Php Post to two tables in Mysql - php

Question

I'm trying to POST to two tables at the same time. I'm trying to get the DonorID to display in to another table under $description. I'm able to just write any text in the $description, but I need it to be dynamic not static, which is what the text is. I have two tables; the first is accounting and the second is donations. I'm trying to alter the $description='Donation from Donor'; and have the donor that made the transaction be listed where the Donor is. Any suggestions would be greatly appreciated.
Here is my code:
<?php
$dbserver = "localhost";
$dblogin = "root";
$dbpassword = "";
$dbname = "";
$date=$_POST['date'];
$firstname=$_POST['firstname'];
$lastname=$_POST['lastname'];
$middleinitial=$_POST['middleinitial'];
$organization=$_POST['organization'];
$donorid=$_POST['donorid'];
$paymenttype=$_POST['paymenttype'];
$nonmon=$_POST['nonmon'];
$event=$_POST['event'];
$Income=$_POST['Income'];
$account='Revenue';
$description='Donation from Donor';
$transactiontype='Income';
$Expense='0.00';
$con = mysql_connect("$dbserver","$dblogin","$dbpassword");
if (!$con)
{
die('Could not connect to the mySQL server please contact technical support
with the following information: ' . mysql_error());
}
mysql_select_db("$dbname", $con);
$sql = "INSERT INTO donations (date, firstname, middleinitial, lastname,
organization, donorid, paymenttype, nonmon, Income, event)
Values
('$date','$firstname','$middleinitial','$lastname','$organization',
'$donorid','$paymenttype','$nonmon','$Income','$event')";
$sql2 = "INSERT INTO accounting (date, transactiontype, account,
description, Income, Expense)
VALUES ('$date','$transactiontype','$account','$description','$Income','$Expense')";
mysql_query($sql2);
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
echo "1 record added";
mysql_close($con);
header( 'Location: http://localhost/donations.php' ) ;
?>

Answer 1

As i said i would personaly use mysqli for new project, here a sample of you code with mysqli:
$dbserver = "localhost";
$dblogin = "root";
$dbpassword = "";
$dbname = "";
$date=$_POST['date'];
$firstname=$_POST['firstname'];
$lastname=$_POST['lastname'];
$middleinitial=$_POST['middleinitial'];
$organization=$_POST['organization'];
$donorid=$_POST['donorid'];
$paymenttype=$_POST['paymenttype'];
$nonmon=$_POST['nonmon'];
$event=$_POST['event'];
$Income=$_POST['Income'];
$account='Revenue';
$description='Donation from Donor';
$transactiontype='Income';
$Expense='0.00';
//opening connection
$mysqli = new mysqli($dbserver, $dblogin, $dbpassword, $dbname);
if (mysqli_connect_errno())
{
printf("Connection failed: %s\n", mysqli_connect_error());
exit();
}
$sql = "INSERT INTO `donations` (`date`, `firstname`, `middleinitial`, `lastname`, `organization`, `donorid`, `paymenttype`, `nonmon`, `Income`, `event`) Values ('$date','$firstname','$middleinitial','$lastname','$organization', '$donorid','$paymenttype','$nonmon','$Income','$event')";
$sql2 = "INSERT INTO `accounting` (`date`, `transactiontype`, `account`, `description`, `Income`, `Expense`) VALUES ('$date','$transactiontype','$account','$description','$Income','$Expense')";
$query1 = $mysqli->query($sql) or die($mysqli->error.__LINE__);
$query2 = $mysqli->query($sql2) or die($mysqli->error.__LINE__);
//closing connection
mysqli_close($mysqli);
header( 'Location: http://localhost/donations.php' ) ;
UPDATE
you can add donorid simply placing both vars in the query like:
$sql2 = "INSERT INTO `accounting` (`date`, `transactiontype`, `account`, `description`, `Income`, `Expense`) VALUES ('".$date."','".$transactiontype."','".$account."','".$donorid . " " . $description."','".$Income."','".$Expense."')";
this way i just separate donorid and description with a space but you can add anything you want to in plain text:
'".$donorid . " - " . $description."'

Answer 2

After this
$sql = "INSERT INTO donations (date, firstname, middleinitial, lastname,
organization, donorid, paymenttype, nonmon, Income, event)
Values
('$date','$firstname','$middleinitial','$lastname','$organization',
'$donorid','$paymenttype','$nonmon','$Income','$event')";
put
mysql_query($sql);
Please execute the query.

Answer 3

Things I see is ..
First your just executing your $sql2 but not the other $sql statement
Another is while inserting you declared some columns name that is a mysql reserved word (date column)
you should have `` backticks for them..
Refer to this link MYSQL RESEERVED WORDS
additional note: Your query is also vulnerable to sql injection
SQL INJECTION
How to prevent SQL injection in PHP?

Answer 4

Just write after insert on trigger on first table to insert data into another table.

Answer 5

You will have to split $sql2 to 2
1st :-
$sql2 = "INSERT INTO accounting (description) SELECT * FROM donations WHERE donorid='$donorid'"
then another one
"UPDATE accounting SET date='', transactiontype='', account ='', Income='', Expense ='' WHERE description=(SELECT * FROM donations WHERE donorid='$donorid')"
that will take all the information from donoation for the given donorid and list it under description in accounting