so that if the 2nd insert fails the first will also be rolled back.

First of all you need using mysql multi query.

$sql="INSERT INTO order_details (part_id, quantity, price, status_id,order_id)
  VALUES
  ('$_POST[part_id]','$_POST[quantity]','$_POST[price]','$_POST[status_id]','$_POST[order_id]'),
  ('$_POST[part_id2]','$_POST[quantity2]','$_POST[price2]','$_POST[status_id2]','$_POST[order_id2]'),
  ('$_POST[part_id3]','$_POST[quantity3]','$_POST[price3]','$_POST[status_id3]','$_POST[order_id3]');
  $sql.= INSERT INTO order (platform)
  VALUES
  ('$_POST[platform]')";


 if (mysqli_multi_query($link, $sql)) {
  //do action 
 }

Thank you for both of you. It worked and here is how I did it:

<?php
  include '../db/connect.php';
  $sql="INSERT INTO order_details (part_id, quantity, price, status_id,order_id)
  VALUES

 ('$_POST[part_id]','$_POST[quantity]','$_POST[price]','$_POST[status_id]','$_POST[order_id]'),
 ('$_POST[part_id2]','$_POST[quantity2]','$_POST[price2]','$_POST[status_id2]','$_POST[order_id2]'),
 ('$_POST[part_id3]','$_POST[quantity3]','$_POST[price3]','$_POST[status_id3]','$_POST[order_id3]')";

 $sql1="INSERT INTO order (platform)
 VALUES
 ('$_POST[platform]')";

 if (!mysqli_query($con,$sql))
 {
  die('Error: ' . mysqli_error($con));
 }
 if (!mysqli_query($con,$sql1))
 {
  die('Error: ' . mysqli_error($con));
 }
 echo "record(s) added";  
?>