Insert data in multiple Mysql tables using single php form

Question

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);

?>

NEVER put $_POST data directly in a query. You must use proper escaping to avoid serious SQL injection bugs. It's not clear why you're mixing mysql_query and mysqli as mysqli has the bind_param method that neatly avoids all of this mess. — tadman
– tadman, Commented Mar 12, 2014 at 23:19

dwurf · Accepted Answer · 2014-03-12 23:15:36Z

1

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.

edited Mar 12, 2014 at 23:15

answered Mar 12, 2014 at 23:09

dwurf

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Comments

Putra L Zendrato · Accepted Answer · 2014-03-13 01:19:18Z

0

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 
 }

answered Mar 13, 2014 at 1:19

Putra L Zendrato

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Comments

Madan Bhandari · Accepted Answer · 2016-11-07 11:10:59Z

0

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";  
?>

edited Nov 7, 2016 at 11:10

Madan Bhandari

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answered Mar 13, 2014 at 16:47

user3412978

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